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Unformatted text preview: 1 AAE 203 Notes Martin Corless November 23, 2009 2 Contents
1 Introduction 2 Units and Dimensions 2.1 Introduction . . . . . . . . . . . 2.2 Units . . . . . . . . . . . . . . . 2.2.1 SI system of units . . . . 2.2.2 US system of units . . . 2.2.3 Unit conversions . . . . 2.3 Dimensions . . . . . . . . . . . 2.3.1 Dimensional systems . . 2.4 Dimensions of derived quantities 2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 9 9 9 10 10 10 11 11 12 15 17 17 20 24 25 26 26 27 29 29 33 42 42 43 47 50 51 51 51 53 54 3 Vectors 3.1 Introduction . . . . . . . . . . . . . . . . . . . 3.2 Vector addition . . . . . . . . . . . . . . . . . 3.2.1 Basic properties of vector addition . . 3.2.2 Addition of several vectors . . . . . . . 3.2.3 Subtraction of vectors . . . . . . . . . 3.3 Multiplication of a vector by a scalar . . . . . 3.3.1 Unit vectors . . . . . . . . . . . . . . . 3.4 Components . . . . . . . . . . . . . . . . . . . 3.4.1 Planar case . . . . . . . . . . . . . . . 3.4.2 General case . . . . . . . . . . . . . . . 3.5 Products of Vectors . . . . . . . . . . . . . . . 3.5.1 The angle between two vectors . . . . . 3.5.2 The scalar (dot) product of two vectors 3.5.3 Cross (vector) product of two vectors . 3.5.4 Triple products . . . . . . . . . . . . . 4 Kinematics of Points 4.1 Derivatives . . . . . . . . . . . 4.1.1 Scalar functions . . . . 4.1.2 Vector functions . . . . 4.1.3 The frame derivative of . . . a .... .... .... vector 3 ..... ..... ..... function 4 4.2 Basic deﬁnitions . . . . . . . . . . 4.2.1 Position . . . . . . . . . . 4.2.2 Velocity and acceleration . Rectilinear motion . . . . . . . . Planar motion . . . . . . . . . . . 4.4.1 Cartesian coordinates . . . 4.4.2 Projectiles . . . . . . . . . 4.4.3 Polar coordinates . . . . . General threedimensional motion 4.5.1 Cartesian coordinates . . . 4.5.2 Cylindrical coordinates . . 4.5.3 Spherical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 58 59 65 66 66 67 70 74 74 74 74 75 75 75 79 79 81 92 93 4.3 4.4 4.5 5 Kinematics of Reference Frames 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 A classiﬁcation of reference frame motions . . . . . . . . . 5.3 Motions with simple rotations . . . . . . . . . . . . . . . . 5.3.1 Angular Velocity . . . . . . . . . . . . . . . . . . . 5.4 The Basic Kinematic Equation (BKE) . . . . . . . . . . . 5.4.1 Polar coordinates revisited . . . . . . . . . . . . . . 5.4.2 Proof of the BKE for motions with simple rotations 6 General Reference Frame Motions 103 6.1 Angular velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 6.2 The basic kinematic equation (BKE) . . . . . . . . . . . . . . . . . . . . . . 106 7 Angular Acceleration 117 7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 8 Kinematic Expansions 125 8.1 The velocity expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 8.2 The acceleration expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 8.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 9 Particle Dynamics 9.1 Introduction . . . . . . . . . . . . . . . 9.2 Newton’s second law . . . . . . . . . . 9.3 Static equilibrium . . . . . . . . . . . . 9.4 Newton’s third law . . . . . . . . . . . 9.5 Forces . . . . . . . . . . . . . . . . . . 9.6 Gravitational attraction . . . . . . . . 9.6.1 Two particles . . . . . . . . . . 9.6.2 A particle and a spherical body 9.6.3 A particle and the earth . . . . 9.7 Contact forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 137 139 140 141 142 142 142 142 143 145 CONTENTS 9.7.1 Strings . . . . . . . . . . . . . . . ¯ 9.8 Application of ΣF = ma . . . . . . . . . ¯ 9.8.1 Free body diagrams . . . . . . . . 9.8.2 A systematic procedure . . . . . . 9.9 Forces due to smooth surfaces and curves 9.9.1 Smooth surfaces . . . . . . . . . . 9.9.2 Smooth curves . . . . . . . . . . 9.10 Rough surfaces, rough curves and friction 9.10.1 Rough curves . . . . . . . . . . . 9.10.2 Springs . . . . . . . . . . . . . . . 9.11 Dashpots . . . . . . . . . . . . . . . . . . 9.12 exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 145 154 154 154 155 155 159 163 170 171 173 177 179 179 185 185 187 190 193 194 196 199 199 206 206 207 210 214 217 217 217 223 224 227 229 230 230 230 232 237 237 238 242 10 Equations of Motion 10.1 Single degree of freedom systems . . . . . . 10.2 Numerical simulation . . . . . . . . . . . . . 10.2.1 First order representation . . . . . . 10.2.2 Numerical simulation with MATLAB 10.3 Multi degree of freedom systems . . . . . . . 10.4 Central force motion . . . . . . . . . . . . . 10.4.1 Equations of motion . . . . . . . . . 10.4.2 Some orbit mechanics . . . . . . . . . 11 Statics of Bodies 11.1 The moment of a force . . . . . . . . . . . . 11.2 Bodies . . . . . . . . . . . . . . . . . . . . . 11.2.1 Internal forces and external forces . . 11.2.2 Internal forces . . . . . . . . . . . . . 11.3 Static equilibrium . . . . . . . . . . . . . . . 11.3.1 Free body diagrams . . . . . . . . . . 11.4 Examples in static equilibrium . . . . . . . . 11.4.1 Scalar equations of equilibrium . . . 11.4.2 Planar examples . . . . . . . . . . . 11.4.3 General examples . . . . . . . . . . . 11.5 Force systems . . . . . . . . . . . . . . . . . 11.5.1 Couples and torques . . . . . . . . . 11.6 Equivalent force systems . . . . . . . . . . . 11.7 Simple equivalent force systems . . . . . . . 11.7.1 A force and a couple . . . . . . . . . 11.7.2 Force systems which are equivalent to 11.7.3 Force systems which are equivalent to 11.8 Distributed force systems . . . . . . . . . . . 11.8.1 Body forces . . . . . . . . . . . . . . 11.8.2 Surface forces . . . . . . . . . . . . . 11.8.3 Connections in 2D . . . . . . . . . . ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... a couple . . single force ....... ....... ....... ....... 6 11.8.4 Connections in 3D . . . . . . . . . . . 11.9 More examples in static equilibrium . . . . . . 11.9.1 Two force bodies in static equilibrium . 11.10Statically indeterminate problems . . . . . . . 11.11Internal forces . . . . . . . . . . . . . . . . . . 11.12Exercises . . . . . . . . . . . . . . . . . . . . . 12 Momentum 12.1 Linear momentum . . . . . . . . . . . . . 12.2 Impulse of a force . . . . . . . . . . . . . . 12.2.1 Integral of a vector valued function 12.2.2 Impulse . . . . . . . . . . . . . . . 12.3 Angular momentum . . . . . . . . . . . . . 12.4 Central force motion . . . . . . . . . . . . 13 Work and Energy 13.1 Kinetic energy . . . . . . . . . . . . . . . 13.2 Power . . . . . . . . . . . . . . . . . . . 13.3 A basic result . . . . . . . . . . . . . . . 13.4 Conservative forces and potential energy 13.4.1 Weight . . . . . . . . . . . . . . . 13.4.2 Linear springs . . . . . . . . . . . 13.4.3 Inverse square gravitational force 13.5 Total mechanical energy . . . . . . . . . 13.6 Work . . . . . . . . . . . . . . . . . . . . 14 Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 244 261 264 269 269 271 271 272 272 273 276 278 279 279 280 281 283 283 283 283 284 285 287 Chapter 1 Introduction HI! In Mechanics, we are concerned with bodies which are at rest or in motion. Kinematics. In kinematics we study motion without consideration of the causes of motion. This course is mainly concerned with kinematics of points. Basic concepts: position, time Derived concepts: velocity, acceleration; angle, angular velocity, angular acceleration Statics Before full consideration of Dynamics, we look at Statics. Here bodies are ”at rest” and we examine the forces on the bodies. Basic concept: force Derived concepts: moment Dynamics Basic concept: mass Basic Laws Newtons First Law Newtons Second Law Newtons Third Law 7 8 CHAPTER 1. INTRODUCTION Chapter 2 Units and Dimensions 2.1 Introduction
length, time, mass, force There are four fundamental quantities in mechanics: The ﬁrst three are scalar quantities and the fourth is a vector quantity. All other quantities in mechanics can be derived from these fundamental quantities. For example, area is length by length, speed can be expressed as the ratio of length over time, and angle can be expressed as the ratio of length over length. Actually, the four fundamental quantities are not independent, they are related by Newton’s second law. Hence one can choose any three of these quantities as basic quantities and consider the fourth as a derived quantity. 2.2 Units When representing a physical quantity by a scalar or a vector, one must also specify units, for example, l = 10 ft . The units of any quantity in mechanics can be expressed in terms of the units of any three of the four fundamental quantities. We will look at the two systems of units in common use, the SI system and the US system. If a quantity is dimensionless its units are independent of the units chosen for the basic quantities. As we shall see shortly, one such quantity is angle. The two commonly used units for angles are radians and degrees. They are related by 180 degrees = π radians where π is the ratio of the circumference of any circle to its diameter; it is approximately given by π ≈ 3.146 . 9 10 CHAPTER 2. UNITS AND DIMENSIONS 2.2.1 SI system of units In the SI (or metric) system of units, the quantities mass, length and time are considered basic and force is derived. quantity mass length time force unit unit symbol kilogram kg meter m second s newton N As a consequence of Newton’s second law, one newton is deﬁned to be the magnitude of the force required to give 1 kg of mass an inertial acceleration of magnitude 1ms−2 , that is, 1N = 1kg m s−2 . The units of any other quantity in mechanics can be expressed in terms of the units of the basic quantities, that is, kilograms, meters and seconds. 2.2.2 US system of units In the US system of units, the quantities force, length and time are considered basic and mass is derived. quantity force length time mass unit pound foot second slug unit symbol lb ft sec slug As a consequence of Newton’s second law, one slug is the mass which has an inertial acceleration of magnitude 1ft sec−2 when subject to a force of magnitude 1 lb, that is, 1 lb = 1 slug ft sec−2 . Hence, 1 slug = 1 lb sec2 ft−1 The units of any other quantity in mechanics can be expressed in terms of the units of the basic quantities, that is, pounds, feet and seconds. 2.2.3 Unit conversions You should already know how to do this. 2.3. DIMENSIONS 11 2.3 Dimensions To every quantity in mechanics, we associate a dimension. Dimension indicates quantity type. We sometimes use symbols to indicate dimension. These symbols for the fundamental quantities are given in the following table. quantity force mass length time dimension symbol F M L T Note that the concept of dimension is not the same as unit. One foot is not the same as one meter, however, both have the same dimension, namely, length. 2.3.1 Dimensional systems The dimensions of the four fundamental quantities are related by Newton’s second law, speciﬁcally, F = MLT −2 . Hence we can choose any three dimensions as basic dimensions and consider the fourth dimension as a derived dimension. Usually, one chooses M, L, T or F, L, T as basic dimensions. Absolute dimensional system. In the absolute dimensional system, mass, length and time are considered basic and force is derived. The dimension of any quantity in mechanics is expressed as M α Lβ T γ where α, β and γ are real numbers. For example, F = MLT −2 . Gravitational dimensional system. In the gravitational dimensional system, force, length and time are considered basic and mass is derived. The dimension of any quantity in mechanics is expressed as F α Lβ T γ where α, β and γ are real numbers. For example, M = F L−1 T 2 . 12 CHAPTER 2. UNITS AND DIMENSIONS 2.4 Dimensions of derived quantities The dimension of any quantity Q in mechanics can be obtained using the following simple rules. We will use the notation dim[Q] to indicate the dimension of quantity Q. The ¯ dimension of a vector quantity Q is considered to be the same as that of its magnitude, that ¯ ] = dim[Q]. ¯ is, dim[Q Dimensions of numbers. A “pure” number Q is considered dimensionless. We indicate this by dim[Q] = 1 Dimensions of products and quotients. If Q1 and Q2 are any two quantities, then dim[Q1 Q2 ] = dim[Q1 ] dim[Q2 ] Example 1 (Angle) and dim[Q1 /Q2 ] = dim[Q1 ]/ dim[Q2 ] . In radians, the angle θ is given by θ = S/R. Since S and R are Figure 2.1: Angle lengths, we have dim[θ] = dim[S/R] = dim[S ]/ dim[R] = L/L = 1 . Since dim[θ] = 1, we consider angles dimensionless. Example 2 (cos and sin ) Since cos θ = a/c, where a and c and lengths, we have dim[cos θ] = dim[a/c] = dim[a]/ dim[c] = L/L = 1 . In a similar fashion, dim[sin θ] = dim[b/c] = dim[b]/ dim[c] = L/L = 1 . and dim[tan θ] = dim[b/a] = dim[b]/ dim[a] = L/L = 1 . Dimensions of powers. If Q is any quantity and α is any real number, then dim[Qα ] = dim[Q]α . 2.4. DIMENSIONS OF DERIVED QUANTITIES 13 Figure 2.2: cos, sin and tan √ Example 3 What is the dimension of the quantity Q = gh where h represents a height and g is a√ gravitational acceleration constant? 1 Since gh = [gh] 2 , we can use the power and product rules to ﬁrst obtain that dim g h = dim (gh) 2 = (dim[g ] dim[h]) 2 .
1 1 Since h represents a height, we have dim[h] = L; since g is an acceleration we also have dim[g ] = LT −2 . Thus dim Notice that √ g h = [(LT −2 )(L)] 2 = LT −1 .
1 gh has the dimension of speed. Dimensions of sums. It does not make sense to add quantities of diﬀerent dimensions, so, we have the following rule: Only quantities of the same dimensions should be added or subtracted. Thus, if Q1 and Q2 are are two quantities of the same dimension, then dim[Q1 + Q2 ] = dim[Q1 ] = dim[Q2 ] Dimensions and derivatives. dim Since dy = dim[y ]/ dim[x] dx d2 y d = 2 dx dx dy dx and dim[Q1 − Q2 ] = dim[Q1 ] = dim[Q2 ] . , it follows from two applications of the above rule that dim that is, dim d2 y dy = dim / dim[x] = (dim[y ]/ dim[x]) / dim[x] , 2 dx dx d2 y = dim[y ]/ dim[x]2 . dx2 14 Dimensions and integrals. dim CHAPTER 2. UNITS AND DIMENSIONS y dx = dim[y ] dim[x] Dimensions and equations. We say that an equation is dimensionally homogeneous if every term in the equation has the same dimension. We have the following rule: All equations (in mechanics) must be dimensionally homogeneous. Example 4 The expression for planar acceleration in polar coordinates is given by ˙e ¨ a = (¨ + r θ2 )ˆr + (r θ + 2rθ )ˆθ ¯ r ˙˙ e where er and eθ are dimensionless unit vectors. Let us check to see if every term in this ˆ ˆ equation has the dimension of acceleration, that is, LT −2 . Example 5 Later we shall meet the inverse square gravitational law which is expressed as F= GMm r2 where F is a force magnitude, M and m are masses while r is a distance. Here we shall determine the dimension of G. 2.5. EXERCISES 15 2.5 Exercises Exercise 1 Obtain expressions for the dimensions of the following quantities using (a) the absolute dimensional system, and (b) the gravitational dimensional system. Here x and y are lengths, t is time, m is some mass, a is an acceleration and F represents a force. (a) − 10 (b) d2 dt2 dx dt
x2 x1 2 a dx
2 dy + dt (c) x(t) 0 F (η ) dη Exercise 2 Determine whether or not the equation d dt
x 0 F dx = 1 dm 2 v + mva 2 dt is dimensionally homogeneous where F is a force, x is a displacement, v is a speed, a is an acceleration, m is some mass, and t is time. Exercise 3 If m denotes a mass, g an acceleration magnitude, x a length, F a force magnitude and t time, determine whether or not the following equation is dimensional homogeneous. 2 x dx d2 x mgx = F dη + m +2 dt dt 0 If not homogeneous, state why. Exercise 4 You have just spent the whole evening deriving the following expression for an acceleration in an AAE 203 problem: ¨ ˙ ˙ˆ a = (lθ + dθΩ)ˆ1 + dΩθs2 ¯ s where l and d represent lengths, θ represents an angle, and Ω represents a rotation rate. Your roommate looks at the expression and without doing any kinematic calculations, says you are wrong. Could she/he be right? Justify your answer. Exercise 5 Determine the dimension of h in order for the following equation to be dimensionally correct. ¨h θ + sin θ = 0 l where θ represents an angle and l represents a length. 16 CHAPTER 2. UNITS AND DIMENSIONS Exercise 6 Determine the dimension of k in order for the following equation to be dimensionally correct. mx + kx = 0 ¨ where x represents a displacement and m represents a mass. Exercise 7 Determine the dimension of ρ in order for the following equation to be dimensionally homogeneous. 1 ˙ mV = − ρV 2 CD S − W sin γ + T cos α 2 where W and T represent forces, m is a mass, V is a speed, α is an angle, S is an area and CD is dimensionless. Exercise 8 Given that F is a force, x is a displacement, θ is an angle, and v is a speed, determine the dimensions of the quantities I and k in order that the following equation be dimensionally homogeneous.
x 0 1 F dx = I 2 dθ dt 2 1 + kv 2 2 Exercise 9 Determine whether or not the following equation is dimensionally homogeneous. ¨ ml2 θ + kx + mgsinθ = 0 where x and l represent lengths, θ represents an angle, m is a mass k is a spring constant and g is an acceleration. Chapter 3 Vectors
3.1 Introduction √ A scalar is a real number, for example, 1, −1/2, 2. Some physical quantities can be represented by a single scalar, for example, time, length, and mass. These quantities are called scalar quantities. Other physical quantities cannot be represented by a single scalar, for example, force and velocity. These quantities have attributes of magnitude and direction. In saying that a motorcycle is traveling south at a speed of 70 mph, we are specifying the velocity of the cycle in terms of magnitude (70 mph) and direction (south). To represent velocity, we need a mathematical concept which has the above two attributes. A vector is such a concept. Mathematically, we deﬁne a vector to be a directed line segment. Graphically, we usually indicate a vector by a line segment with an arrowhead, for example, The magnitude of a vector is the length of the line segment while the direction of the vector is determined by the orientation of the line segment and the sense of the arrowhead. Sometimes a vector is indicated by a segment of a circle with an arrowhead; in this case the direction of the vector is determined by the righthand rule. In the ﬁgure below, the direction of each vector is perpendicular to and out of the page. 17 18 CHAPTER 3. VECTORS All vectors, except unit vectors (we will meet these later), are represented by a symbol ¯ 0, ¯ ¯ with an overhead bar, for example, V , ¯ ♣. If A and B are the endpoints of a vector V and ¯ ¯ the direction of V is from A to B , we sometimes write V = AB . 3.1. INTRODUCTION
B V V = AB 19 A ¯ ¯ ¯ ¯ If V = AB , we call A the tail or point of application of V and B is called the head of V . ¯ lies is called the line of action of V . ¯ The line on which V line of action B head V A tail ¯ ¯ The magnitude or length of a vector V (denoted V  and sometimes by V ) is the distance ¯ ¯ ¯ ¯ ¯ between the endpoints of V . Two vectors V and W are said to be parallel (denoted V // W ), ¯ and W are parallel. ¯ if the lines of action of V ¯¯ ¯ ¯ Two vectors V , W , are deﬁned to be equal, that is, W = V , if they have the same magnitude and direction. Thus one can completely specify a vector by specifying its magnitude and direction.
