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No Announcements class Friday HW# 14 and 15 due Wed, May 12 Newtons Law of Motion (Equation of Motion) Todays Objectives: Students will be able to: a) Write the equation of motion for an accelerating body. b) Draw the free-body and kinetic diagrams for an accelerating body. Sections 13.1-3 In-Class Activities: Reading quiz Applications Newtons laws of motion Newtons law of gravitational attraction Equation of motion for a particle or system of particles Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 13 1 Reading Quiz 1. Newtons second law can be written in mathematical form as F = ma. Within the summation of forces F, ________ are(is) not included. A) external forces C) internal forces B) weight D) All of the above 2. The equation of motion for a system of n-particles can be written as Fi = miai = maG, where aG indicates _______. A) summation of each particles acceleration B) acceleration of the center of mass of the system C) acceleration of the largest particle D) None of the above Applications The motion of an object depends on the forces acting on it. A parachutist relies on the atmospheric drag resistance force to limit his velocity. Knowing the drag force, how can we determine the acceleration or velocity of the parachutist at any point in time? Engr222 Spring 2004 Chapter 13 2 Applications - continued A freight elevator is lifted using a motor attached to a cable and pulley system as shown. How can we determine the tension force in the cable required to lift the elevator at a given acceleration? Is the tension force in the cable greater than the weight of the elevator and its load? Newtons Laws of Motion The motion of a particle is governed by Newtons three laws of motion. First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero. Second Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a magnitude proportional to the resultant force. Third Law: Mutual forces of action and reaction between two particles are equal, opposite, and collinear. Engr222 Spring 2004 Chapter 13 3 Newtons Laws of Motion - continued The first and third laws were used in developing the concepts of statics. Newtons second law forms the basis of the study of dynamics. Mathematically, Newtons second law of motion can be written F = ma where F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive scalar m is called the mass of the particle. Newtons second law cannot be used when the particles speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom). Newtons Law of Gravitational Attraction Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton postulated the law governing this gravitational force as F = G(m1m2/r2) where F = force of attraction between the two bodies, G = universal constant of gravitation , m1, m2 = mass of each body, and r = distance between centers of the two bodies. When near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body. Engr222 Spring 2004 Chapter 13 4 Mass and Weight It is important to understand the difference between the mass and weight of a body! Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by Newtons second law of motion (m = F/a). The weight of a body is not absolute, since it depends on the gravitational field in which it is measured. Weight is defined as W = mg where g is the acceleration due to gravity. Units: SI System vs. FPS System SI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earths surface, g can be taken as g = 9.81 m/s2. W (N) = m (kg) g (m/s2) => N = kgm/s2 FPS System: In the FPS system of units, weight is a base unit and mass is a derived unit. Weight is typically specified in pounds (lb), and mass is calculated from m = W/g. If g is specified in units of ft/s2, then the mass is expressed in slugs. On the earths surface, g is approximately 32.2 ft/s2. m (slugs) = W (lb)/g (ft/s2) => slug = lbs2/ft Engr222 Spring 2004 Chapter 13 5 Equation of Motion The motion of a particle is governed by Newtons second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written F = FR = ma where FR is the resultant force, which is a vector summation of all the forces. To illustrate the equation, consider a particle acted on by two forces. First, draw the particles freebody diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force ma acting in the same direction as the resultant force FR. Inertial Frame of Reference This equation of motion is only valid if the acceleration is measured in a Newtonian or inertial frame of reference. What does this mean? For problems concerned with motions at or near the earths surface, we typically assume our inertial frame to be fixed to the earth. We neglect any acceleration effects from the earths rotation. For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars. Engr222 Spring 2004 Chapter 13 6 System of Particles The equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems. As in statics, there are internal forces and external forces acting on the system. What is the difference between them? Using the definitions of m = mi as the total mass of all particles and aG as the acceleration of the center of mass G of the particles, then maG = miai . The text shows the details, but for a system of particles: F = maG where F is the sum of the external forces acting on the entire system. Key Points 1) Newtons second law is a Law of Nature-experimentally proven and not the result of an analytical proof. 