V W W= V 20 CHAPTER 3. VECTORS 3.2 Vector addition
¯ ¯ V +W ¯ ¯ The sum or resultant of two vectors V and W is denoted by We present two equivalent deﬁnitions of vector addition, the triangle law and the parallelogram rule. ¯ ¯ ¯ ¯ Triangle law. Place the tail of W at the head of V . Then V + W is the vector from tail ¯ ¯ of V to the head of W . W
+W V C W V A V B ¯ ¯ ¯ ¯ In other words, if V = AB and W = BC , then V + W = AC . ¯ ¯ Parallelogram rule. Place the tails of V and W together. Complete the parallelogram ¯ ¯ ¯¯ with sides parallel to V and W . Then V + W lies along a diagonal of the parallelogram with ¯ and W . ¯ tail at the tails of V
C W W
V +W D W V A V B ¯ ¯ ¯¯ In other words, if V = AB and W = AC , then V +W = AD where ABDC is a parallelogram. One may readily show that the above two deﬁnitions are equivalent. 3.2. VECTOR ADDITION Some trigonometry Recall angle sine cosine Pythagorean theorem 21 c2 = a2 + b2 Cosine law: c2 = a2 + b2 − 2ab cos θ Sine Law sin α sin β sin γ = = a b c 22 CHAPTER 3. VECTORS Example 6 (Vector addition, cosine law, sine law.) ¯ ¯ ¯ ¯ ¯ ¯ Given two vectors V and W as shown with V  = 1 and W  = 2. Find V + W . Solution. We use the triangle law for vector addition as illustrated. Using the cosine law on the above triangle, we obtain that ¯ ¯ ¯ ¯ ¯¯ V + W 2 = V 2 + W 2 − 2V  W  cos(120◦ ) 1 = (1)2 + (2)2 − 2(1)(2) − = 7. 2 Hence, ¯ ¯ V + W  = √ 7. Applying the sine law to the above triangle yields sin θ sin(120◦ ) =¯ ¯ ¯. W  V + W  Hence, √ ¯ sin(120◦ )W  ( 3/2)(2) √ sin θ = = = ¯ ¯ V + W  7 3 = 40.89◦ . θ = sin−1 7 3 7 which yields 3.2. VECTOR ADDITION So, ¯ ¯ V + W is a vector of magnitude √ 23 7 with direction as shown where θ = 40.89◦ (Recall that a vector can be completely speciﬁed by specifying its magnitude and direction.) • A zero vector is a vector of zero magnitude. ·¯ 0 ¯ ¯ ¯ • The negative of V (denoted −V ) is a vector which has the same magnitude as V but opposite direction. V V 24 CHAPTER 3. VECTORS 3.2.1 Basic properties of vector addition ¯¯ 1) (Commutativity.) For every pair V , W of vectors, we have ¯ ¯ ¯ ¯ V +W =W +V . (This follows from parallelogram rule) ¯¯¯ 2) (Associativity.) For every triplet U , V , W of vectors, we have ¯¯ ¯ ¯ ¯ ¯ (U + V ) + W = U + (V + W ) . V W U W V) + (U + (V + W) W W
W V+ U+V V U U U+ V ¯ 3) There is a vector ¯ such that for every vector V we have 0 ¯0¯ V +¯=V . 3.2. VECTOR ADDITION ¯ ¯ 4) For every vector V there is a vector −V such that ¯ ¯ V + (−V ) = ¯ 0 25 The above four properties are called the group properties of vector addition. The next property follows from the triangle law. It is called the triangle inequality. ¯¯ 5) For any pair V , W of vectors, we have ¯ ¯ ¯ ¯ V + W  ≤ V  + W  3.2.2 Addition of several vectors Several vectors are added in the following fashion. Starting with the second vector, one simply places the tail of the vector at the head of the preceding vector. The sum of all the vectors is the vector from the tail of the ﬁrst vector to the head of the last vector. 4 V 3 + V 4 V V V2 V
1 V 1 + V 2 + 3 ¯ ¯ ¯ ¯ ¯ V = V1+V2+V3+V4 26 CHAPTER 3. VECTORS 3.2.3 Subtraction of vectors
¯ ¯ ¯ ¯ V − W = V + (−W ) . ¯ ¯ ¯ ¯ The diﬀerence of two vectors V and W is denoted by V − W and is deﬁned by V V V W W W Figure 3.1: Subtraction of vectors ¯ ¯ ¯ ¯ Note that, if the tails of V are W are placed together then, V − W is the vector from the ¯ to the head of W . ¯ head of V 3.3 Multiplication of a vector by a scalar
¯ kV ¯ The product of a vector V and a scalar k is another vector and is denoted by ¯ ¯ The product k V is deﬁned to be a vector whose magnitude is k V . If k > 0, the direction ¯ is the same as V ; if k < 0 the direction of k V is opposite to that of V . If k = 0, the ¯ ¯ ¯ of k V ¯ product k V is zero. V 2V 2 V Basic properties of scalar multiplication ¯ ¯ ¯ ¯ 1) k (V + W ) = k V + k W ¯ ¯ ¯ 2) (k + l)V = k V + lV ¯ ¯ 3) 1V = V 3.3. MULTIPLICATION OF A VECTOR BY A SCALAR ¯ ¯ 4) k (lV ) = (kl)V 27 The above four properties along with the group properties of vector addition are called the ﬁeld properties of vectors. The next property is actually part of the above deﬁnition of scalar multiplication. ¯ ¯ 5) k V  = k V  3.3.1 Unit vectors A unit vector is a vector of magnitude one. In writing unit vectors, we use “hats” instead of bars, for example, u represents a unit vector; hence u = 1. Unit vectors are useful for ˆ ˆ ¯ indicating direction. If V is nonzero, the vector ¯ V uV := ¯ ˆ¯ V  ¯ ¯ is called the unit vector in the direction of V . Clearly, uV has the same direction as V and ˆ¯ one can readily see that uV is a unit vector as follows. ˆ¯ ¯ ¯ 1¯ V  V u V  = ¯ ˆ¯ = ¯  V  = ¯ = 1. V  V  V  Note that ¯ ¯ ˆ¯ V = V  u V ¯ This explicitly represents a vector in terms of its magnitude V  and its direction, the direction being completely speciﬁed by uV . ˆ¯ Some useful facts. The following facts are useful for representing physical quantities by ¯ ¯ vectors. Suppose V and W are any two nonzero vectors. Then the following hold. ¯ ¯ ¯ ¯ 1) If V and W have the same direction, then there is a scalar k > 0 such that W = k V . V W 28 CHAPTER 3. VECTORS ¯ ¯ 2) If the direction of W is opposite to the direction of V , then there is a scalar k < 0 such ¯ = kV . ¯ that W V W ¯ ¯ ¯ ¯ 3) If W is parallel to V , then there is a nonzero scalar k such that W = k V . We now demonstrate why the above facts are true. ¯ ¯ ¯ 1) Since W and V have the same direction, the unit vector in the direction of W is equal ¯ , that is, to the unit vector in the direction of V ¯ ¯ V W =¯; ¯ W  V  hence, ¯ W  ¯ ¯ W= ¯ V V  ¯ ¯ W = kV where ¯ W  k= ¯ V  > 0. or, ¯ ¯ 2) In this case, −W has the same direction as V . Using the previous result, there is a scalar l > 0 such that ¯ ¯ −W = lV . Letting k := −l, we have ¯ ¯ W = kV with k < 0. ¯ ¯ ¯ ¯ 3) Since W is parallel to V , either W and V have the same direction or they have opposite direction. Hence, using the previous two results, ¯ ¯ W = kV where k < 0 or k > 0. 3.4. COMPONENTS 29 3.4 Components So far, our concept of a vector is a geometrical one, speciﬁcally, it is a mathematical object with the properties of magnitude and direction. This representation is useful for initial representation of physical quantities, for example, suppose one wants to describe the velocity of a motorcycle heading south at 70 mph as a vector. However, in manipulating vectors (for example adding them) the geometric representation can become very cumbersome if not impossible. In this section, we learn how to represent any vector as an ordered triplet of scalars, for example (1, 2, 3). This permits us to reduce operations on vectors to operations on scalars. 3.4.1 Planar case Consider ﬁrst the case in which all vectors of interest lie in a single plane, Fact 1 Suppose ¯1 , ¯2 , are any pair of nonzero, nonparallel vectors in a plane. Then, for bb ¯ in the plane, there is a unique pair of scalars, V1 , V2 such that every vector V ¯ V = V1 ¯1 + V2 ¯2 b b The pair (¯1 , ¯2 ) of vectors is called a basis. It deﬁnes a coordinate system. With respect bb ¯ to this basis, V1¯1 and V2¯2 are called the vector components of V ; the scalars V1 and V2 are b b ¯ called the scalar components or coordinates of V . The important thing about a basis is that ¯ it permits one to represent uniquely any vector V in the plane as a pair of scalars (V1 , V2 ). V b2 b1 ¯ Demonstration of Fact 1. Construct a parallelogram with V as diagonal and with sides ¯ ¯ ¯ parallel to ¯1 and ¯2 . Let V1 and V2 be the vectors with tails at the tail of V which make up b b ¯ two sides of the parallelogram as shown. Then V1 is parallel to ¯1 , V2 is parallel to ¯2 and b¯ b from the parallelogram law ¯ ¯ ¯ V = V1 + V2 ¯ ¯ Also, since V1 is parallel to ¯1 and V2 is parallel to ¯2 , there are unique scalars V1 , V2 so that b b ¯ V1 = V1¯1 b and ¯ V2 = V2¯2 . b ¯ ¯ Thus, V may be written as V = V1¯1 + V2¯2 . b b 30
b2 CHAPTER 3. VECTORS V 2 V V1 b1 ¯bb Example 7 (Planar components.) Given the coplanar vectors V , ˆ1 , ˆ2 as shown where ¯ V  = 5, ﬁnd ¯ (i) scalars V1 and V2 such that V = V1ˆ1 + V2ˆ2 , b b (ii) scalars n1 and n2 such that uV = n1ˆ1 + n2ˆ2 . ˆ¯ b b Solution. (i) From the above parallelogram, it should be clear that ¯ ¯ ¯ V = V1 + V2 . 3.4. COMPONENTS Also, θ = 180 − 60 − 45 = 75◦ . Using the sine law, we obtain that sin θ sin 45◦ = ¯ ¯.  V1  V  Hence, ¯  V1  = ¯ sin(75◦ )V  (0.9659)(5) = 6.830 . = 1 ◦ √ sin 45 2 31 ¯ ¯b So, V1 = V1  ˆ1 = 6.830 ˆ1 . Using the sine law again, b sin 60◦ sin 45◦ = ¯ ¯.  V2  V  Hence, sin 60 ¯ ¯ V  =  V2  = sin 45◦ ¯ ¯b So, V2 = V2  ˆ2 = 6.124 ˆ2 and b ¯ V = 6.830 ˆ1 + 6.124 ˆ2 b b ¯ ¯ ¯ Note that, in this example, V1 , V2  > V  . (ii) Since, ¯ V 1 b b uV = ¯ = (6.830 ˆ1 + 6.124 ˆ2 ) ˆ¯ V  5 uV = 1.366 ˆ1 + 1.225 ˆ2 ˆ¯ b b Perpendicular components Consider the special case in which ¯1 = e1 , ¯2 = e2 and e1 , e2 are mutually perpendicular b ˆb ˆ ˆˆ unit vectors.
1 V ^ e2 1 ^ e1
◦ √ 3 2 1 √ 2 (5) √ 53 = √ = 6.124 . 2 we obtain that ¯ ¯ Then, the parallelogram used in obtaining components V1 and V2 is a rectangle and the components are sometimes called rectangular components. 32
^ e2 CHAPTER 3. VECTORS P V V2 θ O V1 A ^ e1 From the parallelogram law of vector addition, it follows from the above ﬁgure that ¯ ¯ ¯ V = V1 + V2 ¯ ¯ Since the vector V1 is parallel to e1 , we must have V1 = V1 e1 for some scalar V1 . In a similar ˆ ˆ ¯2 = V2 e2 where V2 is a scalar. Hence, fashion, V ˆ ¯ V = V1 e1 + V2 e2 . ˆ ˆ ¯ ¯ ¯ Using the Pythagorean theorem on triangle OAP we obtain that V 2 = V1 2 + V2 2 ; hence V= ¯ ¯  V1  2 +  V2  2 ¯ ¯ ¯ ¯ where V := V  is the magnitude of V . Since V1  = V1  and V2  = V2 , we have V= For the situation illustrated, ¯ V1 = V1  = V cos θ, ¯ V2 = V2  = V sin θ. V12 + V22 . ¯ However, the relationships V1 = V cos θ and V2 = V sin θ hold for any direction of V .
^ e2 ^ e2 V V θ θ ^ e1 ^ e1 3.4. COMPONENTS Summarizing, we have the following relationships: ¯ V = V1 e1 + V2 e2 ˆ ˆ V1 = V cos θ, Also, V = V12 + V22 V2 = V sin θ 33 tan θ = V2 /V1 3.4.2 General case Suppose ¯1 , ¯2 and ¯3 are any three nonzero vectors which are not parallel to a common bb b ¯ plane. Then, given any vector V , there exists a unique triplet of scalars, V1 , V2 , V3 such that ¯ V = V1¯1 + V2¯2 + V3¯3 . b b b The triplet of vectors, (¯1 , ¯2 , ¯3 ), is called a basis. It deﬁnes a coordinate system. With bbb ¯ respect to this basis, the vectors V1¯1 , V2¯2 , V3¯3 are the vector components of V and the b b b ¯ scalars V1 , V2 , V3 are the scalar components or coordinates of V . The most important thing ¯ about a basis is that it permits one to represent uniquely any vector V as a triplet of scalars (V1 , V2 , V3 ). In this course, we consider mainly a special case, namely the case in which ¯1 , ¯2 , ¯3 are mutually perpendicular unit vectors. bbb Mutually perpendicular components Let e1 , e2 , e3 be any three mutually orthogonal (perpendicular) unit vectors. We call (ˆ1 , e2 , e3 ) ˆˆˆ eˆˆ ¯ can be uniquely resolved an orthogonal triad. Since (ˆ1 , e2 , e3 ) constitute a basis, any vector V eˆˆ Figure 3.2: An orthogonal triad into components parallel to e1 , e2 , e3 , that is, there are unique scalars V1 , V2 , V3 such that ˆˆˆ ¯ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ ˆ The vectors V1 e1 , V2 e2 , V3 e3 are called rectangular components and the scalars V1 , V2 , V3 are ˆ ˆ ˆ called rectangular scalar components or rectangular coordinates. Also, 34 V= V12 + V22 + V32 CHAPTER 3. VECTORS ¯ where V = V . ¯ ¯ To demonstrate the above decomposition, we ﬁrst decompose V into two components, V3 ¯I where V3 is parallel to e3 and VI is in the plane formed by e1 and e2 . Thus, ¯ ¯ and V ˆ ˆ ˆ ¯ ¯ ¯ V = VI + V3 . Also, using the Pythagorean theorem, we have ¯ ¯ ¯  V  2 =  VI  2 +  V3  2 . ¯ Figure 3.3: Decomposition of V into rectangular components ¯ ¯ ¯ Figure 3.4: Decomposition of V into VI and V3 ¯ ¯ ¯ ¯ ¯ We now decompose VI into two components, V1 and V2 where V1 and V2 are parallel to e1 and e2 , respectively. Thus ˆ ˆ ¯ ¯ ¯ VI = V1 + V2 . Also, by the Pythagorean theorem, we have ¯ ¯ ¯  VI  2 =  V1  2 +  V2  2 . 3.4. COMPONENTS 35 ¯ ¯ ¯ Figure 3.5: Decomposition of V into VI and V3 By combining the above two decompositions, we obtain that ¯ ¯ ¯ ¯ V = V1 + V2 + V3 and ¯ ¯ ¯ ¯  V  2 =  V1  2 +  V2  2 +  V3  2 . ¯ ¯ ¯ ¯ ¯ Since V1 = V1 e1 , V2 = V2 e2 and V3 = V3 e3 with V1 2 = V12 , V2 2 = V22 and V3 2 = V32 , we ˆ¯ ˆ ˆ obtain the desired result, namely, ¯ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ ˆ and ¯ V  = (V12 + V22 + V32 )1/2 . ¯ With the above decomposition, we can regard the vector V as the diagonal of a rectan¯¯ ¯ gular box with edges parallel to e1 , e2 , e3 and with dimensions V1 , V2  and V3 . ˆˆˆ ¯ Figure 3.6: V in a box 36 CHAPTER 3. VECTORS ¯ Example 8 Given V and the orthogonal triad (¯1, u2 , u3 ) as shown, ﬁnd uˆˆ ¯ (i) scalars V1 , V2 , V3 , such that V = V1 u1 + V2 u2 + V3 u3 ˆ ˆ ˆ (ii) scalars n1 , n2 , n3 , such that uV = n1 u1 + n2 u2 + n3 u3 . ˆ¯ ˆ ˆ ˆ Solution: (i) Clearly, ¯ ¯ ¯ ¯ V = V1 + V2 + V3 ¯¯¯ where V1 , V2 , V3 are as shown. 3.4. COMPONENTS Also, ¯ ¯ˆ V1 = V1 u1 = 1ˆ1 = u1 u ˆ ¯ ¯ˆ V2 = V2 u2 = 2ˆ2 u ¯ ¯ˆ V3 = V3 u3 = 3ˆ3 u Thus, ¯ or, V = V1 u1 + V2 u2 + V3 u3 where ˆ ˆ ˆ V1 = 1, V2 = 2, V3 = 3 (ii) Since we have uV = ˆ¯ ¯ V  = V12 + V22 + V32 = (1)2 + (2)2 + (3)2 = √ ¯ V = u1 + 2ˆ2 + 3ˆ3 , ˆ u u 37 14 , ¯ V 1 1 2 3 u u ˆ ˆ ˆ ¯  = √14 (ˆ1 + 2u2 + 3ˆ3) = √14 u1 + √14 u2 + √14 u3 V = 0.2673 u1 + 0.5345 u2 + 0.8018 u3 ˆ ˆ ˆ So, uV = n1 u1 + n2 u2 + n3 u3 where ˆ¯ ˆ ˆ ˆ n1 = 0.2673, n2 = 0.5345, n3 = 0.818 ¯ Example 9 Given the vector V and the orthogonal triad (ˆ1 , u2 , u3) as shown where uˆˆ α = 63.43◦ , β = 53.30◦ , ¯ ﬁnd scalars V1 , V2 , V3 such that V = V1 u1 + V2 u2 + V3 u3 . ˆ ˆ ˆ ¯ V  = 3.742 , 38 Solution: CHAPTER 3. VECTORS Clearly, with and Also, ¯ ¯ ¯ V = VI + V3 ¯ ¯ VI  = V  cos β = 3.742 cos(53.3◦ ) = 2.236 ¯ ¯ V3  = V  sin β = 3.742 sin(53.3◦ ) = 3.000 . ¯ ¯ˆ V3 = V3 u3 = 3ˆ3 u ¯ Considering the decomposition of VI , we have ¯ ¯ ¯ VI = V1 + V2 with ¯ ¯ V1  = VI  cos α = 2.236 cos(63.43) = 1.000 ¯ ¯ˆ V1 =  V1  u 1 = u 1 ˆ and ¯ ¯ V2  = VI  sin α = 2.236 sin(63.43) = 2.000 ¯ ¯ˆ V2 = V2 u2 = 2ˆ2 u 3.4. COMPONENTS Hence, 39 ¯ ¯ ¯ ¯ ¯ ¯ V = VI + V3 = V1 + V2 + V3 = u1 + 2ˆ2 + 3ˆ3 ˆ u u ¯ So, V = V1 u1 + V2 u2 + V3 u3 where ˆ ˆ ˆ V1 = 1 , V2 = 2 , V3 = 3 . ¯ ¯ Note that V is the same as the V considered in the previous example. 40 Example 10 CHAPTER 3. VECTORS Example 11 3.4. COMPONENTS Addition of vectors via addition of scalar components If ¯ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ ˆ and ¯ W = W1 e1 + W2 e2 + W3 e3 ˆ ˆ ˆ 41 then, using the ﬁeld properties of vectors, one readily obtains that ¯¯ V + W = (V1 + W1 )ˆ1 + (V2 + W2 )ˆ2 + (V3 + W3 )ˆ3 e e e Scalar multiplication of a vector via multiplication of its scalar components If ¯ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ ˆ then, using the ﬁeld properties of vectors, one readily obtains that ¯ k V = (kV1 )ˆ1 + (kV2 )ˆ2 + (kV3 )ˆ3 e e e 42 CHAPTER 3. VECTORS 3.5 Products of Vectors Before discussing products of vectors, we need to examine what we mean by the angle between two vectors. 3.5.1 The angle between two vectors ¯ ¯ Suppose V and W are two nonzero vectors. ¯ ¯ ¯¯ We denote the angle between V and W as V W . Properties 1) ¯¯ ¯¯ WV = V W ¯¯ ¯ ¯ 2) If V and W have the same direction then, V W = 0. ¯ ¯ ¯¯ If W is perpendicular to V then, V W = π . 2 ¯ ¯ ¯¯ If W is opposite in direction to V then, V W = π . 3.5. PRODUCTS OF VECTORS ¯¯ 3) In general, 0 ≤ V W ≤ π . 43 π ¯¯ 0< VW < 2 π ¯¯ < VW <π 2 3.5.2 The scalar (dot) product of two vectors ¯ ¯ Suppose V and W are any two nonzero vectors. ¯ ¯ ¯¯ The scalar (or dot) product of V and W is a scalar which is denoted by V · W and is deﬁned by ¯¯ V · W = V W cos θ ¯ ¯ ¯ where V is the magnitude of V , W is the magnitude of W , and θ is the angle between V and ¯ W. ¯ ¯ ¯¯ If either V or W is the zero vector, then, V · W is deﬁned to be zero which also equals V W cos θ for any value of θ. Remarks 1) Since −1 ≤ cos θ ≤ 1, we have ¯¯ −V W ≤ V · W ≤ V W 44 ¯0 ¯ 2) Suppose V = ¯ and W = ¯ Then 0. ¯¯ θ = 0 ⇐⇒ V · W = V W π ¯¯ ⇐⇒ V · W > 0 2 π ¯¯ ⇐⇒ V · W = 0 2 CHAPTER 3. VECTORS 0≤θ< θ= π ¯¯ < θ ≤ π ⇐⇒ V · W < 0 2 ¯¯ θ = π ⇐⇒ V · W = −V W ¯ ¯¯ 3) Since the angle between V and itself is zero, it follows that V · V = V 2 ; hence the ¯ magnitude of V can be expressed as ¯¯ V = (V · V )1/2 . ¯ ¯ 4) If V and W represent physical quantities, ¯¯ ¯ ¯ dim[V · W ] = dim[V ] dim[W ] Basic Properties of the scalar product ¯¯ ¯¯ 1) V · W = W · V ¯¯ ¯ ¯¯ ¯¯ 2) U · (V + W ) = U · V + U · W ¯ ¯ ¯¯ 3) V · (k W ) = k (V · W ) (commutativity) 3.5. PRODUCTS OF VECTORS The scalar product and rectangular components Suppose (ˆ1 , e2 , e3 ) is any orthogonal triad. eˆˆ 45 Facts (1) ei · ej = ˆˆ For example, e1 · e1 = e1 2 = 1 ˆˆ ˆ e1 · e2 = e1  e2  cos ˆˆ ˆˆ ¯ (2) If V = V1 e1 + V2 e2 + V3 e3 , then ˆ ˆ ˆ ¯ˆ V1 = V · e1 , ¯ˆ V2 = V · e2 , ¯ˆ V3 = V · e3 π 2 1 i=j 0 i=j = δij
△ =0 Proof. Consider V1 . Using the properties of the dot product, we obtain that ¯ˆ V · e1 = (V1 e1 + V2 e2 + V3 e3 ) · e1 ˆ ˆ ˆˆ = V1 (ˆ1 · e1 ) + V2 (ˆ2 · e1 ) + V3 (ˆ3 · e1 ) eˆ eˆ eˆ = V1 (1) + V2 (0) + V3 (0) = V1 . Similarly for V2 and V3 . (3) If ¯ ¯ˆ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ then ¯¯ V · W = V1 W1 + V2 W2 + V3 W3 and ¯ W = W1 e1 + W2 e2 + W3 e3 , ˆ ˆ ˆ 46 CHAPTER 3. VECTORS Proof. Using the properties of the dot product and the previous result, we obtain that ¯¯ ¯ V · W = V · (W1 e1 + W2 e2 + W3 e3 ) ˆ ˆ ˆ ¯ˆ ¯ˆ ¯ˆ = W1 (V · e1 ) + W2 (V · e2 ) + W3 (V · e3 ) = W1 V1 + W2 V2 + W3 V3 . ¯ ¯ (4) If θ is the angle between V and W then, cos θ = ¯ ¯ where V = V  and W = W . Proof. Recall that ¯¯ V W cos θ = V · W = V1 W1 + V2 W2 + V3 W3 . Hence, cos θ = Also note that θ = cos−1 Example 12 Suppose ¯ V = e1 + e2 + e3 ˆ ˆ ˆ Then and ¯ W = e1 − e3 . ˆ ˆ V1 W1 + V2 W2 + V3 W3 VW V1 W1 + V2 W2 + V3 W3 . VW V1 W1 + V2 W2 + V3 W3 VW ¯¯ V · W = (1)(1) + (1)(0) + (1)(−1) = 0. ¯¯ ¯ ¯ Since V · W is zero, V is perpendicular to W . ¯ V = e1 + e2 + e3 ˆ ˆ ˆ and ¯ W = e1 + e3 . ˆ ˆ Example 13 Suppose Then ¯¯ V · W = (1)(1) + (1)(0) + (1)(1) = 2 V = (1)2 + (1)2 + (1)2 = 3 W = (1)2 + (0)2 + (1)2 = 2 and cos θ = Hence θ = cos−1 (1/3) = 1.230 rad = 70.53◦ ¯¯ V ·W = 1/ 3 VW (3.1) (3.2) (3.3) 3.5. PRODUCTS OF VECTORS 47 3.5.3 Cross (vector) product of two vectors ¯ ¯ Suppose V and W are any two nonzero nonparallel vectors. ¯ ¯ ¯ ¯ The cross (or vector) product of V and W is a vector which is denoted by V × W and is deﬁned by ¯ ¯ V × W = V W sin θ n ˆ ¯ ¯ ¯ where V is the magnitude of V , W is the magnitude of W , θ is the angle between V and ¯ and n is the unit vector which is normal (perpendicular) to both V and W and whose ¯ ¯ W ˆ direction is given by the righthand rule. Figure 3.7: Cross product ¯ ¯ ¯ ¯ If V and W are parallel or either of them equals ¯ then, V × W is deﬁned to be the zero 0 vector. Remarks (1) Since 0 ≤ sin θ ≤ 1 for 0 ≤ θ ≤ π , it follows that ¯ ¯ V × W  = V W sin θ and ¯ ¯ V × W  ≤ V W . ¯ ¯ (2) Suppose V and W are both nonzero. Then the following relationships hold. ¯ ¯ ¯ ¯ V ×W =¯ 0 ⇐⇒ V is parallel to W ¯ ¯ V × W  = V W ⇐⇒ ¯ ¯ V is perpendicular to W ¯ ¯ (3) If V and W represent physical quantities, then ¯ ¯ ¯ ¯ dim [V × W ] = dim[V ] dim[W ] . 48 CHAPTER 3. VECTORS Figure 3.8: Sine function Basic Properties of the cross product. ¯ ¯ ¯ ¯ 1) W × V = −V × W (not commutative) ¯ ¯ ¯ ¯¯ ¯ 2) U × (V + W ) = (U × V ) + (U × W ) ¯ ¯ ¯ ¯ ¯ ¯ 3) V × (k W ) = (k V ) × W = k (V × W ) and ¯¯ ¯ ¯¯ ¯ ¯ (U + V ) × W = (U × V ) + (U × W ) Cross product and rectangular components An orthogonal triad, (ˆ1 , e2 , e3 ), is said to be righthanded if eˆˆ e3 = e1 × e2 ˆ ˆ ˆ From now on, we shall consider only righthanded orthogonal triads. 3.5. PRODUCTS OF VECTORS
ˆ e3 = ˆ 1 × e2 eˆ ˆ e2 ˆ e2 49 ˆ e1 ˆ e3 ˆ e3 ˆ e1 Facts (1) These relationships are illustrated below. e1 × e1 = ¯ ˆ ˆ 0 e2 × e1 = −e3 ˆ ˆ ˆ e3 × e1 = e2 ˆ ˆ ˆ e1 × e2 = e3 ˆ ˆ ˆ ¯ e2 × e2 = 0 ˆ ˆ e3 × e2 = −e1 ˆ ˆ ˆ e1 × e3 = −e2 ˆ ˆ ˆ e2 × e3 = e1 ˆ ˆ ˆ e3 × e3 = ¯ ˆ ˆ 0 ˆˆˆˆˆ e1 e2 e3 e1 e2 (2) If then ¯ V = V1 e1 + V2 e2 + V3 e3 ˆ ˆ ˆ and ¯ W = W1 e1 + W2 e2 + W3 e3 , ˆ ˆ ˆ ¯ ¯ V × W = (V2 W3 − V3 W2 ) e1 + (V3 W1 − V1 W3 ) e2 + (V1 W2 − V2 W1 ) e3 ˆ ˆ ˆ Proof. Exercise ¯ ¯ (3) the above expression for V × W may also be obtained from e1 e2 e3 ˆ ˆ ˆ ¯ ¯ V × W = det V1 V2 V3 W2 W2 W3 where det denotes determinant. Proof. Exercise 50 CHAPTER 3. VECTORS 3.5.4 Triple products Scalar triple product ¯¯ ¯ The scalar triple product of three vectors U , V and W is the scalar deﬁned by ¯¯ ¯ U · (V × W ) Facts (1) U1 U2 U3 ¯¯ ¯ U · (V × W ) = det V1 V2 V3 W1 W2 W3 (2) ¯¯ ¯ ¯¯¯ U · (V × W ) = (U × V ) · W that is, · and × can be interchanged. Proof. Exercise Vector triple product ¯¯ ¯ The vector triple product of three vectors U , V and W is the vector deﬁned by ¯ ¯ ¯ U × (V × W ) Facts (1) In general, For example, e1 × (ˆ1 × e2 ) = e1 × e3 = −e2 ˆ e ˆ ˆ ˆ ˆ (ˆ1 × e1 ) × e2 = ¯ × e2 = ¯ e ˆ ˆ 0ˆ 0 (2) ¯ ¯¯ ¯¯¯ ¯¯¯ U × (V × W ) = (U · W )V − (U · V )W Proof. Exercise ¯ ¯ ¯ ¯¯ ¯ U × (V × W ) = (U × V ) × W . Chapter 4 Kinematics of Points
In kinematics, we are concerned with motion without being concerned about what causes the motion. If a body is small in comparison to its “surroundings”, we can view the body as occupying a single point at each instant of time. We will also be interested in the motion of points on “large” bodies. The kinematics of points involves the concepts of time, position, velocity and acceleration. 4.1 Derivatives To involve ourselves with kinematics, we need derivatives. 4.1.1 Scalar functions First, consider the situation where v is a scalar function of a scalar variable t. Suppose t1 is a speciﬁc value of t. Then the formal deﬁnition of the derivative of v at t1 is dv v (t) − v (t1 ) (t1 ) = lim t→t1 dt t − t1 Sometimes this is called the ﬁrst derivative of v . Oftentimes, representation is given in Figure 4.1.
dv dt is denoted by v . A graphical ˙ Figure 4.1: Derivative of a scalar function
v The second derivative of v , denoted by d 2 or v, is deﬁned as the derivative of the ﬁrst ¨ dt derivative of v , that is d2 v d dv = . 2 dt dt dt
2 51 52 CHAPTER 4. KINEMATICS OF POINTS In evaluating derivatives, we normally do not have to resort to the above deﬁnition. By knowing the derivatives of commonly used functions (such as cos, sin, and polynomials) and using the following properties, one can usually compute the derivatives of most commonly encountered functions. Properties. The following hold for any two scalar functions v and w . (a) d dv dw (v + w ) = + dt dt dt (b) (Product rule) dv dw d (vw ) = w + v dt dt dt (c) (Quotient rule) Whenever w (t) = 0, dv dt w (d) (Chain rule) d dv dw (v (w )) = dt dw dt Example 14 Consider the function given by f (t) = cos(t2 ). Then f (t) = v (w (t)) where v (w ) = cos w and w (t) = t2 . =
dv w dt − v dw dt w2 Applying the chain rule, we obtain that df dv dw f˙ = = = (− sin w )(2t) . dt dw dt Hence, f˙ = −2t sin(t2 ) . Exercises
Exercise 10 Compute the ﬁrst and second derivatives of the following functions. (a) θ(t) = cos(20t) (b) f (t) = et
2 (c) x(t) = sin(et ) (d) h(t) = e2t cos(10t) Exercise 11 Compute the derivative of the following functions. sin(10t) (a) y (t) = 1 + t2 (b) z (t) = t2 e3t sin(4t) 4.1. DERIVATIVES 53 4.1.2 Vector functions ¯ Consider now the situation where V is a vector function of a scalar variable t. Suppose t1 is ¯ a speciﬁc value of t. Then the formal deﬁnition of the derivative of V at t1 is ¯ ¯ ¯ V (t) − V (t1 ) dV (t1 ) = lim t→t1 dt t − t1 Oftentimes,
¯ dV dt ˙ ¯ is denoted by V . ¯ ¯ Properties. The following hold for any two vector functions V and W and any scalar function k . (a) ¯ ¯ d¯ dV dW ¯ (V + W ) = + dt dt dt (b) ¯ d¯ dk ¯ dV (k V ) = V +k dt dt dt (c) ¯ ¯ d¯ ¯ dV ¯ ¯ dW (V · W ) = ·W +V · dt dt dt (d) ¯ ¯ d¯ dV ¯ ¯ ¯ dW (V × W ) = ×W +V × dt dt dt (e) (Chain rule) ¯ d¯ dk dV (V (k )) = dt dt dk Derivatives and components. Usually we evaluate the derivative of a vector function by diﬀerentiating its scalar components. Suppose e1 , e2 , e3 is a set of constant basis vectors ˆˆˆ and ¯ V (t) = V1 (t)ˆ1 + V2 (t)ˆ2 + V3 (t)ˆ3 . e e e Then, using properties (a) and (b) above, we obtain that ¯ dV dV1 dV2 dV3 = e1 + ˆ e2 + ˆ e3 ˆ dt dt dt dt or ˙ ¯ ˙ˆ ˙ˆ ˙ˆ V = V1 e1 + V2 e2 + V3 e3 54 CHAPTER 4. KINEMATICS OF POINTS 4.1.3 The frame derivative of a vector function We deﬁne a reference frame (or frame of reference) to be an righthanded orthogonal triad of unit vectors which have the same point of application. Figure 4.2 illustrates several reference frames. Usually we use a single symbol to reference frame; thus the reference frame consisting Figure 4.2: Reference frames ˆˆˆ of the vectors f1 , f2 , f3 will be referred to as the reference frame f . ¯ Consider a timevarying vector V . If one observes this vector from diﬀerent reference ¯ frames then, one will observe diﬀerent variations of V with time. For that reason, we have the following deﬁnition. Figure 4.3: The frame derivative of a vector ¯ The derivative of V in f is denoted by
f
f dV ¯ dt and is deﬁned by ¯ dV dV1 ˆ dV2 ˆ dV3 ˆ = f1 + f2 + f3 dt dt dt dt ¯ where V1 , V2 , V3 are the scalar components of V relative to f , that is, ˆ ˆ ˆ ¯ V = V1 f1 + V2 f2 + V3 f3 . Oftentimes,
f dV ¯ dt ˙ is denoted by f V . Thus,
f ˙ ¯ ˙ˆ ˙ˆ ˙ˆ V = V1 f1 + V2 f2 + V3 f3 4.1. DERIVATIVES 55 ¯ ¯ Properties. The following hold for any two vector functions V and W and any scalar function k . (a)
f f¯ f¯ d¯ dV dW ¯ (V + W ) = + dt dt dt (b)
f f¯ dk ¯ dV d¯ (k V ) = V +k dt dt dt (c)
f f¯ f¯ dV ¯ d¯ ¯ ¯ dW (V · W ) = ·W +V · dt dt dt (d)
f f¯ f¯ dV dW d¯ ¯ ¯ ¯ (V × W ) = ×W +V × dt dt dt (e) (Chain rule) f ¯ d¯ dk f dV (V (k )) = dt dt dk 56 Example 15 (Frame derivative) CHAPTER 4. KINEMATICS OF POINTS 4.1. DERIVATIVES Exercises 57 ˆˆˆ Exercise 12 Consider the reference frames f = (f1 , f2 , f3 ) and g = (ˆ1 , g2 , g3 ) as illustrated gˆˆ ¯ is given by where θ = 2t rads. Suppose the vector Z ¯ Z = 2tg1 + t2 g2 . ˆ ˆ In terms of t and the units vectors of g , ﬁnd expressions for the following quantities. ˙ ¯ (a) g Z ˙ ¯ (b) f Z ˙ˆ ¯¯ ¯ (c) g Z + ω × Z where ω = θg3 ¯ Compare the answers for parts (b) and (c). 58 CHAPTER 4. KINEMATICS OF POINTS 4.2 Basic deﬁnitions Besides time, there are three additional basic concepts in the kinematics of points, namely, position, velocity, and acceleration. 4.2.1 Position Consider any two points O and P . We deﬁne the position of P relative to O (denoted r OP ) or the position vector from O to P as the vector from O to P , that is, rOP := OP Figure 4.4: Position vector, r OP Clearly, the position of a point O relative to itself is the zero vector, that is, r OO = ¯ . 0 It should also be clear that the position of O relative to P is the negative of the position of P relative to O , that is, r P O = −r OP . Figure 4.5: r P O ¯ Composition of position vectors. For any three points O, P, Q, we have rOQ = rOP + rP Q This follows from the triangle law of vector addition and is illustrated in Figure 4.6. From the above relationship, we also have r P Q = r OQ − rOP . (4.1) 4.2. BASIC DEFINITIONS 59 Figure 4.6: Composition of position vectors Consider now several points, P1 , P2 , . . . , Pn . Then, by repeated application of result (4.1), we obtain r P1 Pn = rP1 P2 + r P2 P3 + . . . + rPn−1 Pn . This is illustrated in Figure 4.7. Figure 4.7: Composition of several position vectors 4.2.2 Velocity and acceleration Suppose we are observing the motion of some point P from a reference frame f . We ﬁrst demonstrate the following result. If O and O ′ are any two points which are ﬁxed in reference frame f , then
f d OP f d O′P r= r dt dt To see this, ﬁrst note that r OP = r OO + rO P . Hence, d OP f d OO′ f d O′ P r= r +r dt dt dt ′ OO ′ Since points O and O are ﬁxed in f , the vector r is a ﬁxed vector in f , hence
f f
′ ′ d OO′ r =0 dt 60 CHAPTER 4. KINEMATICS OF POINTS Figure 4.8: Independence of velocity on origin and the desired result follows. We deﬁne the velocity of P in f (denoted f v P ) by
fP f v := d OP r dt where O is any point ﬁxed in f . The speed of P in f is  f v P , the magnitude of the velocity of P in f . Figure 4.9: The velocity of P in f We deﬁne the acceleration of P in f (denoted f aP ) by
fP f a := dfP v dt 4.2. BASIC DEFINITIONS Note that
fP f2 61 a= d OP r dt2 In the next section, we consider some special types of motions. First we have some examples to illustrate the above concepts. Example 16 (Pendulum with moving support) 62 Example 17 (Bug on bar on cart) CHAPTER 4. KINEMATICS OF POINTS 4.2. BASIC DEFINITIONS 63 Exercises
Exercise 13 The two link planar manipulator is constrained to move in the plane deﬁned by the vectors e1 and e2 of reference frame e. Point O is ﬁxed in e. ˆ ˆ Find expressions for e v P and e aP in terms of θ1 , θ2 , their ﬁrst and second time derivatives, ¯ ¯ l1 , l2 and e1 , e2 , e3 . ˆˆˆ Exercise 14 The small ball P moves in the straight slot which is ﬁxed in the disk. Relative to reference frame e, point O is ﬁxed and the disk rotates about an axis through O which is parallel to e3 and perpendicular to the plane of the disk. ˆ 64 CHAPTER 4. KINEMATICS OF POINTS Find expressions for e v P and e aP in terms of l, r , θ, r , θ, r, θ and unit vectors ﬁxed in ¯ ¯ ˙ ˙¨¨ the disk. 4.3. RECTILINEAR MOTION 65 4.3 Rectilinear motion The simplest type of motion is rectilinear. The motion of point P in a reference frame f is called rectilinear if P always moves in a straight line ﬁxed in f . If we choose a reference Figure 4.10: Rectilinear motion point O which is along the line and ﬁxed in f and if we choose one direction along the line as a positive direction, then the location of P can be uniquely speciﬁed by specifying the displacement x of P from O . Thus we can describe rectilinear motion with a single scalar. In Figure 4.11, the displacement x is considered positive when P is to the right of O . Figure 4.11: Displacement x Since we are only dealing with the motion of one point P relative to a single reference frame f , we simplify notation here and let r be the position of P relative to O , v be the ¯ ¯ velocity of P in f and a be the acceleration of P in f . ¯ Figure 4.12: e ˆ If we introduce a unit vector e which along the line of motion and pointing in the positive ˆ direction for displacement along the line, then r = xe. Since O is ﬁxed in f , we have ¯ ˆ
f v= ¯ dr f d ¯ = (xe) = xe . ˆ ˙ˆ dt dt 66 CHAPTER 4. KINEMATICS OF POINTS The last equality above follows from the fact that e is a constant vector in reference frame ˆ f . We also have that f dv f d ¯ a= ¯ = (xe) = xe . ˙ˆ ¨ˆ dt dt So, summarizing, we have r = xe , ¯ ˆ where v=x ˙ and a = v = x. ˙¨ (4.2) v = ve , ¯ ˆ a = ae ¯ ˆ 4.4 Planar motion The motion of point P in a reference frame f is called planar if P always moves in a plane which is ﬁxed in f . Figure 4.13: Planar motion If we choose a reference point O which is in the plane and ﬁxed in f , then the location of P can be uniquely speciﬁed by specifying the position of P relative to O . For the rest of this section, we let r be the position of P relative to O , v be the velocity of P in f and a be ¯ ¯ ¯ the acceleration of P in f . Hence v= ¯ dr ¯ dt and a= ¯ dv ¯ dt where it is understood that the above diﬀerentiations are carried out relative to frame f . In general, planar motion can be described with two scalar coordinates. We now consider two diﬀerent coordinate systems for describing planar motion, namely cartesian coordinates and polar coordinates. 4.4.1 Cartesian coordinates Choose any two mutually perpendicular lines in the plane passing through O and ﬁxed in f . By choosing a positive direction for each line, The location of point P can be uniquely determined by the cartesian coordinates x and y as illustrated in Figure 4.14. We now compute expressions for r, v and a in terms of cartesian coordinates. To this ¯¯ ¯ end, we introduce unit vectors e1 , e2 ﬁxed in the plane as illustrated in Figure 4.15. Then, ˆˆ 4.4. PLANAR MOTION 67 Figure 4.14: Cartesian coordinates x and y Figure 4.15: e1 and e2 ˆ ˆ the position of P relative to O can be expressed as r = xe1 + y e2 ¯ ˆ ˆ Since e1 and e2 are ﬁxed vectors in f , diﬀerentiating the above expression in f yields the the ˆ ˆ velocity of P in f : dr ¯ d v= ¯ = (xe1 + y e2 ) = xe1 + y e2 . ˆ ˆ ˙ˆ ˙ˆ dt dt Diﬀerentiating once more yields the acceleration of P in f : a= ¯ So, summarizing, we have r = xe1 + y e2 ¯ ˆ ˆ v = v1 e1 + v2 e2 ¯ ˆ ˆ a = a1 e1 + a2 e2 ¯ ˆ ˆ dv ¯ d = (xe1 + y e2 ) = xe1 + y e2 . ˙ˆ ˙ˆ ¨ˆ ¨ˆ dt dt where where v1 = x ˙ a1 = v1 = x ˙ ¨ and and v2 = y ˙ a2 = v2 = y ˙ ¨ (4.3) 4.4.2 Projectiles As an application of cartesian coordinates, let us consider the motion of a projectile near the surface of YFHB (your favorite heavenly body, for example, the earth or the dark side 68 CHAPTER 4. KINEMATICS OF POINTS Figure 4.16: Projectile motion of the moon). Suppose that a projectile P is launched from point O on YFHB at a launch angle θ and with launch speed v relative to YFHB. Modeling YFHB as ﬂat and neglecting all forces other than gravitational forces, then relative to YFHB, P move in a vertical plane and its acceleration is given by a = gg ¯ ˆ where g is the unit vector in the direction of the local vertical and g is the gravitational ˆ acceleration of YFHB. We will show this fact later in the course. Introduce reference frame e ﬁxed in YFHB with origin at O as shown. Then, the position of P is completely described by the cartesian coordinates x, y where y is the height of P above the surface of YFHB and we call x the horizontal range. Let v and a be the velocity ¯ ¯ and acceleration, respectively, of P in e. Then, Hence, it follows from (4.3) that x=0 ¨ a = −g e2 . ¯ ˆ and y = −g . ¨ (4.4) Choosing t to be zero when P is launched from O , the velocity of P at launch, that is v(0), ¯ is given by v(0) = v cos θ e1 + v sin θ e2 . ¯ ˆ ˆ Hence, it follows from (4.3) that x(0) = v cos θ ˙ and y (0) = v sin θ . ˙ (4.5) Integrating relationships (4.4) from 0 to t and using the initial conditions (4.5), we obtain that x(t) = v cos θ ˙ and y (t) = v sin θ − gt . ˙ (4.6) Since x(0) = 0 and y (0) = 0 we can integrate (4.6) from 0 to t to obtain x(t) = (v cos θ)t and y (t) = (v sin θ)t − 1 gt2 . 2 (4.7) Note that if we use the ﬁrst equation above to express t in terms of x and then substitute this expression for t into the second equation, we obtain g y = (tan θ)x − x2 . (4.8) 2v 2 cos θ2 This equation tells us that the trajectory of the projectile is parabolic. 4.4. PLANAR MOTION Maximum height. 69 If th is the time at which P reaches its maximum height h, we must Figure 4.17: Maximum projectile height have y (th ) = 0. Hence, using (4.6), we obtain that ˙ y (th ) = v sin θ − gth = 0 . ˙ Solving for th yields th = v sin θ . g Since h = y (th ), substitution for th into the second equation in (4.7) yields h= v 2 sin2 θ 2g (4.9) Range at impact. Letting l be the horizontal range when P impacts YFHB and tl the corresponding time, we have y (tl ) = 0. Hence Figure 4.18: Range at impact 1 y (tl ) = (v sin θ)tl − gt2 = 0 . 2l This last equation has two solutions for tl , namely tl = 0 and tl = 2v sin θ . g It is the second solution we want. Note that this is twice the time that the projectile took to reach maximum height. We now obtain that l = x(tl ) = 2v 2 sin θ cos θ . g 70 Noting that 2 sin θ cos θ = sin(2θ) we have l= CHAPTER 4. KINEMATICS OF POINTS v 2 sin(2θ) g (4.10) It should be clear from the last expression, that if one wants to maximize the range of the projectile for a given launch speed, then one must choose the launch angle θ to be 45◦ . 4.4.3 Polar coordinates There are some situations in which it is more convenient to use polar coordinates instead of cartesian coordinates to describe planar motion. We shall see this later when we look at the motion of a satellite in orbit about YFHB (your favorite heavenly body). To describe the position of point P relative to O , we ﬁrst introduce a halfline which is ﬁxed in reference frame f , lies in the plane of motion of P and which starts at O . Then, the polar coordinates which describe the position of P are (r, θ) where r is the distance between O and P and θ is the angle between the line segment OP and the chosen reference halfline; θ is considered positive when counterclockwise. We now compute expressions for r, v and a in terms of ¯¯ ¯ polar coordinates Figure 4.19: Polar coordinates Figure 4.20: e1 and e2 ˆ ˆ 4.4. PLANAR MOTION 71 Introduce unit vectors e1 , e2 ﬁxed in the plane as illustrated in Figure 4.20. Then, the ˆˆ position of P relative to O can be expressed as r = rCθ e1 + rSθ e2 . ¯ ˆ ˆ Since e1 and e2 are ﬁxed vectors (in f ), diﬀerentiating the above expression (in f ) yields the ˆ ˆ the velocity of P ( in f ): ˙e ˙e v = (rCθ − r θSθ )ˆ1 + (rSθ + r θCθ )ˆ2 . ¯ ˙ ˙ Diﬀerentiating once more (groan!) yields the acceleration of P (in f ): ¨ ˙ ¨ ˙ a = (¨Cθ − 2rθSθ − r θSθ − r θ2 Cθ )ˆ1 + (¨Sθ + 2rθCθ + r θCθ − r θ2 Sθ )ˆ2 ¯ r ˙˙ e r ˙˙ e To obtain much simpler expressions for r , v , and a, we introduce two new unit vectors ¯¯ ¯ er and eθ as illustrated in Figure 4.21 Considering the relationships between er , eθ and e1 , e2 ˆ ˆ ˆˆ ˆˆ Figure 4.21: er and eθ ˆ ˆ (see Figure 4.22) we have Figure 4.22: er = Cθ e1 + Sθ e2 ˆ ˆ ˆ and eθ = −Sθ e1 + Cθ e2 . ˆ ˆ ˆ Recalling our expression for r, we obtain that ¯ r = r (Cθ e1 + Sθ e2 ) ¯ ˆ ˆ = r er . ˆ 72 CHAPTER 4. KINEMATICS OF POINTS This is as expected since, r is the magnitude of r and er is the unit vector in the direction ˆ of r . ¯ Rearranging our expression for v , we see that ¯ ˙ v = r(Cθ e1 + Sθ e2 ) + r θ(−Sθ e1 + Cθ e2 ) ¯ ˙ ˆ ˆ ˆ ˆ ˙eθ . = rer + r θ ˆ ˙ˆ This is much simpler! Rearranging our expression for a, we see that ¯ ˙ ¨ a = (¨ − r θ2 )(Cθ e1 + Sθ e2 ) + (r θ + 2rθ)(−Sθ e1 + Cθ e2 ) ¯ r ˆ ˆ ˙˙ ˆ ˆ ˙e ¨ = (¨ − r θ2 )ˆr + (r θ + 2rθ)ˆθ . r ˙˙ e Much simpler! Summarizing, we obtain the following expressions for position, velocity, and acceleration in terms of polar coordinates. r = r er ¯ ˆ v = vr er + vθ eθ ¯ ˆ ˆ ˆ ˆ a = ar er + aθ eθ ¯ Circular motion ˙ vθ = r θ ¨ aθ = r θ + 2rθ ˙˙ where where vr = r ˙ ˙ ar = r − r θ2 ¨ and and Figure 4.23: Polar coordinates are a natural to describe circular motion. Consider a point P in moving in a circle as illustrated in Figure 4.23. If we choose O as the center of the circle, then r is simple the radius of the circle and is constant; hence r=0 ˙ and r = 0. ¨ In describing circular motion, one sometimes introduces a new variable ˙ ω := θ . 4.4. PLANAR MOTION 73 Using the above expression for velocity in polar coordinates, we obtain that the velocity of P is given by v = v eθ ¯ ˆ where v = rω . Thus, the velocity of P is always tangential to the circle. Figure 4.24: Velocity for circular motion ¨ Noting that θ = ω and using the above above expression for acceleration in polar coor˙ dinates, we obtain that the acceleration of P is given by a = ar er + aθ eθ ¯ ˆ ˆ where ar = −rω 2 and aθ = r ω ˙ So the acceleration has both a radial and a tangential component. Since v = rw , we may express the acceleration as a = ar er + aθ eθ ¯ ˆ ˆ where ar = − v2 r and aθ = v ˙ Uniform circular motion. Suppose P is moving counterclockwise in a circle at constant speed v . Then v=0 ˙ and v2 . r Thus, the acceleration of P is always towards the center of the circle. Sometimes this is called centripetal acceleration. The above expression also holds for clockwise motion. a = ar er ¯ ˆ where ar = − 74 CHAPTER 4. KINEMATICS OF POINTS Figure 4.25: Acceleration for uniform circular motion 4.5
4.5.1 General threedimensional motion
Cartesian coordinates
r = xe1 + y e2 + z e3 ¯ ˆ ˆ ˆ 4.5.2 Cylindrical coordinates
ρ, θ, z 4.5.3 Spherical coordinates
r, θ, φ Chapter 5 Kinematics of Reference Frames
5.1 Introduction So far we have considered the kinematics of particles and points. Here we consider the kinematics of rigid bodies and reference frames. A rigid body is a body which has the property that the distance between every two particles of the body is constant with time. Although a rigid body is an idealized concept, it is a very useful concept in studying the motion of real bodies such as aircraft and spacecraft. To study the kinematics of a rigid body, we need only look at the kinematics of a reference frame in which the body is ﬁxed; we call this a body ﬁxed reference frame. Figure 5.1: Body ﬁxed frame As we shall see shortly, the study of reference frame motion is also very useful in looking at the motion of points. 5.2 A classiﬁcation of reference frame motions
g = (ˆ1 , g2 , g3 ) gˆˆ 75 Consider the motion of a reference frame g relative to another reference frame f. Suppose 76 and G is the origin of g. CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Figure 5.2: The motion of g in f 5.2. A CLASSIFICATION OF REFERENCE FRAME MOTIONS 77 Translation. The motion of g in f is a translation or g translates in f if the directions of g1 , g2 , g3 are constant in f. ˆˆˆ Figure 5.3: A translation Thus, a translation can be completely characterized by the motion of the origin of g ; hence the kinematics of translations can be completely described by the kinematics of points. A translation is said to be a rectilinear translation if the motion of the origin of g is rectilinear. Figure 5.4: A rectilinear translation Rotation. The motion of g in f is a rotation or g rotates in f if the origin of g is ﬁxed in f. The motion of g in f is a simple rotation if there if a line L containing the origin of g which is ﬁxed in both f and g. With regard to the above deﬁnition, we call L the axis of rotation and say that g rotates about L. 78 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Figure 5.5: A simple rotation Fact. Any rotation can be decomposed into at most three simple rotations. To illustrate the above fact consider the motion of g in f in the following picture where g is ﬁxed in the bar. The motion of g in f is a rotation but it is not a simple rotation. Suppose one introduces a reference frame d which is ﬁxed in the disc. Then the motion of g in f can be considered a composition of the motion of d in f followed by the motion of g in d. The latter two motions are simple rotations. Thus the motion of g in f is a composition of two simple rotations. Figure 5.6: A composition of two simple rotations General Reference Frame Motions. into a translation and a rotation. Any reference frame motion can be decomposed The above fact is illustrated by the motion of g in f in the following picture where g is ﬁxed in the wheel. The motion of g in f is neither a translation nor a rotation. Suppose one introduces reference frame a which translates in f and whose origin is at the wheel center. Then the motion of g in f can be considered a composition of the motion of a in f and the motion of g in a, that is, it is a composition of a translation and a simple rotation. 5.3. MOTIONS WITH SIMPLE ROTATIONS 79 Figure 5.7: A composition of a translation and a rotation 5.3 Motions with simple rotations Recall that the motion of a reference frame g in a reference frame f can be decomposed into a translation and a rotation. In this section, we consider special motions, namely, motions whose rotational part is a simple rotation. As illustrated in Figure 5.8, we consider the motion of g in f; this motion being a composition of a translation and a simple rotation. Figure 5.8: A motion whose rotation is simple As a consequence of the above motion, there is a line containing G (the origin of g ) which is ﬁxed in g and has ﬁxed orientation in f . We can call this the axis of rotation for the ˆ motion. In what follows we shall assume that f3 and g3 are chosen so that they are parallel ˆ ˆ to the axis of rotation. So, g moves in such a manner that g3 is always parallel to f3 . The ˆ next concept is the most important concept in the kinematics of reference frames. 5.3.1 Angular Velocity ˆ Let θ be the signed angle between g1 and f1 where θ is considered positive when g1 is counterˆ ˆ ˆ ˆ clockwise of f1 (as viewed from the head of f3 ). Then the angular velocity of g in f is denoted 80 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Figure 5.9: Angular velocity by f ω g and is deﬁned by ¯