2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object. 3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = ma, i.e., W = mg. 4) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems! Engr222 Spring 2004 Chapter 13 7 Procedure for the Application of the Equation of Motion 1) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used. 2) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particles inertial force, ma. Resolve this vector into its appropriate components. 4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5) It may be necessary to apply the proper kinematic relations to generate additional equations. Example Given: A crate of mass m is pulled by a cable attached to a truck. The coefficient of kinetic friction between the crate and road is k. Find: Draw the free-body and kinetic diagrams of the crate. Plan: 1) Define an inertial coordinate system. 2) Draw the crates free-body diagram, showing all external forces applied to the crate in the proper directions. 3) Draw the crates kinetic diagram, showing the inertial force vector ma in the proper direction. Engr222 Spring 2004 Chapter 13 8 Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the crate: y Example - continued The weight force (W) acts through the crates center of mass. T is the tension x force in the cable. The normal force (N) is perpendicular to the surface. The friction force (F = uKN) acts in a direction F = uKN opposite to the motion of the crate. N 3) Draw the kinetic diagram of the crate: W = mg T 30 ma The crate will be pulled to the right. The acceleration vector can be directed to the right if the truck is speeding up or to the left if it is slowing down. Concept Quiz 1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is k. mg mg A) N kN B) N v kN C) mg kmg N D) None of the above. Engr222 Spring 2004 Chapter 13 9 Concept Quiz - continued 2. Packaging for oranges is tested using a machine that exerts ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the y correct FBD and kinetic diagram for this condition. A) Rx W may B) max Rx W x = Ry = Ry max C) = Ry may D) W may Ry = max Group Problem Solving Given: Each block has a mass m. The coefficient of kinetic friction at all surfaces of contact is . A horizontal force P is applied to the bottom block. Find: Draw the free-body and kinetic diagrams of each block. Plan: 1) Define an inertial coordinate system. 2) Draw the free-body diagrams for each block, showing all external forces. 3) Draw the kinetic diagrams for each block, showing the inertial forces. Engr222 Spring 2004 Chapter 13 10 Group Problem Solving - continued 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of each block: y x Block B: WB = mg T NB y x NB P Block A: WA = mg FfB = NB FfA = NA FfB = NB The friction forces oppose the motion of each block relative to the surfaces on which they slide. 3) Draw the kinetic diagram of each block: NA Block B: maB = 0 maA Block A: Attention Quiz 1. Internal forces are not included in an equation of motion analysis because the internal forces are_____. A) B) C) D) equal to zero equal and opposite and do not affect the calculations negligibly small not important F v 2. A 10 lb block is initially moving down a ramp with a velocity of v. The force F is applied to bring the block to rest. Select the correct FBD. A) 10 F k10 N B) 10 F k10 N C) 10 F kN N Engr222 Spring 2004 Chapter 13 11 Textbook Problem 13- Announcements HW 14 and 15 due Wed, May 12 Engr222 Spring 2004 Chapter 13 12 Equations of Motion: Rectangular Coordinates Section 13.4 Todays Objectives: Students will be able to apply Newtons second law to determine forces and accelerations for particles in rectilinear motion. In-Class Activities: Reading quiz Applications Equations of motion using rectangular (Cartesian) Coordinates Concept quiz Group problem solving Attention Quiz Reading Quiz 1. In dynamics, the friction force acting on a moving object is always A) in the direction of its motion. C) a static friction. B) a kinetic friction. D) zero. 2. If a particle is connected to a spring, the elastic spring force is expressed by F = ks . The s in this equation is the A) B) C) D) spring constant. unstretched length of the spring. difference between stretched and unstretched length. stretched length of the spring. Engr222 Spring 2004 Chapter 13 13 Applications If a man is pushing a 100 