f ˙ˆ ˙ˆ ω g = θf3 = θg3 ¯ Note that f ω g is a vector and is parallel to the axis of rotation of the motion. For ¯ ˙ practical purposes its direction can be determined by the righthand rule. The quantity θ is ˙ called a rate of rotation and the angular speed of g in f is deﬁned as  f ω g  = θ. ¯ fg If g translates in f , then ω = ¯ ¯ 0. dim [ f ω g ] = T −1 ¯ units: rad s−1 , rev min−1 Fact. Suppose B is a rigid body and g and h are any two reference frames ﬁxed in B. Then, for any reference frame f , we have f ω h = f ω g The above fact leads to the following deﬁnition ¯ ¯ for a rigid body B. Deﬁnition 1 The angular velocity of rigid body B in f is
f ω B := f ω g ¯ ¯ where g is any reference frame ﬁxed in B. Example 18 When viewed from above, a vinyl record R rotates clockwise at a the rate 1 ω = 33 3 rev/min. If reference frame f is ﬁxed in the base of the record player, then
f ˆ ω R = −ω f3 ¯ where (33 1 )(2π rad) 1 3 ω = 33 rev min−1 = = 3.491rad sec−1 . 3 60sec 5.4. THE BASIC KINEMATIC EQUATION (BKE) 81 Figure 5.10: Vinyl time 5.4 The Basic Kinematic Equation (BKE) ¯ Suppose Z is any vector function of a scalar variable t and f and g are any two reference ˙ ¯ ¯ frames. Suppose that we are interested in f Z , the derivative of Z in f , but for some reason Figure 5.11: BKE ˙ ˙ ¯ ¯ ¯ or other, it is more convenient to obtain g Z , the derivative of Z in g . Can we relate g Z to ˙ f¯ Z ? For the special motions considered in this section, the following theorem yields such a desired relationship. ¯ Theorem 1 (Basic Kinematic Equation (BKE)) If Z is any vector function of a scalar variable t, then ˙ ˙ f¯ ¯ ¯ Z = g Z + f ωg × Z ¯ Before looking at a proof of this result, let us look at some examples which illustrate the use of the BKE. These are problems which we have previously solved without using the BKE. 82 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Given: θ(t) = π/2 + t2 rad Example 19 (Pendulum with moving support) l = 1ft , ˙ h = −2ft/sec (constant) , Find: e v P and e aP at t = 0 sec. ¯ ¯ Figure 5.12: Pendulum with moving support Solution: 5.4. THE BASIC KINEMATIC EQUATION (BKE) Example 20 (Bug on bar on cart) 83 Given: The bug P is crawling along the bar which rotates counterclocwise at a rate ω relative to the cart. Figure 5.13: Pendulum with moving support Find: Nice expressions for e v P and e aP where reference frame e is ﬁxed in the wall. ¯ ¯ Solution: 84 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES 5.4. THE BASIC KINEMATIC EQUATION (BKE) 85 86 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES 5.4. THE BASIC KINEMATIC EQUATION (BKE) 87 88 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES 5.4. THE BASIC KINEMATIC EQUATION (BKE) 89 90 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES 5.4. THE BASIC KINEMATIC EQUATION (BKE) 91 92 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES 5.4.1 Polar coordinates revisited Recall that if we are studying the planar motion of a point P in a reference frame f , it is sometimes convenient to describe the motion of the point with polar coordinates, r and θ as illustrated. We previously derived expressions for v , the velocity of P in f , and a, the ¯ ¯ Figure 5.14: Polar coordinates acceleration of P in f , in terms of polar coordinates. Without the BKE, the derivation of these expressions was tedious. We now rederive these expressions using the BKE. This is one illustration of the usefulness of the BKE. Figure 5.15: A useful reference frame for polar coordinates We ﬁrst introduce a new reference frame g which consists of the three mutually perpendicular unit vectors er , eθ , e3 as illustrated. Note that ˆˆˆ
fg ˙ˆ ω = θe3 . ¯ The position of P relative to O is given by r = r er . ¯ ˆ To obtain the velocity of P in f , we use the BKE between reference frames f and g with ¯¯ Z = r to yield f g d d v= ¯ (r er ) = ˆ (r er ) + f ω g × (r er ) . ˆ ¯ ˆ dt dt 5.4. THE BASIC KINEMATIC EQUATION (BKE) Since er is ﬁxed in g , ˆ 93 d (r er ) = r er . ˆ ˙ˆ dt
f g Also, Hence, ˙ˆ ˙ˆ ω g × (r er ) = (θe3 ) × (r er ) = r θeθ ¯ ˆ ˆ ˙ˆ v = rer + r θeθ . ¯ ˙ˆ To obtain the acceleration of P in f , we use the BKE between reference frames f and g ¯¯ with Z = v to yield f ¯ dv g dv f g ¯ a= ¯ = + ω ×v. ¯ ¯ dt dt Since er and eθ are ﬁxed in g , ˆ ˆ
g dv g d ¯ ˙ˆ ¨e = (r er + r θeθ ) = r er + (r θ + r θ)ˆθ ˙ˆ ¨ˆ ˙˙ dt dt Also,
f Hence, ˙ˆ ˙ˆ ˙ˆ ω g × v = (θ e3 ) × (rer + r θeθ ) = r θeθ − r θ2 er . ¯ ¯ ˙ˆ ˙ ˙ˆ ˙e ¨ a = (¨ − r θ2 )ˆr + (r θ + 2rθ)ˆθ . ¯ r ˙˙ e 5.4.2 Proof of the BKE for motions with simple rotations
ˆˆˆ f = (f1 , f2 , f3 ) and g = (ˆ1 , g2, g3 ) gˆˆ Consider two reference frames ˆ ˆ and suppose that f3 and g3 are chosen so that g3 always has the same direction as f3 . ˆ ˆ Figure 5.16: Proof of BKE ¯ Consider any vector Z . Since g1 , g2 , g3 constitute a basis, there is a unique triplet of ˆˆˆ scalars Z1 , Z2 , Z3 such that ¯ Z = Z1 g1 + Z2 g2 + Z3 g3 . ˆ ˆ ˆ (5.1) 94 By deﬁnition, CHAPTER 5. KINEMATICS OF REFERENCE FRAMES g ˙ ¯ ˙ˆ ˙ˆ ˙ˆ Z = Z1 g1 + Z2 g2 + Z3 g3 . (5.2) Utilizing (5.1) and (5.2), we obtain
f ˙ ¯ Z= f d (Z 1 g 1 + Z 2 g 2 + Z 3 g 3 ) ˆ ˆ ˆ dt f f f dg1 ˆ dg2 ˆ dg3 ˆ ˙ˆ ˙ˆ ˙ˆ = Z1 g1 + Z2 g2 + Z3 g3 + Z1 + Z2 + Z3 dt dt dt f f f dg1 ˆ dg2 ˆ dg3 ˆ ˙ ¯ = g Z + Z1 + Z2 + Z3 . dt dt dt (5.3) To compute f dg ˆi dt we need to express gi in terms of the unit vectors of f . ˆ ˆ ˆ g1 = Cθ f1 + Sθ f2 ˆ ˆ ˆ g2 = −Sθ f1 + Cθ f2 ˆ ˆ g3 = f3 ˆ Hence,
f dg1 ˆ ˙ˆ ˙ˆ ˙ˆ = −θSθ f1 + θCθ f2 = θg2 dt f dg2 ˆ ˙ˆ ˙ˆ ˙ˆ = −θCθ f1 − θSθ f2 = −θg1 dt f dg3 ˆ =¯ 0 dt (5.4a) (5.4b) (5.4c) Looking at equations (5.4) and noting that
fg ˙ˆ ˙ˆ ω = θ f3 = θ g3 , ¯ we obtain
f dg1 ˆ ˙ˆ ˙ˆ ω × g1 = (θg3 ) × g1 = θg2 = ¯ ˆ ˆ dt f ˆ fg ˙g3 ) × g2 = −θg1 = dg2 ˙ˆ ω × g2 = (θ ˆ ¯ ˆ ˆ dt f dg3 ˆ fg ˙ˆ ω × g3 = (θg3 ) × g3 = ¯ = ¯ ˆ ˆ 0 dt fg (5.5a) (5.5b) (5.5c) It follows from (5.5) that
f Z1 f f dg1 ˆ dg2 ˆ dg3 ˆ + Z2 + Z3 = Z1 ( f ω g × g1 ) + Z2 ( f ω g × g2 ) + Z3 ( f ω g × g3 ) ¯ ˆ ¯ ˆ ¯ ˆ dt dt dt = f ω g × (Z 1 g 1 + Z 2 g 2 + Z 3 g 3 ) ¯ ˆ ˆ ˆ fg ¯ = ω ×Z. ¯ (5.6) 5.4. THE BASIC KINEMATIC EQUATION (BKE) Substitute (5.6) into (5.3) to obtain
f 95 ˙ ˙ ¯ ¯ ¯ Z = g Z + f ωg × Z ¯ In a later section, we shall see that the BKE holds for arbitrary motions of g in f. 96 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Exercises
Exercise 15 The two link planar manipulator is constrained to move in the plane deﬁned by the vectors e1 and e2 of reference frame e. Point O is ﬁxed in e. ˆ ˆ Find nice expressions for e v P and e aP in terms of θ1 , θ2 , their ﬁrst and second time ¯ ¯ derivatives, l1 , l2 and appropriate unit vectors. 5.4. THE BASIC KINEMATIC EQUATION (BKE) 97 Exercise 16 The small ball P moves in the straight slot which is ﬁxed in the disk. Relative to reference frame e, point O is ﬁxed and the disk rotates about an axis through O which is parallel to e3 and perpendicular to the plane of the disk. ˆ Find nice expressions for e v P and e aP . ¯ ¯ 98 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Exercise 17 The Thandle rotates at a constant rate Ω about a line ﬁxed in reference frame e. Your favorite bug P is strolling along one leg of the handle. Find nice expressions for e v P and e aP . ¯ ¯ 5.4. THE BASIC KINEMATIC EQUATION (BKE) 99 Exercise 18 Relative to reference frame e, the rigid frame A rotates at a constant rate rate Ω about a line passing through point O . Point P is moving along a line ﬁxed in frame A. Find nice expressions for e v P and e aP . ¯ ¯ 100 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Exercise 19 Relative to reference frame f , the disk of radius R is in simple rotation about an axis passing through point A; it oscillates according to φ(t) = cos(t2 ) . The particle P is attached to point B on the disk by a taut string of constant length l. Find nice expressions for e v P and e aP . ¯ ¯ Exercise 20 The Thandle rotates at a constant rate of Ω = 100 rpm about a line passing through point O and ﬁxed in reference frame e. Your favorite bug B is strolling along one leg of the handle with a constant speed of w = 5 mph. Given that l = 5 inches, ﬁnd e aB when the bug reaches A. ¯ Exercise 21 The pipe rotates at a constant rate of ω = 150 rpm about a vertical line passing through point O and ﬁxed in the grass. The small ball P is moving along the pipe with a constant speed of ν = 60 ft/min. Given that φ = 30 deg, ﬁnd the acceleration of the ball relative to the grass when it reaches O . 5.4. THE BASIC KINEMATIC EQUATION (BKE) 101 Exercise 22 Relative to reference frame e, the large disk B rotates at a constant rate Ω about an axis which passes through point O and is parallel to e3 . The small disk A rotates ˆ relative to B at a constant rate ω about an axis which passes through point Q and is parallel to e3 . Your favorite bug P is strolling along a radial line ﬁxed in A. ˆ Find nice expressions for e v P and e aP . ¯ ¯ Exercise 23 Relative to reference frame e, point P moves in the e1 − e2 plane with velocity ˆˆ v . Assuming v = 0, let uv be the unit vector in the direction of v ; thus ¯ ¯ ˆ ¯ v = v uv ¯ ˆ Also let uφ be the unit vector which is 90 degrees counterclockwise from uv . ˆ ˆ Show that eP ˙ˆ a = vuv + v φuφ ¯ ˙ˆ 102 CHAPTER 5. KINEMATICS OF REFERENCE FRAMES Chapter 6 General Reference Frame Motions
Here we consider the general motion of a reference frame g as seen by another reference f . Figure 6.1: General reference frame motions 6.1 Angular velocity The angular velocity of a reference g in another reference frame f (denoted f ω g ) can be ¯ rigorously deﬁned. Also the following property can be shown to hold. If f , g , and h are any three reference frames, then
f ωh = ¯ f ω g + gω h ¯ ¯ In other words, angular velocities add up like position vectors. Example 21 Propeller on pitching aircraft 103 104 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS Example 22 (Pitching and rolling aircraft.) Here θ is the pitch angle of the aircraft and φ is the roll angle. Using the above property, one can obtain the following general property If f 1 , f 2 , . . . , f n is any ﬁnite number of reference frames, then
f1 ωf = ¯ n f1 ωf + ¯ 2 f2 ωf +···+ ¯ 3 f (n−1) ωf ¯ n This relationship is very useful for practical computation of angular velocities. Recall that every rotation can be decomposed into at most three simple rotations. Since we know how to compute angular velocities for simple rotations, we can use the above relationship to compute angular velocities for general rotations. 6.1. ANGULAR VELOCITY Example 23 105 106 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS 6.2 The basic kinematic equation (BKE) ¯ The basic kinematic equation (BKE) holds for any motion of g in f . Speciﬁcally, if Z is any Figure 6.2: The Basic Kinematic Equation (BKE) vector function of time t, then
f dZ ¯ dt = gdZ ¯ dt ¯ + fω g × Z ¯ 6.2. THE BASIC KINEMATIC EQUATION (BKE) 107 Example 24 (Point on rotating pendulum) Given: The angular speed Ω is constant. Find: Nice expressions for e v P and e aP where e is ﬁxed in the ground. ¯ ¯ Figure 6.3: Example 24 Solution: 108 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS Example 25 Given: The bug P is at point B when θ = 90◦ . Also, the speed w and the angular speeds ω and Ω are constant. Find: e v P and e aP when θ = 90◦ where e is ﬁxed in the ground. ¯ ¯ Figure 6.4: Example 25 6.2. THE BASIC KINEMATIC EQUATION (BKE) Example 26 109 110 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS 6.2. THE BASIC KINEMATIC EQUATION (BKE) 111 112 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS 6.2. THE BASIC KINEMATIC EQUATION (BKE) 113 114 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS 6.2. THE BASIC KINEMATIC EQUATION (BKE) 115 116 CHAPTER 6. GENERAL REFERENCE FRAME MOTIONS Chapter 7 Angular Acceleration
Consider the motion of a reference frame g as seen by another reference frame f . Figure 7.1: Angular acceleration Angular acceleration is the time rate of change of angular velocity. The angular acceleration of g in f (denoted f α g ) is deﬁned as the time rate of change (in f ) of the angular velocity of g in f , that is, f df g fg α := ω dt If B is any rigid body and f is any reference frame, we denote the angular acceleration of B in f by f αB and deﬁne it to be equal to f αb where b is any reference frame ﬁxed in B. 117 118 CHAPTER 7. ANGULAR ACCELERATION Figure 7.2: Angular acceleration of a rigid body 119 Example 27 (Two bars on a cart) Figure 7.3: Two bars on a cart • As illustrated in the previous example, for motions with simple rotations, the angular acceleration is always parallel to the angular velocity. In general, this is not true for general rotations; see the next example. Some properties of angular acceleration. In general, one cannot add angular accelerations like angular velocities, that is,
f αg = f αh + h αg FALSE This is illustrated in the next example. 120 CHAPTER 7. ANGULAR ACCELERATION Example 28 (Rotating pendulum) Find: An expression for the angular acceleration of the bar relative to the ground. Figure 7.4: Rotating pendulum Solution: 121 Example 29 (Propeller on pitching aircraft) Given: The angular speed ω of the propeller relative to the aircraft is constant. Figure 7.5: Propeller on pitching aircraft Find: An expression for the angular acceleration of the propeller relative to the earth. Solution: 122 • Another property:
f CHAPTER 7. ANGULAR ACCELERATION αg = g df g ω dt The above relationship says that one can obtain f α g by diﬀerentiating f ω g in the g frame instead of the f frame. In other words, the rate of change of f ω g is the same for the frames f and g . To see this, use the deﬁnition of f ω g and the BKE to obtain:
f αg = df g ω dt g df g = ω+ dt g df g = ω dt f f ωg× f ωg 123 Example 30 (Arm on centrifuge) Given: The angular speed Ω of the centrifuge relative to the ground is constant. Figure 7.6: Arm on rotating centrifuge • The rotation of c relative to e: e ω c = Ωˆ3 = Ωˆ3 . ¯ e c d e c ed ω= ¯ (Ωˆ3 ) = ¯ . e 0 dt dt ˙ˆ ˙ˆ ω a = θc2 = θa2 . ¯ Recalling that Ω is constant, we have
e αc = ¯ e • The rotation of a relative to c: Hence
c a c c d c a ed ˙ ¨ˆ ω= ¯ θ c2 = θ c2 . ˆ α= ¯ dt dt • The rotation of a relative to e:
e ˙ˆ ˙ˆ ω a = e ω c + c ω a = Ωˆ3 + θ c2 = Ωˆ3 + θc2 . ¯ ¯ ¯ e c Noticing that e ω a can be expressed in terms of the vectors of c, we use the BKE to obtain ¯ that e de a ea α= ¯ ω ¯ dt c de a e c e a = ω +ω ×ω ¯ ¯ ¯ (using BKE) dt c d˙ ˙ˆ = θ c2 + Ωˆ3 + (Ωˆ3 ) × (θc2 + Ωˆ3 ) ˆ c c c dt ˙ˆ ¨ˆ = −Ωθ c1 + θc2 . Note that
e αa = e αc + c αa . ¯ ¯ ¯ 124 CHAPTER 7. ANGULAR ACCELERATION 7.1 Exercises Exercise 24 Exercise 25 Chapter 8 Kinematic Expansions Figure 8.1: Kinematic expansions 8.1 The velocity expansion Suppose we are interested in the velocity of a point P as seen by an observer in a reference f , but, it is much easier to obtain the velocity of P relative to another reference frame g . Can we relate f v P to g v P ? Yes, we can and that relationship is given by the velocity expansion ¯ ¯ (VE):
f vP = ¯ f vG + ¯ f ω g × r GP + g v P ¯ ¯ ¯ (8.1) where G is any point which is ﬁxed in reference frame g . Note that in applying the velocity expansion between f and g we must choose some convenient point G which is ﬁxed in g . Quite often, G is the origin of g . Examples and proof of VE. 125 126 CHAPTER 8. KINEMATIC EXPANSIONS Given: θ(t) = π/2 + t2 rad Example 31 (Pendulum with moving support) l = 1ft , Find: e v P at t = 0 sec. ¯ ˙ h = −2ft/sec (constant) , Figure 8.2: Pendulum with moving support Solution: 8.1. THE VELOCITY EXPANSION Example 32 (Bug on bar on cart) 127 Given: The bug P is crawling along the bar which rotates counterclocwise at a rate ω relative to the cart. Figure 8.3: Bug on bar on cart Find: A nice expression for e v P a where reference frame e is ﬁxed in the wall. ¯ Solution: 128 CHAPTER 8. KINEMATIC EXPANSIONS Example 33 (Point on rotating pendulum) Given: The angular speed Ω is constant. Find: A nice expression for e v P where e is ﬁxed in the ground. ¯ Figure 8.4: Example 33 Solution: 8.2. THE ACCELERATION EXPANSION 129 8.2 The acceleration expansion The acceleration expansion (AE) does for accelerations what the velocity expansion does for velocities. It is given by fP a= ¯ fG a + f ω g × ( f ω g × r GP ) + ¯ ¯ ¯ ¯ f αg × r GP + 2 f ω g × g v P + g a P ¯ ¯ ¯ ¯ ¯ (8.2) where G is any point which is ﬁxed in reference frame g . Examples and proof of AE. 130 CHAPTER 8. KINEMATIC EXPANSIONS Given: θ(t) = π/2 + t2 rad Example 34 (Pendulum with moving support) l = 1ft , Find: e aP at t = 0 sec. ¯ ˙ h = −2ft/sec (constant) , Figure 8.5: Pendulum with moving support Solution: 8.2. THE ACCELERATION EXPANSION Example 35 (Bug on bar on cart) 131 Given: The bug P is crawling along the bar which rotates counterclocwise at a rate ω relative to the cart. Figure 8.6: Bug on bar on cart Find: A nice expression for e aP where reference frame e is ﬁxed in the wall. ¯ Solution: 132 CHAPTER 8. KINEMATIC EXPANSIONS Example 36 (Point on rotating pendulum) Given: The angular speed Ω is constant. Find: Using the acceleration expansion, ﬁnd a nice expression for e aP where e is ﬁxed in the ¯ ground. Figure 8.7: Example 36 Solution: 8.2. THE ACCELERATION EXPANSION 133 Example 37 Given: The bug P is at point B when θ = 90◦ . Also, the speed w and the angular speeds ω and Ω are constant. Find: Using the kinematic expansions, ﬁnd expressions for e v P and e aP when θ = 90◦ where ¯ ¯ e is ﬁxed in the ground. Figure 8.8: Example 37 Solution: 134 CHAPTER 8. KINEMATIC EXPANSIONS 8.3 Exercises Exercise 26 Exercise 27 8.3. EXERCISES Exercise 28 135 Exercise 29 The disk rotates about a vertical axis through point O at a constant rate Ω. The small bug P moves relative to a slot ﬁxed in the disk at a constant speed v . Considering a reference frame g ﬁxed in the ground, ﬁnd expressions for g v P and g aP at the instant P ¯ ¯ reaches point B . 136 CHAPTER 8. KINEMATIC EXPANSIONS Chapter 9 Particle Dynamics
9.1 Introduction So far, we have considered motion without considering the causes of motion. We consider now what causes the motions of bodies. We initially look at the simplest types of bodies, namely, particles. Recall that a particle is a body that occupies a single point in space at each instant of time. Of course, this is a convenient mathematical idealization. However, it is very useful in modelling a physical body whose size is very small in comparison to other signiﬁcant sizes in the situation under consideration. Consider the motion of the earth about the sun. In this situation, a good ﬁrst approximation of the earth would be a particle. However, in studying the motion of an aircraft near the surface of the earth, a particle model of the earth is no longer useful. The point a particle occupies is called its position (or location). To study how motions are caused we need the concepts of mass and force. The mass of a body is “a measure of its resistance to change in motion.” The mass of a body is the same throughout the universe. Do not confuse the concepts of mass and weight. Mass is a positive scalar quantity. Its dimension is indicated by M and the SI and US units are kilogram (kg) and slug (slug), respectively. Forces are the interactions between bodies. Every force is due to the interaction between two bodies. For a given body, the forces exerted on it by other bodies cause its motion. The eﬀect of a force depends not only its magnitude and direction but also on where it is applied. So, we model forces with bound vectors. A bound vector is a vector which is associated with a speciﬁc point of application. Figure 9.1: Bound vector The point of application of a force acting on a particle is the position of the particle. The dimension of force is indicated by F . The SI and US units of force are the newton (N) and pound (lb), respectively, for example, ¯ F = (ˆ1 + 2ˆ2 + 3ˆ3 ) lb e e e 137 138 or CHAPTER 9. PARTICLE DYNAMICS ¯ F = (2ˆ1 + 3ˆ3 ) N . e e 9.2. NEWTON’S SECOND LAW 139 9.2 Newton’s second law
¯¯ ¯ F 1, F 2, . . . , F N , Consider a particle which is subject to a bunch of forces, ¯ as illustrated in Figure 9.2. Let ΣF be the resultant force acting on the particle, that is, it is the sum of all the forces acting on the particle. So, ¯ ¯ ¯ ¯ ΣF = F 1 + F 2 + . . . + F N =
N j =1 ¯ Fj Figure 9.2: Newton’s second law Sometimes Newton’s second law is stated as follows: In an inertial reference frame, the acceleration of a particle is always proportional to the resultant force on the particle. This proportionality constant m is called the mass of the particle. So, we have ¯ ΣF = ma ¯ where a is the acceleration of the particle in an inertial reference frame. The problem with ¯ the above statement is that it introduces the undeﬁned notion of an inertial reference frame. Another way to approach Newton’s second law is ﬁrst to deﬁne an inertial reference frame ¯ as any reference frame for which ΣF = ma always holds and then simply state Newton’s ¯ second law as: There exists an inertial reference frame. Practical inertial reference frames. For many everyday problems and problems in mechanical and civil engineering, a reference frame in which the earth is ﬁxed can be considered inertial. However, in many aerospace situations, for example, a satellite orbiting the earth, one must use as inertial a reference frame with origin at the center of the earth and with respect to which the earth rotates at a rate of one revolution per day. If one is studying the motion of the earth relative to the sun, one must consider a reference ﬁxed in the sun as inertial. 140 CHAPTER 9. PARTICLE DYNAMICS If f is an inertial reference frame and g is a reference frame which translates with constant velocity in f , then g aP = f aP ; hence g is also inertial. ¯ ¯ ¯ ΣF = ma is not good for speeds close to the speed of light. In these situations, one must ¯ resort to relativistic mechanics. ¯ From ΣF = ma, we must have ¯ F = MLT −2 1N = 1 kg m s−2 1 slug = 1 lb sec2 ft−1 Suppose ˆ1 , ˆ1 , ˆ1 are three basis vectors which are not necessarily orthogonal to each bbb other and are not necessarily ﬁxed in an inertial frame. Considering components relative to this basis, we have a = a1ˆ1 + a2ˆ2 + a3ˆ3 ¯ b b b and ¯ ΣF = (ΣF1 ) ˆ1 + (ΣF2 ) ˆ2 + (ΣF3 ) ˆ3 b b b where, for i = 1, 2, 3, the scalar ΣFi is the sum of the components in the ˆi direction of all b the forces acting on the particle. Hence we obtain the following three scalar equations: ΣF1 = ma1 ΣF2 = ma2 ΣF3 = ma3 Basically, these equations state that the resultant force in the ith direction equals the mass times the acceleration in that direction. 9.3 Static equilibrium A particle P is in static equilibrium if it is at rest in some inertial reference frame; that is, fP v = ¯ where f is inertial ¯ 0 Result. If a particle is in static equilibrium, then the sum of all the forces acting on the particle is zero, that is, ¯0 ΣF = ¯ Proof. Since the particle is at rest in an inertial reference frame, its acceleration a in that ¯ ¯ frame is zero. The above result now follows from ΣF = ma ¯ . Suppose ˆ1 , ˆ1 , ˆ1 are three basis vectors which are not necessarily orthogonal to each bbb other and are not necessarily ﬁxed in an inertial frame. Considering components relative to this basis, we have ¯ ΣF = (ΣF1 ) ˆ1 + (ΣF2 ) ˆ2 + (ΣF3 ) ˆ3 b b b 9.4. NEWTON’S THIRD LAW 141 where, for i = 1, 2, 3, the scalar ΣFi is the sum of the components in the ˆi direction of all b the forces acting on the particle. Hence we obtain the following three scalar equations: ΣF1 = 0 ΣF2 = 0 ΣF3 = 0 Basically, these equations state that the resultant force in the ith direction equals zero. 9.4 Newton’s third law Newtons Third Law has two parts: ¯ (a) If a body A exerts a force F on another body B, then the second body B exerts a force ¯ on the ﬁrst body A which is equal in magnitude but opposite in direction to F . ¯ −F Sometimes this is loosely stated as “action and reaction are equal but opposite”. Figure 9.3: Newton’s third law: ﬁrst part ¯ ¯ (b) If A and B are particles then, the forces F and −F are along the line joining the two particles. Figure 9.4: Newton’s third law: second part 142 CHAPTER 9. PARTICLE DYNAMICS 9.5 Forces As mentioned before, forces are interaction between bodies. A body is a collection of matter (solid, liquid, gas or a mixture of these states) which at each instant of time occupies some region of space. In the following discussion of forces, we consider forces between arbitrary bodies; these bodies are not necessarily particles or rigid bodies. We can divide forces into two types (a) Contact forces are due to direct contact between bodies. One example is friction. (b) Noncontact forces are exerted by bodies which are at a distance from each other and are not necessarily in contact. Examples include gravitational attraction and electromagnetic forces. 9.6
9.6.1 Gravitational attraction
Two particles Universal law of gravitation. If two particles of masses m1 and m2 are a distance r apart then, each attracts the other with a force of magnitude F =G m1 m2 r2 acting along the line joining the two particles where G = 6.673 × 10−11 m3 kg−1 s−2 is a universal constant. Figure 9.5: Universal law of gravitation The constant G is called the universal constant of gravitation. 9.6.2 A particle and a spherical body Consider a particle of mass m and a spherical body of mass M and suppose that the particle is outside of the spherical body. Suppose also that the density at each point in the sphere only depends on the radial distance of that point from the center of the sphere. Then, applying the universal law of gravitation between the particle and every particle of the sphere and 9.6. GRAVITATIONAL ATTRACTION 143 integrating over the sphere one can show that the gravitational attraction of the sphere on the particle is equivalent to that of a particle of mass M located at the center of the sphere. Thus, the resultant force exerted by the sphere on the particle has magnitude F =G Mm r2 where r is the distance of the particle from the center of the sphere. This force is along the line joining the particle to the sphere center and is directed towards the sphere center. Figure 9.6: Gravitational attraction of a sphere on a particle 9.6.3 A particle and the earth Suppose we model the earth as a spherical body. Then, the gravitational attraction of the earth on a particle above earth is equivalent to that of a particle of mass M⊕ located at the center of earth where M⊕ is the mass of the earth and is given by M⊕ = 5.976 × 1024 kg . Thus, the magnitude F of the resultant force exerted by the earth on a particle above the earth is given by M⊕ m F =G 2 r where r is the distance of the particle from the center of the earth. This force is along the line joining the particle to the center of the earth and is directed towards the center of the earth. Weight. If a particle is close to or on the surface of the earth then, r ≈ R⊕ where R⊕ is the mean radius of the earth and is given by R⊕ = 6.371 × 106 m . 144 CHAPTER 9. PARTICLE DYNAMICS Figure 9.7: Gravitational attraction of the earth on a particle Thus, near the surface of the earth, the gravitational attraction of the earth on the particle is a force of magnitude W = mg where g= GM⊕ . 2 R⊕ The scalar is called the weight of the particle. Substituting in the values for G, R⊕ and M⊕ we obtain g = 9.82 m s−2 = 32.2 ft s−2 The constant g is called the Earths surface gravitational constant. Figure 9.8: Weight As before the gravitational attraction of the earth is towards the center of the earth. This direction deﬁnes the local downward vertical direction which we usually indicate by the unit vector g . ˆ 9.7. CONTACT FORCES 145 9.7 Contact forces We idealize a contact force by idealizing the body exerting that force. 9.7.1 Strings We idealize ropes, cables, etc., as strings. A string is a onedimensional body. When it is taut and attached to another body, it exerts a force whose direction is tangential to the string and into the string at the point of attachment. If the string is not taut then, the force exerted by the string is zero. The magnitude of the force exerted by a string is called the tension in the string. Thus, a string pulls, but never pushes. Also, the direction of the force it exerts is completely determined by the string geometry. For a straight string, the force exerted by the string is parallel to the string and into the string. ¯ Mathematically, we can represent a force T due to a string as ¯ T = Tu ˆ where T is the tension in the string and the unit vector u is tangential to the string and into ˆ the string at the point of attachment. Figure 9.9: Straight strings Figure 9.10: Curved strings 146 CHAPTER 9. PARTICLE DYNAMICS Example 38 Given: The block of weight W = 10 lb is supported by two cables as shown. Figure 9.11: Example 38 Find: the tension in each cable at the block. Solution: 9.7. CONTACT FORCES Example 39 Given: The ball of mass m = 5 kg is supported by two cables as shown. 147 Figure 9.12: Example 39 Find: the tension in each cable at the ball. Solution: 148 CHAPTER 9. PARTICLE DYNAMICS Example 40 Given: The ball is supported by a string and moves in a circle of radius R = 0.25 m at a speed of 10 m/s. Figure 9.13: Example 40 Find: φ Solution: 9.7. CONTACT FORCES 149 Example 41 Given: The ball is supported by two strings and moves in a circle of radius R = 1/2 m with φ = 45◦ . Figure 9.14: Example 41 Find: Minimum value of v . Solution: 150 CHAPTER 9. PARTICLE DYNAMICS 9.7. CONTACT FORCES 151 152 CHAPTER 9. PARTICLE DYNAMICS 9.7. CONTACT FORCES 153 154 CHAPTER 9. PARTICLE DYNAMICS 9.8
9.8.1 ¯ Application of ΣF = ma ¯
Free body diagrams In order to reliably model the forces on a given body, a free body diagram (FBD) is drawn. A free body diagram of a body is a picture containing the body and all the external forces acting on the body. All other bodies are replaced by the forces they exert on the given body. ¯ A FBD should be drawn before applying ΣF = ma. ¯ The FBD should contain all available information on the forces. All forces should be labeled. 9.8.2 A systematic procedure ¯ The following is a systematic procedure for applying ΣF = ma to obtain scalar equations. ¯ (a) Obtain the inertial acceleration a. In static problems, this is trivial because a = 0. ¯ ¯ (b) Model all the forces on the body. A free body diagram is needed at this step. (c) Introduce a set of basis vectors and resolve the acceleration and all the forces into components relative to this basis. ¯ (d) Apply ΣF = ma. ¯ (e) Obtain scalar equations. Sometimes step (c) is performed as part of step (e). 9.9. FORCES DUE TO SMOOTH SURFACES AND CURVES 155 9.9
9.9.1 Forces due to smooth surfaces and curves
Smooth surfaces Consider a smooth small block on a smooth ﬂat table. What can you say about the direction Figure 9.15: Block on table of the force exerted by the table on the ball? Figure 9.16: Force exerted by table on block Consider YFI (your favorite insect) on YFBC (your favorite beverage can). Assuming Figure 9.17: YFI on YFBC the can is cylindrical and there is no friction between YFI and YFBC, what can you say about the direction of the force exerted by YFBC on YFI? 156 CHAPTER 9. PARTICLE DYNAMICS Figure 9.18: Force exerted by YFBC on YFI Consider now the general situation of a particle in contact with a smooth surface. By a surface, we mean a 2dimensional geometric object such as a plane or the “surface” of a cylinder. At every point on a surface (except at corners), there is a well deﬁned line which passes through the point and is normal (perpendicular) to the surface at the point. Figure 9.19: Particle on a surface The force exerted by a smooth surface on a particle in contact with the surface is normal to the surface at the point of contact and is directed towards the particle. (Surfaces don’t suck.) We can represent this by ¯ N = Nu ˆ N ≥0 where the unit vector u is normal to the surface at the location of the particle and is directed ˆ towards the particle . Thus the direction of the force exerted by a smooth surface is com Figure 9.20: Normal force due to a surface pletely determined by the surface geometry and the location of the particle. This force is called a normal force. 9.9. FORCES DUE TO SMOOTH SURFACES AND CURVES 157 Example 42 Given: The block of mass m = 2 kg lies on a smooth inclined plane (Θ = 30◦ ) and is attached to the wall via a string. Find: The tension in the string. Figure 9.21: Example 42 Solution: 158 CHAPTER 9. PARTICLE DYNAMICS Example 43 Given: The block of mass m lies on a smooth inclined plane which rotates about a vertical axis at a constant rate ω . The block is attached to a point on the rotation axis via a string of length l. Find: An expression for the minimum value of ω . Figure 9.22: Example 43 Solution: 9.9. FORCES DUE TO SMOOTH SURFACES AND CURVES 159 9.9.2 Smooth curves Consider a smooth bead constrained to move along a smooth straight wire. Figure 9.23: Smooth bead on smooth wire What can you say about the force exerted by the wire on the bead? The force exerted by the wire on the bead is in the plane which is perpendicular or normal to the wire. Figure 9.24: Force exerted by wire on bead Consider a smooth small ball constrained to move inside a smooth circular tube . What Figure 9.25: Smooth ball in smooth tube can you say about the force exerted by the tube on the ball? The force exerted by the tube on the ball is in the plane which is perpendicular or normal to the line which is tangent to the tube at the ball’s location. 160 CHAPTER 9. PARTICLE DYNAMICS Figure 9.26: Force exerted by tube on ball Consider now the general situation of a particle in contact with a curve. By a curve, we mean a onedimensional geometric object such as a straight line or a circle. At every point on a curve (except at corners), there is a well deﬁned plane which passes through the point and is normal (perpendicular) to the line which is tangent to the curve at the point. We say that this plane is normal to the curve. Figure 9.27: Particle in contact with a curve ¯ The force N exerted by a smooth curve on a particle is normal to the curve at the position of the particle, that is, it lies in the plane which is normal to the curve at the location of the particle. We can represent this by ¯ N = N2 u2 + N3 u3 ˆ ˆ where u2 , u3 are any two independent unit vectors which are normal to the curve at the ˆˆ location of the particle Note that in contrast to smooth surfaces, the direction of the force exerted by a smooth curve is unknown. This force is called a normal force. 9.9. FORCES DUE TO SMOOTH SURFACES AND CURVES 161 Figure 9.28: Normal force due to a curve 162 CHAPTER 9. PARTICLE DYNAMICS Example 44 Given: The smooth ball of mass m is being pulled by the string (of varying length l) relative to the smooth tube at a constant speed v . The tube rotates about a vertical axis at a constant angular speed of ω . Find: Expressions for (a) the tension in the string (b) the force exerted by the tube on the ball. Figure 9.29: Example 44 Solution: 9.10. ROUGH SURFACES, ROUGH CURVES AND FRICTION 163 9.10 Rough surfaces, rough curves and friction Consider a small block on a rough ﬂat table. What can we say about the force exerted by Figure 9.30: Small block on a rough table the table on the block? Figure 9.31: Forces due to a rough table Consider now YFI on a a rough YFBC. What can we say about the force exerted by the Figure 9.32: Bug on a rough can can on the bug? 164 CHAPTER 9. PARTICLE DYNAMICS Figure 9.33: Forces due to a rough can Consider now the general situation of a particle on a rough surface. The total force exerted Figure 9.34: Particle on a rough surface by a rough surface on a particle is represented by ¯ ¯ N + Ff ¯ • The normal force N is normal to the surface at P and is towards P . ¯ • The friction force F f is tangential to the surface at P . Figure 9.35: Forces due to a rough surface If we introduce a bunch of basis vectors u1 , u2 , u3 where u1 , u2 are tangential to the surface ˆˆˆ ˆˆ at P and u3 is normal to the surface at P , then ˆ
f f ¯ F f = F1 u1 + F2 u2 ˆ ˆ ¯ N = N u3 ˆ N3 ≥ 0 9.10. ROUGH SURFACES, ROUGH CURVES AND FRICTION Coulomb friction 165 A very common type of friction is Coulomb friction or dry friction. It usually occurs between two dry solid bodies in contact with each other. How do we walk? Why do motorcycles move? Consider the general situation of a particle which is constrained to remain on a rough surface and suppose the friction force between the particle and the surface is due to Coulomb friction. Figure 9.36: Particle moving relative to surface The Coulomb friction force depends on whether there is relative motion between the particle and the surface. To describe this, introduce a reference frame e in which the surface is ﬁxed and let v be the velocity of the particle in e, that is, v is the velocity of the particle relative ¯ ¯ to the surface, that is relative to e. ¯ where N = N  is the magnitude of the normal force exerted by the surface on the particle. The nonnegative constant µ = µs is called a coeﬃcient of static friction. It depends the surface properties of the objects in contact. Some examples are: rubber on asphalt: µs = 0.85 rubber on ice: µs = 0.1 Static friction. Consider ﬁrst the case in which there is no motion of the particle relative to the surface, that is, v = ¯ Then the only additional statement that we can make about ¯ 0. ¯ f exerted by the surface on the particle is that its magnitude F f must the friction force F satisfy the inequality, F f ≤ µN , In static friction, the friction force is determined by the other forces on the particle and by the acceleration of the particle. In general, Ff =
f f (F1 )2 + (F2 )2 f In one of the components is zero then the expression for F f is simpler, for example, if F2 is zero then, f F f = F1  . 166 CHAPTER 9. PARTICLE DYNAMICS Sliding friction. Consider now the case in which there is motion of the particle relative to the surface, that is the particle is sliding on the surface and v = ¯ In this case the direction ¯ 0. ¯ f must be opposite to that of the velocity v . Also the magnitude F f of the friction force F ¯ of the friction force must satisfy the equality F f = µN The nonnegative constant µ = µk is called a coeﬃcient of kinetic friction. It depends on the surface properties of the objects in contact. Usually, µk < µs Why do you achieve maximum braking in a road vehicle by applying the brakes to the point where the wheels are just about to slip? 9.10. ROUGH SURFACES, ROUGH CURVES AND FRICTION 167 Example 45 Given: The block is in static equilibrium on the rough plane which is inclined at an angle of θ to the horizontal. The coeﬃcient of static friction between the block and the plane is µ. Figure 9.37: Example 45 Find: An expression for the maximum value of θ. Solution: 168 CHAPTER 9. PARTICLE DYNAMICS Example 46 Given: The weight of the block is 9 lb and the coeﬃcient of static friction between the block and the rough surface is µ = 0.5. Figure 9.38: Example 46 Find: out whether or not this block can be in static equilibrium? Solution: 9.10. ROUGH SURFACES, ROUGH CURVES AND FRICTION 169 Example 47 Given: The block is at rest relative to the rough inclined plane; this plane rotates at a constant rate Ω about a vertical axis. The coeﬃcient of static friction between the block and the plane is µ. Consider γ = 45◦ and µ = 1/2. Figure 9.39: Example 47 Find: Expressions for the range of possible values of Ω. Solution: 170 Linear viscous friction CHAPTER 9. PARTICLE DYNAMICS Sometimes, the friction between two lubricated bodies can be modelled as linear viscous friction. This type of friction force is opposite in direction to the velocity v and is proportional ¯ to v . Mathematicaly, this can be expressed as ¯ ¯ F f = −cv ¯ where the constant c is nonnegative. This constant c is called a linear damping coeﬃcient. Note that, unlike coulomb friction, this type of friction is zero when the velocity v is zero. ¯ 9.10.1 Rough curves Similar to rough surfaces. Friction force is 9.10. ROUGH SURFACES, ROUGH CURVES AND FRICTION 171 9.10.2 Springs A common component in many machines and vehicles is a spring. We model a spring as a deformable one dimensional body of some length l. Usually springs are straight; but they can also be curvy. Graphical representations of springs are given in Figure 9.40. Figure 9.40: Springs Every spring has a rest length, free length or unstretched length l0 . This is the length of the spring when it is not exerting any forces; thus it is not subject to any forces. When extended beyond its rest length, it pulls and we say that the spring is in tension. When compressed under its rest length, it pushes and we say the spring is in compression. Figure 9.41: Unstretched length, tension, compression The force exerted by a spring attached to another body is tangential to the spring at the point of contact of the spring with the body. For a straight spring, the force is parallel to the spring. Thus the line of action of the force exerted by a spring is completely determined by the spring geometry. Thus we can represent a spring force by a single scalar S . Vectorially, we can represent a spring force by ¯ S = Su ˆ where u is a unit vector which is is tangential to the spring at its point of attachment and ˆ points into the spring; see Figure 9.42. In this ﬁgure, a positive value of S corresponds to the spring being in tension, while a negative value of S corresponds to the spring being in compression. The scalar S depends only on the deformation of the spring from its unstretched state. Let x be the change in length of the spring from its rest length, that is, x = l − l0 where l 172 CHAPTER 9. PARTICLE DYNAMICS Figure 9.42: Spring force is the current length of the spring and l0 is the unstretched length of the spring. Then, x represents the amount by which the spring is extended (x > 0) or contracted (x < 0). Also S is positive when x is positive and S is negative when x is negative. The graph of S versus Figure 9.43: Spring deﬂection x for a spring is called the characteristic curve of the spring; see Figure 9.44. Figure 9.44: Spring characteristic curve The simplest characteristic characteristic curve is linear, that is, S = kx for some positive scalar k . This is sometimes referred to as Hookes Law and is illustrated in Figure 9.45. The proportionality constant k is called the spring constant for the spring. It has dimension F L−1 and possible units are N/m or lbs/ft. 9.11. DASHPOTS 173 Figure 9.45: Linear spring Nonlinear springs. A softening spring, is a spring for which the slope of its characteristic curve decreases with increasing deﬂection magnitude; see Figure 9.46. Figure 9.46: Softening spring A hardening spring is a spring for which the slope of its characteristic curve increases with increasing deﬂection magnitude; see Figure 9.47. Figure 9.47: Hardening spring 9.11 Dashpots By a dashpot we mean a onedimensional massless deformable body with the property that the force it exerts depends only on its rate of extension or compression. Dashpots are useful for modeling many types of damping devices and the damping behavior of vehicle suspension components. 174 CHAPTER 9. PARTICLE DYNAMICS Example 48 (Parallel springs) Consider two springs of the same free length in parallel as shown in Figure 9.48. Figure 9.48: Springs in parallel 9.11. DASHPOTS Example 49 (Series springs) Consider two springs in series as shown in Figure 9.48. 175 Figure 9.49: Springs in series 176 CHAPTER 9. PARTICLE DYNAMICS Example 50 Given: W = 1 lb, k = 2 lb ft−1 , µ = 1/2 and θ = 30◦ . Figure 9.50: Example 50 Find: The range of possible values of the spring deﬂection x. 9.12. EXERCISES 177 9.12 exercises Exercise 30 The rough inclined plane is rotating about a vertical axis at a constant rate Ω. The small block of mass 0.1 kg rests on the inclined plane. The coeﬃcient of static friction between the block and the plane is µ = 1/2. If θ = 45◦ and d = 100 mm, determine the minimum and maximum values of Ω. Exercise 31 The small ball of mass m = 0.1 kg rests on a hemisphere of radius R = 0.25m and is attached to point O via a string. Find the tension in the string. 178 CHAPTER 9. PARTICLE DYNAMICS Chapter 10 Equations of Motion In designing a spacecraft or aircraft, we like to know how it is going to behave before ﬂying it. If aircraft and pilots were expendable like darts, one could probably design them totally by experimental trial and error, that is, dream up a design and ﬂy the vehicle to see if it is cool or sucks. Then based on the outcome, make modiﬁcations and “ﬂy” again. This is called the Beavis and Butthead approach to aerospace vehicle design. Many men–women and much expense would have been incurred in trying to land on the moon by this method. So, before ﬂying an aerospace vehicle, we want to be able to predict its behavior as accurately as possible. This we do by developing a mathematical model of the vehicle. The most common model is a set of diﬀerential equations which describe the motion of the vehicle. These are called equations of motion. Of course, the concept of an equation of motion applies to any system described by the laws of mechanics. Actually, the idea of describing the behavior of a physical system using diﬀerential equation extends to all braches of engineering and science. It has even been used in economics. Let us begin with some simple systems. 10.1 Single degree of freedom systems Example 1 (I’m falling) Consider particle P in vertical free fall near the surface of Y F HB . ¯ Neglecting any ﬂuid resistance, application of ΣF = ma in a vertical direction yields ¯ x = −g ¨ (10.1) where g is the gravitational acceleration constant of YFHB. This equation is a second order ordinary diﬀerential equation. It has the property that, given any initial displacement x0 and any initial rate of displacement v0 at any initial time t0 , it has a unique solution for x(t). Speciﬁcally, integrating (10.1) with respect to time over the interval [0, t] yields x(t) − x(0) = ˙ ˙ hence, x(t) = v0 − gt ˙ 179
t 0 −g dτ 180 CHAPTER 10. EQUATIONS OF MOTION Figure 10.1: I’m falling where v0 = x(0). Integrating again yields ˙ x(t) − x(t0 ) = v0 t − that is, x(t) = x0 + v0 t − g2 t 2 g2 t (10.2) 2 where x0 = x(0). From this last equation it should be clear that if x0 and v0 are speciﬁed then x(t) is determined for all t, that is, the motion of P is completely determined by its initial position and velocity. Also all motions of P are given by this expression. For the above reasons, equation (10.1) is called a (scalar) equation of motion (EOM) for P . Its solutions describe the manner in which x changes with time. Since the position of P is completely speciﬁed by x, this equation describes all possible motions of P . Exercise 1 Show that if x(t0 ) = x0 x(t0 ) = v0 ˙ g (t − t0 )2 2 then x(t) is given by (10.2) with t replaced by t − t0 on the righthandside of (10.2), that is, x(t) = x0 + v0 (t − t0 ) − Example 2 (The simple harmonic oscillator) The “small” block P of mass m moves without friction along a straight horizontal line ﬁxed in YFHB. It is connected to point A by a linear spring of spring constant k and free length l0 . Show that the motion of P is described by mx + kx = 0 ¨ ¯ Solution. We shall apply ΣF = ma to P . ¯ Choose reference frame e, ﬁxed in YFHB, as inertial. Then, a = e aP = xe1 ¯ ¯ ¨ˆ 10.1. SINGLE DEGREE OF FREEDOM SYSTEMS 181 Figure 10.2: Simple harmonic oscillator Figure 10.3: Kinematics of the simple harmonic oscillator ¯ Application of ΣF = ma yields ¯ −W e2 + N e2 − kx e1 = mx e1 ˆ ˆ ˆ ¨ˆ The e1 components of the above equation yield: ˆ e1 : −kx = mx ˆ ¨ Hence, mx + kx = 0 ¨ (10.3) The above equation is an equation of motion for P . It is a linear, second order, ordinary, diﬀerential equation. It has the property that, given any initial displacement x0 and any Figure 10.4: Free body diagram of P 182 CHAPTER 10. EQUATIONS OF MOTION Figure 10.5: Simple harmonic oscillator with an attitude initial rate of displacement v0 at some initial time t0 , it has a unique solution for x(t) at any other time t. In fact, with t0 = 0, the solution is given by: x(t) = x0 cos(ωt) + (v0 /ω ) sin(ωt) where and Note that we can rewrite (10.3) as x + ω2x = 0 ¨ (10.7) x0 = x(0) ; ω=
△ △ (10.4) v0 = x(0) ˙ k /m △ (10.5) (10.6) Exercise 2 Show that the above expression for x(t) is the solution to (10.3) with initial conditions (10.5). Exercise 3 (Simple harmonic oscillator with an attitude) Show that the motion of P is described by mx + kx + g sin θ = 0 ¨ Example 3 (The simple pendulum) The simple pendulum consists of a particle P attached to a YFHB ﬁxed point via a taut string of length l. It is constrained to move in a vertical plane. The position of P is completely speciﬁed by θ, the angle between the string and a vertical line. Show that the motion of P is governed by: ¨g θ + sin θ = 0 l where g is the gravitational acceleration constant of YFHB. 10.1. SINGLE DEGREE OF FREEDOM SYSTEMS 183 Figure 10.6: Kinematics of the simple pendulum Figure 10.7: Free body diagram of pendulum bob The simple pendulum ¯ Solution. We shall apply ΣF = ma to P . ¯ Choose reference frame e, ﬁxed in YFHB, as inertial. Then, ¨ˆ ˙ˆ a = e aP = lθs1 + lθ2 s2 ¯ ¯ where reference frame s is ﬁxed in the string. ¯ Now applying ΣF = ma yields ¯ 184 CHAPTER 10. EQUATIONS OF MOTION ¨ˆ ˙ˆ −mg e2 + T s2 = m(lθ s1 + lθ2 s2 ) ˆ ˆ where m is the mass of P . Since e2 = sin θ s1 + cos θ s2 ˆ ˆ ˆ we obtain ¨ s1 : −mg sin θ ˆ = mlθ ˙ s2 : −mg cos θ + T = mlθ2 ˆ (10.8) The ﬁrst equation yields ¨g θ + sin θ = 0 l (10.9) ¯ Note that application of ΣF = ma in the above example yielded two equations. Suppose ¯ you were not given the EOM in the problem statement; why would you choose the ﬁrst of equations (10.8) as the EOM? In general an EOM should only depend on the coordinate of interest (θ in this case), its ﬁrst and second derivatives, and system parameters such as masses, spring constants, etc. Things like normal forces or string tensions should not appear in the ﬁnal EOM. In the above example, one could use the second of the equations in (10.8) to solve for the string tension T as a function of P ’s motion. In general, if a particle is ¯ constrained to move along a curve (a circle in this example) application of ΣF = ma in the ¯ the direction tangential to the curve yields the equation of motion. The above EOM is a second order nonlinear diﬀerential equation. It has the property that for each set of initial conditions: θ(0) = θ0 ˙ ˙ θ(0) = θ0 there is a unique solution θ(t) satisfying these conditions. However, since the equation is nonlinear, you cannot use the techniques (Laplace etc.) learned in MA 262, MA 265, MA 266 to solve it. Although it is possible to obtain exact solutions to this equation, in general one cannot exactly solve nonlinear diﬀerential equations. To solve them one has to resolve to approximate numerical techniques . A very special solution to (10.9) is the equilibrium solution θ(t) ≡ 0 ˙ which corresponds to initial conditions θ(0) = θ (0) = 0. Suppose we are interested in the behaviour of the system near this equilibrium solution. For “small” θ, sin θ ≈ θ 10.2. NUMERICAL SIMULATION and the EOM (10.9) can be approximated by ¨g θ+ θ=0 l This looks familar, especially if we write it as ¨ θ + ω2θ = 0 where ω= g /l 185 (10.10) (Recall (10.7)). All of the systems considered so far were described by a single second order diﬀerential equation of the form F (¨, x, x, t) = 0 x˙ This is because we only needed one coordinate to completely describe each of these systems. Such systems are called single degree of freedom systems. In a later section we look at multi degree of freedom systems. 10.2
10.2.1 Numerical simulation
First order representation
y1 , y2 , . . . , yn By appropriate deﬁnition of state variables one can rewrite any system of ordinary diﬀerential equations as a bunch of ﬁrst order ordinary diﬀerential equations of the general form: y1 = f1 (t, y1 , y2 , . . . , yn ) ˙ y2 = f2 (t, y1 , y2 , . . . , yn ) ˙ . . . yn = fn (t, y1 , y2 , . . . , yn ) ˙ Note that there is one equation for each state variable. Example 51 The simple harmonic oscillator. Letting y1 = x this system has the ﬁrst order representation: y˙1 = y2 y˙2 = − k y1 m y2 := x ˙ 186 Example 52 The simple pendulum. With y1 := θ this system has the ﬁrst order representation: CHAPTER 10. EQUATIONS OF MOTION ˙ y2 := θ y˙1 = y2 g y˙2 = − sin y1 l 10.2. NUMERICAL SIMULATION 187 10.2.2 Numerical simulation with MATLAB >> help ode45 ODE45 Solve differential equations, higher order method. ODE45 integrates a system of ordinary differential equations using 4th and 5th order RungeKutta formulas. [T,Y] = ODE45(’yprime’, T0, Tfinal, Y0) integrates the system of ordinary differential equations described by the Mfile YPRIME.M, over the interval T0 to Tfinal, with initial conditions Y0. [T, Y] = ODE45(F, T0, Tfinal, Y0, TOL, 1) uses tolerance TOL and displays status while the integration proceeds. INPUT: F  String containing name of usersupplied problem description. Call: yprime = fun(t,y) where F = ’fun’. t  Time (scalar). y  Solution columnvector. yprime  Returned derivative columnvector; yprime(i) = dy(i)/dt. t0  Initial value of t. tfinal Final value of t. y0  Initial value columnvector. tol  The desired accuracy. (Default: tol = 1.e6). trace  If nonzero, each step is printed. (Default: trace = 0). OUTPUT: T  Returned integration time points (columnvector). Y  Returned solution, one solution columnvector per toutvalue. The result can be displayed by: plot(tout, yout). See also ODE23, ODEDEMO. Example 53 Lets say we want to numerically simulate the simple pendulum over the time interval 0 ≤ t ≤ 20 sec with parameters l=1 and initial conditions θ(0) = π/2 rad g=1 ˙ θ(0) = 0 We ﬁrst write the equations in ﬁrst order form; recall example 52. Next we create an Mﬁle (lets call it pendulum.m) with the following lines. function ydot = pendulum(t,y) ydot(1) = y(2) ydot(2) = sin(y(1)) 188 We now simulate in MATLAB CHAPTER 10. EQUATIONS OF MOTION >>[t,y]=ode45(’pendulum’,0,25,[pi/2; 0]) To get a plot: >>plot(t,y)
2 1.5 1 0.5 0 0.5 1 1.5 2 0 2 4 6 8 10 12 14 16 18 20 Now lets say we only want a plot of y1 vs t. >>plot(t,y(:,1))
2 1.5 1 0.5 0 0.5 1 1.5 2 0 2 4 6 8 10 12 14 16 18 20 Now suppose we want to plot y1 vs. y2 . This is called a phase plane or state plane plot. >>plot(y(:,1),y(:,2)) 10.2. NUMERICAL SIMULATION 189 1.5 1 0.5 0 0.5 1 1.5 2 1.5 1 0.5 0 0.5 1 1.5 2 190 CHAPTER 10. EQUATIONS OF MOTION 10.3 Multi degree of freedom systems Example 4 (The cannonball: ballistics in drag) Consider a cannonball P of mass m in ﬂight in a vertical plane near the surface of the earth. Suppose we model the aerodynamic force on P as a force of magnitude D (v ) acting opposite to the velocity v of the cannonball ¯ relative to the earth and only dependent on the corresponding speed v := v . Show that the ¯ motion of P is governed by p ˙ ˙ h mv ˙ mv γ ˙ = = = = v cos γ v sin γ −mg sin γ − D (v ) −mg cos γ (10.11) where γ is the ﬂight path angle and p and h are the horizontal range and altitude of P , respectively Figure 10.8: Cannonball Solution. First note that v= ¯ = d OP (¯ ) r dt e d (p e1 + h e2 ) ˆ ˆ dt
e ˙ˆ = pe1 + he2 ˙ˆ Also, v = v cos γ e1 + v sin γ e2 ¯ ˆ ˆ Comparing these two expressions for v yields ¯ p = v cos γ ˙ ˙ h = v sin γ ¯ Now we apply ΣF = ma to P . ¯ 10.3. MULTI DEGREE OF FREEDOM SYSTEMS 191 Reference frame u Introduce reference frame u = (ˆv , uγ , u3 ) uˆˆ where uv is the unit vector in the direction of v , uγ is the unit vector which is 90 degrees ˆ ¯ˆ counterclockwise from uv , and u3 = e3 . Then v = v uv ; e ω u = γ u3 ; and application of the ˆ ˆ ˆ ¯ ˆ ¯ ˙ˆ BKE between frames u and e yields a := ¯ =
eP u a= ¯ d (¯) v dt e d (¯) + e ω u × v v ¯ ¯ dt = v uv + v γ uγ ˙ˆ ˙ˆ Free body diagram of P ¯ Application of ΣF = ma yields: ¯ uv : ˆ uγ : ˆ mv = −mg sin γ − D (v ) ˙ mv γ = −mg cos γ ˙ In the above example, the EOMs consisted of four ﬁrst order diﬀerential equations. We could have obtained two second order diﬀerential equations in the coordinates p and h; however, they are not as nice as (10.11). 192 CHAPTER 10. EQUATIONS OF MOTION Exercise 4 For the above example, obtain two second order diﬀerential equations which describe the motion of P . Exercise 5 (Sprung together) Consider a system consisting of two particles P1 and P2 , connected together by a linear spring and constrained to move along a smooth horizontal line ﬁxed in YFHB. Show that the motion of this system can be described by m1 q1 + k (q1 − q2 ) = 0 ¨ m2 q2 − k (q1 − q2 ) = 0 ¨ Sprung together 10.4. CENTRAL FORCE MOTION 193 10.4 Central force motion A force is called a central force if its line of action always passes through an inertially ﬁxed point. We call that point the force center. A particle is said to undergo central force motion if the only force acting on it is a central force. Central force motion Example 5 The table oscillator Consider particle P which is constrained to move on the surface of a smooth horizontal table and is attached to inertially ﬁxed point O by a linear spring of spring constant k and free length l0 . The table oscillator ¯ Consideration of ΣF = ma in a vertical direction shows that the normal force cancels ¯ out the weight force. Free body diagram of P Hence, the original free body diagram of P is equivalent to the next free body diagram. Thus P undergoes a central force motion where the central force is the spring force. 194 CHAPTER 10. EQUATIONS OF MOTION Equivalent free body diagram of P Example 6 Some orbit mechanics Consider a body P of mass m in orbit about YFHB B of mass M . Some orbit mechanics Modelling B as a sphere whose mass density depends only on the distance from its center, the gravitational attaction of B on P is a force directed towards the center of B. If we consider the situation in which B is relatively massive in comparison to P , that is, M ≫m then we may regard the center of B as inertially ﬁxed. Hence P undergoes a central force motion. Examples of this include: B earth earth moon sun P you satelite you earth 10.4.1 Equations of motion We shall see later that every central force motion is a planar motion and the plane of motion must contain the force center. The plane is determined by an initial position and initial velocity of the particle. We use polar coordinates (r, θ) to describe central force motion. 10.4. CENTRAL FORCE MOTION 195 Kinematics of central force motion The inertial acceleration of P is given by ˙ˆ ¨ a = (¨ − r θ2 ) er + (r θ + 2rθ) eθ ¯ r ˙˙ ˆ Free body diagram of P ¯ Applying ΣF = ma yields ¯ −F er = ma ˆ ¯ ¯ Looking at the er and eθ components of ΣF = ma yields the following two EOMs: ˆ ˆ ¯ ˙ r − r θ2 + F/m = 0 ¨ ¨ r θ + 2r θ ˙˙ =0 196 CHAPTER 10. EQUATIONS OF MOTION 10.4.2 Some orbit mechanics Consider the motion of a small body P of mass m about a much larger spherical body B of mass M . We can regard the center of the spherical body as inertially ﬁxed; hence the motion of the smaller body is a central force motion with F= GMm r2 Some orbit mechanics Recalling the above EOMs for general central force motion, the motion of P is described by ˙ r − r θ2 + µ/r 2 = 0 ¨ ¨ r θ + 2r θ ˙˙ =0 where µ := GM . All solutions of the above two diﬀerential equations are conic sections; that is they satisfy an equation of the form: a r= 1 + b cos(θ − c) The constants a, b, and c depend on the motion. b=0 0<b<1 b=1 b>1 circle ellipse parabola hyperbola For the moment, we will only look at circular orbits. AAE 340 contains a closer look at all orbits. AAE 532 (Orbit mechanics) is a whole course devoted to orbit mechanics. Circular orbits Let’s look for solutions corresponding to circular orbits, that is, r (t) ≡ R where R is constant; hence r = r = 0. The above EOMs reduce to ¨˙ ˙ −Rθ2 + µ/R2 = 0 ¨ Rθ =0 10.4. CENTRAL FORCE MOTION ˙ The second equation implies that θ is constant, that is, ˙ θ(t) ≡ ω where ω is constant. The ﬁrst equation now implies that R3 ω 2 = µ hence ω= Geostationary orbits µ/R3 or R = (µ/ω 2)1/3 197 (10.12) Suppose one wants to position a satellite so that it always remains above a ﬁxed point on the earth. To study this motion, we need to take an inertial reference frame in which the earth rotates about its northsouth axis at the rate: ωe = 1rev/24 hour = (2π rad) (24)(60)(60) sec = 7.272 × 10−5 rad/sec Inertial reference frame Since the satellite must move in a plane which contains the center of the earth, it must be located above the equator. 198 CHAPTER 10. EQUATIONS OF MOTION Geostationary orbit Using (10.12), the satellite must be located at the following distance from the center of the earth: R= =
2 GMearth /ωe 1/3 1/3 (6.673 × 10−11 )(5.976 × 1024 ) (7.272 × 10−5)2 = 42, 247 km Hence, the satellite must be located at a height h = R − Rearth = 42, 247 − 12, 755/2 = 35, 870 km Chapter 11 Statics of Bodies
Prior to this, we have considered the statics of particles. In this chapter, we consider the statics of general bodies. First, we need a new concept which is basic in the study of the statics and dynamics of bodies, namely, the moment of a force. 11.1 The moment of a force The moment of a force about a point is its turning eﬀect about that point. As an example, think of a person pushing or pulling on one end of a joystick and consider the turning eﬀect of this force about the pivot point at the other end of the joystick. The formal deﬁnition of a moment is as follows. ¯ The moment of a force F about a point Q is deﬁned by ¯ MQ = r × F ¯¯ ¯ where r = QP (the vector from Q to P ) and P is the point of application of F . ¯
F ¯ •
r ¯ P •Q
Figure 11.1: Moment of a force • Recalling the deﬁnition of the cross product, it follows that ¯ M Q = M Q n where M Q = rF sin θ ˆ 199 200 CHAPTER 11. STATICS OF BODIES Here r is the distance from Q to the point of application of the force, F is the magnitude of ¯ the force, θ is the angle between r and F , and n is the unit vector which is normal to both ¯ ˆ ¯ and whose sense is given by the righthand rule; see Figure 11.2. Note that M Q is r and F ¯ ¯ the magnitude of M Q .
F θ •
r ˆ n •Q
Figure 11.2: Moment of a force again • Units: newtonmeter (N·m) or footpound (ft·lb). 11.1. THE MOMENT OF A FORCE Example 54 201 202 CHAPTER 11. STATICS OF BODIES The next fact tells us that in evaluating the moment of a force, we can choose the position vector to terminate at any point on the line of action of the force. ¯ Fact 2 If P is any point on the line of action of F , then ¯ MQ = r × F ¯¯ where r = QP (the vector from Q to P ). ¯ Figure 11.3: Any point on the line of action will do. Proof. By deﬁnition, ¯ ¯ M Q = QP ∗ × F ¯ where P ∗ is the point of application of F . Since QP ∗ = QP + P P ∗ ¯ ¯ ¯ M Q = QP × F + P P ∗ × F ¯ Since the points P and P ∗ are on the line of action of F , the vector P P ∗ is along the line of ¯ ; hence action of F ¯0 PP∗ × F = ¯ and ¯ ¯ M Q = QP × F we have 11.1. THE MOMENT OF A FORCE Example 55 203 204 CHAPTER 11. STATICS OF BODIES The following fact is useful for determining moments by inspection, especially in planar problems. Fact 3 If d is the distance from a point Q to the line of action of a force of magnitude F, then the magnitude M Q of the moment of that force about Q is given by M Q = dF . Figure 11.4: M Q = dF ¯ Proof. Let P be the point on the line of action of F which is closest to Q. Then ¯ the vector QP is perpendicular to F and the magnitude of this vector is d. Using the ﬁrst deﬁnition of the cross product, we have ¯ ¯ M Q = M Q  = QP × F  ¯ = QP F  sin(90◦ ) = dF Note that the direction of the moment is determined by the righthand rule. The next result is an immediate consequence of the previous fact. ¯ Fact 4 Suppose F is a nonzero force. Then its moment about a point Q is zero if and only ¯ if Q is on the line of action of F . 11.1. THE MOMENT OF A FORCE Example 56 205 206 CHAPTER 11. STATICS OF BODIES 11.2 Bodies A physical body is any material object. It can be solid, liquid, gas or a combination of these. A piece of a body is just another body. A collection of bodies can also be regarded as a single body. Mathematically, a body is something with two properties: i) At each instant of time, it occupies a region of space. ii) It has a mass distribution; to each piece of the body, we can associate a real number, the mass of that piece. • A particle is the simplest type of body; at each instant of time, it occupies a single point. • An arbitrary body can be regarded as a collection of particles. • A rigid body can be deﬁned as a body with the property that the distance between any two particles of the body always remains the same. Note that the results in this section are not restricted to rigid bodies; they apply to a body of any nature. 11.2.1 Internal forces and external forces All the forces acting on a particle act at a single point, namely, the position of the particle. Figure 11.5: Forces on a particle The forces acting on a general body do necessarily not act at a single point. They can act at any point in the region of space the body occupies. Figure 11.6: Forces on a general body Consider a general body B. We can classify the forces associated with B into two types: 11.2. BODIES 207 • Forces internal to the body B. An internal force is a force exerted by one piece of B on another piece of B. • Forces external to the the body B. These are the forces exerted on B by other bodies. 11.2.2 Internal forces By considering a body as a collection of particles and applying Newton’s third law, one can obtain the following result. Theorem 2 (Internal forces) The internal forces of any body satisfy
int ¯ F =¯ 0 int ¯ MQ = ¯ 0 for every point Q where
int int ¯ F is the sum of all the internal forces in the body, and ¯ M Q is the sum of the moments about Q of all the internal forces in the body Proof. Consider a general body B which we will regard as a collection of N particles P1 , P2 , . . . , PN . ¯ Let F jk be the resultant internal force exerted on particle Pj by particle Pk . 208 CHAPTER 11. STATICS OF BODIES Thus the internal force system of B consists of the forces ¯ F jk , j = k, j, k = 1, 2, . . . , N First note that the resultant of all the internal forces is given by
int ¯ F N N =
j =1 ¯ F jk k=1 k=j ¯ + F 12 + ¯ 0 ¯ + F 13 ¯ + F 23 + ¯ 0 . . . ¯ + . . . + F 1N ¯ + . . . + F 2N ¯ + . . . + F 3N . . . ¯ 0 = ¯ 0 ¯ + F 21 ¯ + F 31 . . . ¯ + F 32 . . . ¯ ¯ ¯ + F N1 + F N2 + F N3 + . . . + ¯ ¯ ¯ ¯ ¯ ¯ = (F 12 + F 21 ) + (F 13 + F 31 ) + . . . + (F 1N + F N 1 ) ¯ ¯ ¯ ¯ + (F 23 + F 32 ) + . . . + (F 2N + F N 2 ) ¯ ¯ + (F 3N + F N 3 ) . . . +
N −1 N ¯ 0 = ¯ ¯ (F jk + F kj ) j =1 k =j +1 ¯ ¯ By the ﬁrst part of Newton’s Third Law, F kj = −F jk ; hence ¯ ¯ F jk + F kj = ¯ 0 11.2. BODIES Thus,
int 209 ¯0 F =¯ ¯ ¯ Consider now any corresponding pair of internal forces F jk , F kj . By the second part of Newtons Third Law the line of action of these two forces must be the line joining Pj and Pk ; ¯ ¯ hence the vector Pk Pj is parallel to F jk and F kj . So, ¯ Evaluating the sum of the moments of these two forces about any point Q and using F kj = ¯ jk we get −F ¯ ¯ ¯ r j × F jk + r k × F kj = (¯j − r k ) × F jk ¯ ¯ r ¯ ¯ = Pk Pj × F jk =¯ 0 ¯ Hence, the moments due to the internal forces cancel out in pairs. If int M Q is the sum of the moments of all the internal forces about Q, then using the same computations we used ¯ for int F we must have int ¯ Q M =¯ 0 ¯ Pk Pj × F jk = ¯ 0 • Note that the above result holds regardless of the motion of the body; the body does not have to be in static equilibrium (see next section). 210 CHAPTER 11. STATICS OF BODIES 11.3 Static equilibrium Recall that a particle is in static equilibrium if it is at rest in some inertial reference frame. If we regard an arbitrary body as a collection of particles, we have the following deﬁnition. DEFN. A body is in static equilibrium if every particle of the body is at rest in the same inertial reference frame. Our next result is the most important result in the statics of bodies. It can be obtained by applying Newton’s second law to each particle of a body, summing over all the particles in the body, and using the fact that the internal forces and moments sum to zero. Theorem 3 If a body is in static equilibrium, then for any point Q ¯ F =¯ 0 ¯ MQ = ¯ 0 where ¯ F is the sum or resultant of all the external forces acting on the body ¯ M Q is the sum of the moments or the moment resultant about Q of all the external forces acting on the body 11.3. STATIC EQUILIBRIUM Example 57 (A lever) Show that a R= F b for the lever illustrated in Figure 11.7. 211 Figure 11.7: A lever 212 Example 58 (Another lever) Show that a R= F b for the lever illustrated in Figure 11.8. CHAPTER 11. STATICS OF BODIES Figure 11.8: Another lever 11.3. STATIC EQUILIBRIUM Example 59 (Yet another lever) Show that a R= F b for the lever illustrated in Figure 59. 213 Figure 11.9: Yet another lever 214 CHAPTER 11. STATICS OF BODIES 11.3.1 Free body diagrams
Body + External forces Examples of FBDs Example 60 Figure 11.10: Example 60 11.3. STATIC EQUILIBRIUM Example 61 215 Figure 11.11: Example 61 216 Example 62 CHAPTER 11. STATICS OF BODIES Figure 11.12: Example 62 11.4. EXAMPLES IN STATIC EQUILIBRIUM 217 11.4
11.4.1 Examples in static equilibrium
Scalar equations of equilibrium
¯ ΣF = ¯ 0 Q ¯ ΣM =¯ 0 In general, the two vector equations yield six scalar equations. However, in some cases, the six equations are not independent or some are trivial, for example, 0 = 0. The following are force systems which do not give rise to six independent scalar equations of equilibrium. Force system collinear coplanar parallel parallel to a common plane Max. no. of independent scalar equations 1 3 3 5 11.4.2 Planar examples A planar problem is one in which all the bodies and forces of interest lie in a single plane. Suppose e1 , e2 , e3 is an orthogonal triad of unit vectors with e1 , e2 lying in the plane of ˆˆˆ ˆˆ interest and with e3 perpendicular to the plane. Then all forces and position vectors can be ˆ expressed in terms of e1 and e2 ; hence all moments are parallel to e3 . So, the conditions of ˆ ˆ ˆ static equilibrium give rise to at most three nontrivial scalar equations: e1 : ˆ e2 : ˆ e3 : ˆ F1 = 0 F2 = 0 MQ = 0 218 CHAPTER 11. STATICS OF BODIES Example 63 Given: The block of weight W = 10 lb sits on the massless beam which is supported at A and B by frictionless roller supports. Figure 11.13: Example 63 Find: the reactions on the beam at A and B . 11.4. EXAMPLES IN STATIC EQUILIBRIUM 219 Example 64 Given: The block of weight W = 100 N sits on a massless bar which is supported at A and B by cables. Figure 11.14: Example 64 Find: the tension in each cable at the bar. 220 CHAPTER 11. STATICS OF BODIES Example 65 Given: The Lunacycle of weight W is parked on a hill which is inclined at an angle of θ to the horizontal. Figure 11.15: Example 65 Find: expressions for the magnitude of normal force on each wheel. 11.4. EXAMPLES IN STATIC EQUILIBRIUM Example 66 Given: l = 2m, µ = 1/2 and θ = 60◦ . 221 Figure 11.16: Example 66 Find: max value of distance d. 222 CHAPTER 11. STATICS OF BODIES Example 67 Given: A uniform stepped cylinder with radii R = 1m, r = 0.5m and weight W = 100N is in static equilibrium on a rough horizontal plane and is attached to two cables at points A and B . Also, µ = 1/4. Figure 11.17: Example 63 Find: the maximum value of the cable tension T . 11.4. EXAMPLES IN STATIC EQUILIBRIUM 223 11.4.3 General examples Example 68 Given: The block of weight W = 20 lb is resting on the massless plate which is supported at A, B and C by cables. Figure 11.18: Example 63 Find: the tensions in the cables at points A, B and C . 224 CHAPTER 11. STATICS OF BODIES 11.5 Force systems Here we develop some general properties of force systems. A force system is just a bunch of ¯ ¯ ¯ forces, F 1 , F 1 , . . . , F N as illustrated in Figure 11.19 Figure 11.19: A force system DEFN. The resultant of a force system is the sum of all its forces and is given by: ¯ F :=
N j =1 ¯ Fj = ¯ ¯ ¯ F1 + F2 + ... + FN DEFN. The moment resultant of a force system about a point Q is the sum of the moments of all its forces about Q and is given by: ¯ M Q :=
j N j =1 ¯ rj × F j ¯ = ¯ ¯ ¯ r1 × F 1 + r2 × F 2 + . . . + rN × F N ¯ ¯ ¯ ¯ where r is a vector from Q to a point on the line of action of F j . ¯ Figure 11.20: Moment resultant Suppose one knows the resultant and moment resultant about some point Q′ of a force system. The following result tells us how to compute the moment resultant about another point Q without having to compute all the moments of all the forces about Q. Fact 5 For any two points Q and Q , ¯ MQ = where r = r QQ (the vector from Q to Q ). ¯¯
′ ′ ′ ¯ ¯ M Q + r × ΣF ¯ ′ 11.5. FORCE SYSTEMS Proof. By deﬁnition we have ¯ MQ =
N i=1 225 ¯ rj × F j ¯ and ¯′ MQ = N i=1 ¯ ρj × F j ¯ ¯ where r j is the vector from Q to the point of application of F j and ρj is the vector from Q′ ¯ ¯ ¯j. to the point of application of F Figure 11.21: Q and Q′ However r j = ρj + r where r = r QQ ; hence ¯ ¯ ¯ ¯¯ ¯ MQ = =
i=1 N i=1 N
′ ¯ (¯j + r) × F j ρ ¯ ¯ ρ × Fj + r × ¯ ¯
j N i=1 ¯ Fj = ¯ ¯ M Q + r × ΣF ¯ ′ 226 CHAPTER 11. STATICS OF BODIES Example 69 Here we illustrate the moment formula just developed. Consider the force system in Figure 11.22. Figure 11.22: Force system for Example 69 11.5. FORCE SYSTEMS 227 11.5.1 Couples and torques DEFN. A couple is a pair of forces which have equal magnitude but opposite direction. Figure 11.23: A couple ¯¯ So, if (F 1 , F 2 ) is a couple, then, ¯ ¯ F 2 = −F 1 . • A couple has zero resultant. • The moment resultant of a couple about every point is the same. ¯ DEFN. The torque T of a couple is its moment resultant about any point. Quite often, we are only interested in the torque of a couple and we are not necessarily interested in the two forces that make up the couple. So, we often represent a couple by its ¯ torque T . In graphic representations of torque vectors, we use a double arrowhead instead of the usual single arrowhead. Figure 11.24: Representations of torque 228 CHAPTER 11. STATICS OF BODIES Example 70 Each of the following couples are equivalent. 11.6. EQUIVALENT FORCE SYSTEMS 229 11.6 Equivalent force systems DEFN. Two force systems are equivalent if they have the same resultant and the same moment resultant about some point. So, the system of internal forces in any body is equivalent to a zero force. Also, if a body is in static equilibrium, its system of external forces is equivalent to a zero force. Example 71 Consider the following force system.
3N 2N 2m It is equivalent to any of the following force systems.
5N 6Nm 5N 4Nm 5N 6/5 m 4/5 m 230 CHAPTER 11. STATICS OF BODIES • If two force systems are equivalent, then they have the same moment about every point. ′ ¯ ¯ ¯ This follows from the relationship, M Q = M Q + r × ΣF . ¯ You will see later that if two force systems are equivalent, then they have exactly the same eﬀect when applied to a given rigid body. 11.7 Simple equivalent force systems Sometimes it is very useful to replace a complicated force system by a simpler equivalent force system. 11.7.1 A force and a couple Every force system (regardless of complexity) is equivalent to a force and a couple. To see this, choose any point Q and let ¯ ¯ F = ΣF and ¯ ¯ T = ΣM Q ¯ It can readily seen that the new force system consisting of a force F placed at Q and a couple ¯ of torque T is equivalent to the original force system. Note that Q can be any point. 11.7.2 Force systems which are equivalent to a couple
¯0 ΣF = ¯ Suppose that a force has zero resultant, that is, ¯ ¯ ¯ Then this force system is equivalent to a couple of torque T = ΣM Q . Since ΣF = ¯ this 0, torque is independent of the point Q. 11.7. SIMPLE EQUIVALENT FORCE SYSTEMS Example 72 231 Figure 11.25: A force system which is equivalent to a couple 232 CHAPTER 11. STATICS OF BODIES 11.7.3 Force systems which are equivalent to single force
¯ ΣM Q = ¯ 0 Suppose there is a point Q about which the force system has zero moment resultant, that is, ¯ ¯ Then this force system is equivalent to a single force F = ΣF whose point of application is Q. Example 73 Figure 11.26: A force system which is equivalent to a single force The single force 11.7. SIMPLE EQUIVALENT FORCE SYSTEMS 233 Note that the point of application of a single equivalent force is not unique. One can “slide” a single equivalent force along its line of application and obtain another equivalent force with a diﬀerent point of application. We now demonstrate the following result. A force system is equivalent to a single force if and only if it has nonzero resultant and the moment resultant about some point is perpendicular to the resultant. ¯ Suppose one has a force system which is equivalent to single force F placed at a point O ¯ ¯ ¯ ¯ Q. Let T = M where O is any point. Then it is necessary that T is perpendicular to F . ¯ placed at Q is equivalent to the original force system, This can be seen as follows. Since F ¯ the moment of F about O must equal the moment resultant of the original force system about O , that is, ¯ r×F =T ¯¯ (11.1) ¯ where r is the vector from O to Q. The above expression and the properties tell that T is ¯ ¯. perpendicular to F ¯ ¯ ¯ Now suppose we know that F and T are perpendicular with F = 0. We claim that one can always ﬁnd a vector r such that (11.1) holds. One such vector is given by ¯ ¯¯ F ×T r= ¯ . (11.2) F2 ¯ where F is the magnitude of F . This can be seen as follows. Recall that for any three vectors ¯¯ ¯ U , V and W we have ¯ ¯ ¯ ¯ ¯¯ ¯¯¯ U × (V × W ) = (U · W )V − (U · V )W . Hence 1¯ 1 ¯ ¯¯ ¯ ¯¯ ¯ ¯¯ × F = − 2 F × (F × T ) = − 2 (F · T )F − (F · F )T . F F ¯ ¯ ¯¯ ¯¯ Since, by assumption, F is perpendicular to T we must have F · T = 0. Also, F · F = F 2 . ¯ Hence, we obtain that r × F = T . ¯¯ The following force systems are examples of force systems which are equivalent to a single force. r×F = ¯¯ Concurrent force system Here the line of each force passes through a common point. ¯¯ F ×T F2 234 CHAPTER 11. STATICS OF BODIES Planar force systems with nonzero resultant Example 74 Figure 11.27: A planar force system which is equivalent to a single force 11.7. SIMPLE EQUIVALENT FORCE SYSTEMS Parallel force systems with nonzero resultant* Example 75 235 Figure 11.28: A parallel force system which is equivalent to a single force 236 CHAPTER 11. STATICS OF BODIES Parallel force systems. Choose a reference frame so that all the forces are parallel to e3 . ˆ ¯ j can be expressed as Then every force F ¯ F j = F j e3 ˆ and ¯ ΣF = F e3 ˆ where F :=
i=1 N Fj Let O be the origin of the reference frame and let r j be the vector from O to the point of application of F j ; then r j can be expressed as r j = xj e1 + y j e2 + z j e3 ¯ ˆ ˆ ˆ Hence, and ¯ ΣM O = T1 e1 + T2 e2 ˆ ˆ Letting r ∗ = x∗ e1 + y ∗ e2 + z ∗ e3 ¯ ˆ ˆ ˆ ¯ be the vector from O to the point of application of F we get ¯ r × F = (y ∗ ¯
∗ N i=1 N N N ¯ r j × F j = (y j F j ) e1 − (xj F j ) e2 ¯ ˆ ˆ where T1 :=
i=1 yF j j and T2 := − xj F j .
i=1 F ) e1 − (x ˆ j ∗ i=1 F j ) e2 ˆ ¯ ¯ Since r ∗ × F = ΣM O , we obtain ¯
N N y∗
i=1 N Fj =
i=1 N yj F j xj F j
i=1 x∗
i=1 Fj = Hence x∗ = y∗ =
N i=1 xj F j F
N i=1 yj F j F 11.8. DISTRIBUTED FORCE SYSTEMS 237 11.8 Distributed force systems So far, we have considered forces to act at a single point. Here we look at forces which do not act a single point, but act over a region of space. We divide these forces into body forces and surface forces. 11.8.1 Body forces A body force acts over a threedimensional region of space. The main example of a body force is the gravitational attraction of one body on another. Gravitational forces The gravitational force exerted by one body (the oﬀending body) on another body (the suﬀering body) is a force system which is distributed throughout the entire suﬀering body. Every particle of the suﬀering body is subject to the gravitational attraction of the oﬀending body. When the oﬀending body is YFHB and the suﬀering body is near the surface of YFHB and its dimensions are insigniﬁcant compared to YFHB then, all the gravitational force system is in the same direction, namely, in the direction of the local vertical g . So ˆ ¯ this gravitational force system is a parallel force system. It is equivalent to a single force W placed at the mass center of the suﬀering body and given by ¯ W = Wg ˆ where W is called the weight of the suﬀering body and is given by W = mg where m is the mass of the suﬀering body and g is the gravitational acceleration constant of YFHB. For bodies of uniform mass density, the mass center is at the geometric center. Note that the mass center does not have to be a point on the body; see donut. Figure 11.29: Weight When the suﬀering body is not near the surface of YFHB (think of a spacecraft in orbit around the earth) then, the gravitational forces on the suﬀering body may have a nonzero moment resultant about the mass center of the suﬀering body. 238 CHAPTER 11. STATICS OF BODIES Figure 11.30: Some mass centers 11.8.2 Surface forces A surface force acts over a twodimensional region of space. One example of a surface force is hydrostatic force, that is, the forces exerted on a body when it is immersed in water. Another example is aerodynamic forces, that is the forces on a body moving relative to the air. Hydrostatic forces Archimedes principle, center of buoyancy 11.8. DISTRIBUTED FORCE SYSTEMS Aerodynamic forces 239 If you pick any reference point on an aircraft, the aerodynamic force system is equivalent to ¯ ¯¯ a single force F placed at the point and a moment T . F is independent of the point whereas ¯ is the moment resultant of the force about that point; hence it depends on the point. If T ¯ T is zero, that point is called a center of pressure. In general the center of pressure changes as the aircraft changes its orientation relative to the wind; it is not a ﬁxed point on the aircraft. ¯ There is usually another point which is ﬁxed and has the property that T is constant for that point; this is the aerodynamic center. lift, drag, pitching moment, Figure 11.31: Aerodynamic forces 240 Connection forces CHAPTER 11. STATICS OF BODIES The force exerted by one body on another at a connection between the two bodies is a distributed force system. We usually represent this force system by an equivalent force ¯ ¯ system consisting of a single force R and a couple of torque T . If the connection is smooth and permits translational motion in a speciﬁc direction then, ¯ the force R has no component in that direction. Conversely, if the connection prevents ¯ translational motion in a speciﬁc direction then, F can have a component in that direction. If the connection is smooth and permits rotational motion about a speciﬁc axis then, the ¯ torque T has no component about that axis. Conversely, if the connection prevents rotational ¯ motion about a speciﬁc axis then, T can have a component about that axis. Example 76 (A connection) Figure 11.32: A connection 11.8. DISTRIBUTED FORCE SYSTEMS Example 77 (Another connection) 241 Figure 11.33: Another connection 242 CHAPTER 11. STATICS OF BODIES 11.8.3 Connections in 2D We are assuming no friction at these connections. Figure 11.34: Frictionless connections in 2D 11.8. DISTRIBUTED FORCE SYSTEMS 243 11.8.4 Connections in 3D 244 CHAPTER 11. STATICS OF BODIES 11.9 More examples in static equilibrium Example 78 Given: The load of weight W = 100N is supported by the massless structure. Figure 11.35: Example 78 Find: Reaction forces on the structure at A and B . 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM 245 Example 79 Given: The load of weight W = 150N is supported by the symmetric massless structure. Figure 11.36: Example 79 Find: The tension in the rope. 246 CHAPTER 11. STATICS OF BODIES Example 80 Given: The load of weight W = 10 lb is supported by the massless structure. Figure 11.37: Example 80 Find: Reaction at A on the structure. 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM Example 81 The following system is in static equilibrium with W1 = 50N and W2 = 60N . 247 g ˆ • 45o • A W1 1m 1m • B W2 Find the reactions on the bar at A and B . Solution. Consider ﬁrst the equilibrium of the block of weight W2 .
T
ˆ e 2 W2 ¯ Considering ΣF = 0 we obtain e2 : −W2 + T = 0 ˆ Hence, T = W2 = 60 N Consider now the equilibrium of bar plus block of weight W1 . (11.3) 248 CHAPTER 11. STATICS OF BODIES ˆ e A R2 2 ˆ e1 A R 1 A 1m T 45o RB ˆ e3 1m W1 B ¯ Considering ΣF = 0 we obtain
A e1 : R1 + T cos 45o = 0 ˆ A e2 : R2 + T sin 45o − W1 + RB = 0 ˆ (11.4) (11.5) ¯ Considering ΣM A = 0 we obtain e3 : −W1 + sin 45o T + 2RB = 0 ˆ We can now use these last three equations to solve for RB = (W1 − T sin 45o )/2 = 3.787N A o R1 = −T cos 45 = −42.43N A R2 = −T sin 45o + W1 − RB = 3.787N ¯ RA = −42.43 e1 + 3.787 e2 ˆ ˆ ¯ B = 3.787 e2 N R ˆ N (11.6) 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM Example 82 249 250 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM 251 252 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM Example 83 253 254 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM 255 256 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM Example 84 257 258 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM 259 260 CHAPTER 11. STATICS OF BODIES 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM 261 11.9.1 Two force bodies in static equilibrium Consider a body which is subject to only two forces. Figure 11.38: A two force body ¯0 Suppose that this body is in static equilibrium. Then, consideration of ΣF = ¯ tells us that the sum of these two forces must be zero. Thus, one force is the negative of the other; hence, the two forces have the same magnitude, but, opposite direction. ¯ Consideration of ΣM Q = ¯ where Q is the point of application of one of the forces tells 0 us that this point must also be on the line of application of the other force; since the two forces are parallel, they have the same line of application; since this line must contain the points of application of the two forces, this line is uniquely determined by these two points. Figure 11.39: Two force bodies in static equilibrium So, we have the following result. If a twoforce body is in static equilibrium then, the two forces must be equal in magnitude, opposite in direction, and the line of action of each force is the line passing through the point of application of the two forces 262 Example 85 Given: W = 50 lb. CHAPTER 11. STATICS OF BODIES Figure 11.40: Example 85 Find: The magnitude of the reaction at A on the structure. 11.9. MORE EXAMPLES IN STATIC EQUILIBRIUM Example 86 Given: F = 250 N. 263 Figure 11.41: Example 86 Find: Reactions on the structure at B and C . 264 CHAPTER 11. STATICS OF BODIES 11.10 Statically indeterminate problems A problem is statically determinate if one can solve for all the unknown forces and torques ¯0 using only the conditions of static equilibrium: ΣF = ¯ and ΣM Q = ¯ 0. A problem is statically indeterminate if one cannot solve for all the unknown forces and ¯0 torques using only the conditions of static equilibrium: ΣF = ¯ and ΣM Q = ¯ To solve for 0. all the forces an torques in such a problem, one usually has to use some information on the material properties of the bodies under study. So there are three possibilities for a statics problem: Statically determinate: Same number of unknowns as independent equations. Statically indeterminate: More unknowns than independent equations. Impossible: Less unknowns than independent equations. These possibilities are illustrated in the following examples. 11.10. STATICALLY INDETERMINATE PROBLEMS Example 87 (Statically indeterminate) 265 Figure 11.42: Example 87 266 Example 88 (Statically determinate) CHAPTER 11. STATICS OF BODIES Figure 11.43: Example 89 11.10. STATICALLY INDETERMINATE PROBLEMS Example 89 (Statically determinate) 267 Figure 11.44: Example 89 268 Example 90 (Impossible) CHAPTER 11. STATICS OF BODIES Figure 11.45: Example 90 11.11. INTERNAL FORCES 269 11.11 11.12 Internal forces Exercises Exercise 32 Determine the reaction on the massless structure at pin joint A. Exercise 33 Obtain a simple force system consisting of a single force or a single couple which is equivalent to the force system shown. 270 CHAPTER 11. STATICS OF BODIES Chapter 12 Momentum 12.1 Linear momentum m• P ¯ v Suppose we are observing the motion of a particle of mass m moving with velocity v ¯ relative to some inertial reference frame. • The linear momentum of the particle in the inertial frame is deﬁned by L = mv ¯ dim[L] = MLT−1 = FT units: kg ms−1 or lb sec ¯ Suppose ΣF is the sum of all the forces acting on the particle, i.e., ΣF = F ,F ,...,F
1 2 N j N j =1 F , where are all the forces acting on the particle.
– F1 P • – FN • • –2 F • 271 272 CHAPTER 12. MOMENTUM ¯ Then, using ΣF = ma we can obtain the following result for any inertial reference frame: ¯ • The sum of the forces acting on a particle equals the time rate of change of the linear momentum of the particle, i.e., ˙ F =L Proof. Since m is constant, ˙ ¯ L= d (mv ) ¯ dt d = m (¯) v dt = ma ¯ ¯ where a is the inertial acceleration of the particle. It now follows from ΣF = ma that ¯ ¯ ˙ ΣF = L 12.2
12.2.1 Impulse of a force
Integral of a vector valued function
– Z (t)
© ˆ e3 ˆ e2 ˆ e1 ¯ Suppose Z is a function of a scalar variable t and ¯ Z = Z1 e1 + Z2 e2 + Z3 e3 ˆ ˆ ˆ ¯ Consider any interval [t0 , t1 ]. Relative to reference frame e, the integral of Z over [t0 , t1 ] is deﬁned by
t1 t0 ¯ Z dt := t1 t0 Z1 dt e1 + ˆ t1 t0 Z2 dt e2 + ˆ t1 t0 Z3 dt e3 ˆ 12.2. IMPULSE OF A FORCE Example 91 Suppose Then
1 0 273 ¯ Z (t) = e1 + te2 + sin te3 ˆ ˆ ˆ
1 0 1 0 1 0 ¯ Z dt = 1 dt e1 + ˆ t dt e2 + ˆ sin t dt e3 ˆ 1 = e1 + e2 + (1 − cos(1))ˆ3 ˆ ˆ e 2 • Note that t1 t0 ˙ ¯ ¯ ¯ Z dt = Z (t1 ) − Z (t0 ) 12.2.2 Impulse ¯ • The impulse of a force F over a time interval [t0 , t1 ] is deﬁned by ¯1 Itt0 :=
t1 t0 ¯ F dt Impulsive forces. Large magnitude over a short time interval. They usually occur during impacts and collisions. Suppose we are observing the motion of a particle relative to some inertial reference frame over some time interval [t0 , t1 ]. Let ∆L be the change in the linear momentum of the particle over the time interval, i.e, ∆L = L(t1 ) − L(t0 ) ¯ ¯ Let ΣI be the total impulse acting on the particle over the time interval, i.e, ΣI is the sum of the impulses of all the forces acting on the particle:
N ΣI =
j =1 t1 t0 F dt j Then we have the following result for any inertial reference frame: • Over any time interval, the total impulse acting on a particle equals the change in the linear momentum of a particle, i.e., ¯ ¯ ΣI = ∆L Proof. Recall that ˙ ¯ ΣF = L 274 Integrate over the time interval [t0 , t1 ] to yield:
t1 t0 CHAPTER 12. MOMENTUM ΣF dt = t1 t0 ˙ ¯ L dt We have
t1 t0 ΣF dt = = t1 t0 N j =1 N j =1 t1 t0 j F dt F dt j ¯ = ΣI
t1 t0 ˙ ¯ L dt = L(t1 ) − L(t0 ) = ∆L This yields the desired result. We have the following immediate consequence of the above result. • If the total impulse acting on a particle over any time interval [t0 , t1 ] is zero, then the linear momentum of the particle is conserved over that interval, i.e., ¯ ¯ L(t1 ) = L(t0 ) From this it follows that the velocity is also conserved, i.e., v (t1 ) = v(t0 ) ¯ ¯ This is also true for any compoment, i.e, if the total impulse has zero component in some direction, then the corresponding components of the linear momentum and the velocity are conserved; see next example. 12.2. IMPULSE OF A FORCE Example 92 275 Ball impacts smooth wall. Find the exit speed v . Solution. Suppose the impact between wall and ball occurs over the interval [t0 , t1 ]. Looking at a FBD of the ball during impact, the only force with a vertical component is weight. Its impulse is
t1 t0 mg g dt = mg (t1 − t0 )ˆ ˆ g Ideally, we can choose t1 − t0 arbitrarily small; so the impulse due ro the weight force is negligible. Hence, the vertical component of the total impulse is zero. This implies that the vertical component of the ball’s velocity is preserved during impact, i.e., v cos(30◦ ) = 10 cos(60◦ ) Hence, 10 v = √ ms−1 3 276 CHAPTER 12. MOMENTUM 12.3 Angular momentum Suppose we are observing the motion of a particle of mass m relative to some inertial reference ˙ frame i and Q is any point. Let r be the position of particle relative to Q and let r be its ¯ ¯ time derivative in i. • The angular momentum of the particle about Q is deﬁned by ¯ ˙ H Q = r × mr ¯ ¯ dim[H ] = ML2 T−1 = FLT units: kg m2 s−1 or lb ft sec In the above deﬁnition, Q does not have to be a ﬁxed point. Suppose Q = O where O is a point ﬁxed in the inertial reference frame; then ˙ r=v ¯ where v is the velocity of the particle in the frame; hence ¯ ¯ H O = r × mv ¯ ¯ = r×L ¯¯ ¯ where L is the momentum of the particle in the inertial frame. Example 93 Angular momentum and polar coordinates
Q We have r = r er ¯ ˆ ˙ r = rω eθ ¯ ˆ 12.3. ANGULAR MOMENTUM Hence 277 ¯ H O = mr 2 ω e3 ˆ ¯ Suppose ΣM O is the sum of the moments about point O of all the forces acting on the particle, i.e., ¯ ΣM O :=
N j =1 r×F ¯
N j = r× ¯ F
j =1 j ¯ = r × ΣF ¯ where r is the position of the particle relative to O . We have now the following result for ¯ any inertial reference frame: • The time rate of change of the angular momentum of a particle about a ﬁxed point O is equal to the sum of the moments about O of all the forces acting on the particle, i.e.,
O ˙O M =H Proof. Since m is constant, ¯O ˙ O = dH H dt d = (¯ × mr ) r ¯ dt ˙ ˙ = r × mr + r × m¨ ¯ ¯¯ r ¯ = r × m¨ ¯ r ¯ Since O is ﬁxed in frame i, the vector ¨ equals a, the inertial acceleration of the particle in r ¯ ¯ ¯ i; since i is inertial we have ΣF = ma, hence ¯ ˙ O = r × ΣF ¯ H ¯ ¯ = ΣM O and we obtain the desired result. Example 94 Simple pendulum 278 CHAPTER 12. MOMENTUM 12.4 Central force motion Recall that a particle undergoes central force motion if is subject to a single force F whose line of action always passes through some inertially ﬁxed point O . Chapter 13 Work and Energy 13.1 Kinetic energy Consider the motion of a particle of mass m relative to some inertial reference frame. The kinetic energy of the particle is denoted by T and is given by T= 1 mv 2 2 where v is the speed of the particle and m is the mass of the particle. Note that kinetic energy is a scalar quantity. If v is the velocity of the particle, we may also express T as ¯ T= 1 mv · v . ¯¯ 2 Units SI: Joule (J); J = N m = kg m2 /s2 Other: lb ft, btu Example 95 (Kinetic energy of pendulum) T= 1 2 ˙2 ml θ 2 279 280 CHAPTER 13. WORK AND ENERGY Consider now a rigid body of mass m which is translating with speed v relative to some inertial reference frame. The kinetic energy of the body is simply the sum of the kinetic energies of the particles composing the body. Hence the kinetic energy of a translating rigid body is also given by T = 1 mv 2 where m is the mass of the body. 2 13.2 Power ¯ Consider a force F acting on a particle which is moving with velocity v relative to some ¯ inertial reference frame. The power of the force is denoted by P and is given by ¯¯ P =F ·v Example 96 Normal forces and friction Consider now a rigid body which is translating with velocity v relative to some inertial ¯ ¯ reference frame. Suppose it is subject to a distributed force system whose resultant is F . Then the power of the distributed force system is deﬁned to be the resultant of the powers ¯¯ of the forces which make up the force system. This power is also given by P = F · v . Example 97 Lift and drag 13.3. A BASIC RESULT 281 13.3 A basic result Consider a particle in motion relative to an inertial reference frame. We deﬁne the resultant power of all the forces acting on the particle to be the sum of the powers of all the forces acting on the particle. We have now the following result. The resultant power of all the forces acting on a particle equals the time rate of change of the kinetic energy of the particle. Mathematically, we can represent the above result as ˙ P =T where P is the resultant power of all the forces acting on the particle and T is the kinetic energy of the particle. To prove the above result, recall that in an inertial reference frame we have
N i=1 ¯ F i = ma ¯ ¯ ¯ where a is the inertial acceleration of the particle and F1 , · · · , FN are the forces acting on ¯ i i ¯ ¯¯ the particle. Since the power of force F is F · v where v is the velocity of the particle in ¯ the inertial frame, we have
N P= Also i=1 ¯¯ (F i · v ) = N i=1 ¯ F i · v = ma · v. ¯ ¯¯ 1 1 d1 ˙ ˙ ˙ ˙ mv · v = mv · v + mv · v = mv · v = ma · v ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ T= dt 2 2 2 ˙ It now follows that P = T . 282 CHAPTER 13. WORK AND ENERGY Example 98 (The simple pendulum) Show that ¨g θ + sin θ = 0 l 13.4. CONSERVATIVE FORCES AND POTENTIAL ENERGY 283 13.4 Conservative forces and potential energy Consider a force acting on a particle moving in some inertial reference frame. We say that this force is conservative, if there is a scalar function φ with the following properties: (a) The function φ is only a function of the position of the particle relative to some inertially ﬁxed point. (b) The power of the force is the negative of the time rate of change of φ, that is ˙ P = −φ The function φ is called the potential energy of the force. So, we have the following ststement. The power of a conservative force is the negative of the time rate of change of its potential energy. where P is the power of the force. 13.4.1 Weight
φ = mgh 13.4.2 Linear springs
1 φ = k (l − l0 )2 2 13.4.3 Inverse square gravitational force
φ=− GMm r 284 CHAPTER 13. WORK AND ENERGY 13.5 Total mechanical energy Consider a particle in motion relative to some inertial reference frame. We can divide the forces acting on the particle into conservative forces and nonconservative forces. Associated with each conservative force is a potential energy. Let φ be the sum of the potential energies of all the conservative forces. We call this the total potential energy. The total mechanical energy is the sum of the kinetic energy and the total potential energy, that is E =T +φ where E denotes the total mechanical energy. Let c P be the resultant power of all the conservative forces. Since the power of a conservative force equals the negative of the time rate of change of its potential energy, it follows that the resultant power of the conservative forces equals the negative of the time rate of change of the total potential energy, that is,
c ˙ P = −φ . Let nc P demote the resultant power of all the nonconservative forces. Then the resultant power P of all the forces satisﬁes
c nc P= Recalling that ˙ P = T , we have P+ ˙ P = −φ + nc P. ˙ −φ + Hence or
nc nc ˙ P =T. ˙ ˙ P =T +φ ˙ P=E . nc The resultant power of all the nonconservative forces equals the time rate of change of the total mechanical energy. So, if the resultant power of the nonconservative forces is zero, then the time rate of change of the mechanical energy is zero. In this case the energy E is constant and we say that the total mechanical energy is conserved. Example 99 (The simple pendulum) Show that ¨g θ + sin θ = 0 l 13.6. WORK 285 13.6 Work Basically, work is the time integral of power. More speciﬁcally, consider any time interval [t1 , t2 ]. Then the work done by a force over that interval is denoted by W D is given by WD = where P is the power of the force. We have now the following results: W D = ∆T where ∆T = T (t2 ) − T (t1 ). Also, where ∆E = E (t2 ) − E (t1 ). Example 100
t2 t1 P dt nc W D = ∆E 286 CHAPTER 13. WORK AND ENERGY Chapter 14 Systems of Particles 287 ...
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 Spring '09

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