lb crate, how large a force F must he exert to start moving the crate? What would you have to know before you could calculate the answer? Applications - continued Objects that move in any fluid have a drag force acting on them. This drag force is a function of velocity. If the ship has an initial velocity vo and the magnitude of the opposing drag force at any instant is half the velocity, how long it would take for the ship to come to a stop if its engines stop? Engr222 Spring 2004 Chapter 13 14 Equation of Motion The equation of motion, F = ma, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k) Fx = max , Fy = may , and Fz = maz . or, as scalar equations, Procedure for Analysis Free Body Diagram Establish your coordinate system and draw the particles free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ma vector (sometimes called the inertial force) on a separate diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic spring, a spring force equal to ks should be included on the FBD. Engr222 Spring 2004 Chapter 13 15 Procedure for Analysis - continued Equations of Motion If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is F = ma or Fx i + Fy j + Fz k = m(ax i + ay j + az k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2-D. Procedure for Analysis - continued Kinematics The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion. Any of the tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem. Engr222 Spring 2004 Chapter 13 16 Example Given: WA = 10 lb WB = 20 lb vA = 2 ft/s k = 0.2 Find: vA when A has moved 4 feet. Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion . Using dependent motion equations, derive a relationship between aA and aB and use with the equation of motion formulas. Example - continued Free-body and kinetic diagrams of B: 2T = WB Apply the equation of motion to B: + Fy = ma y WB 2 T = mB aB 20 20 2 T = 32. 2 aB (1) mBaB Engr222 Spring 2004 Chapter 13 17 Example - continued y x Free-body and kinetic diagrams of A: WA T N F = kN = m A aA + Apply the equations of motion to A: + Fx = ma x Fy = ma y = 0 N = W = 10 lb A F T = mA aA 10 2 T = 32 2 a . A (2) F = N = 2 lb k Example - continued Now consider the kinematics. Constraint equation: sA + 2 sB = constant or vA + 2 vB = 0 Therefore aA + 2 aB = 0 aA = -2 aB (3) sA A Datums sB B (Notice aA is considered positive to the left and aB is positive downward.) Engr222 Spring 2004 Chapter 13 18 Example - continued Now combine equations (1), (2), and (3). T = 22 = 7. 33 lb 3 ft a = 17. 16 fts 2 = 17.16 s2 A Now use the kinematic equation: v A 2 2 = v + 2 aA ( sA soA ) oA v 2 = 2 2 + 2 (17.16 )(4 ) A ft v = 11. 9 s A Concept Quiz 1. If the cable has a tension of 3 N, determine the acceleration of block B. A) 4.26 m/s2 down C) 8.31 m/s2 down B) 4.26 m/s2 up D) 8.31 m/s2 up 10 kg k=0.4 4 kg 2. Determine the acceleration of the block. A) 2.20 m/s2 B) 3.17 m/s2 5 kg 30 C) 11.0 m/s2 D) 4.26 m/s2 60 N Engr222 Spring 2004 Chapter 13 19 Group Problem Solving Given: The 400 kg mine car is hoisted up the incline. The force in the cable is F = (3200t2) N. The car has an initial velocity of vi = 2 m/s at t = 0. Find: The velocity when t = 2 s. Plan: Draw the free-body diagram of the car and apply the equation of motion to determine the acceleration. Apply kinematics relations to determine the velocity. Group Solving Problem - continued Solution: 1) Draw the free-body and kinetic diagrams of the mine car: W = mg y x N F ma = Since the motion is up the incline, rotate the x-y axes. = tan-1(8/15) = 28.07 Motion occurs only in the x-direction. Engr222 Spring 2004 Chapter 13 20 Group Problem Solving - continued 2) Apply the equation of motion in the x-direction: + Fx = max => F mg(sin) = max => 3200t2 (400)(9.81)(sin 28.07) = 400a => a = (8t2 4.616) m/s2 3) Use kinematics to determine the velocity: a = dv/dt => dv = a dt v v1 t dv = 0 (8t2 4.616) dt, v1 = 2 m/s, t = 2 s 2 0 v 2 = (8/3t3 4.616t) = 12.10 => v = 14.1 m/s Attention Quiz 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s2 acceleration. A) 2265 N C) 5524 N B) 3365 N D) 6543 N 60 a = 4 m/s2 T 2. A 10 lb particle has forces of F1= (3i + 5j) lb and F2= (-7i + 9j) lb acting on it. Determine the acceleration of the particle. A) (-0.4 i + 1.4 j) ft/s2 C) (-12.9 i + 45 j) ft/s2 B) (-4 i + 14 j) ft/s2 D) (13 i + 4 j) ft/s2 Engr222 Spring 2004 Chapter 13 21 Textbook Problem 13-15 The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originally at rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is s=0.4, and the coefficient of kinetic friction is k=0.3. Announcements New schedule Mon, May 10 Wed, May 12 Fri, May 14 Mon, May 17 Wed, May 19 13.1-4 13.5 13.6 14.1-3 Mid-term #2 HW #14,15 due HW #16 due HW #17 due Mid-term #2 covers 12.6-10, 13.1-6 Fri, May 21 Mon, May 24 14.5-6 HW #18 due Back on schedule Note: We will not cover section 14.4 (scratch HW#19) Engr222 Spring 2004 Chapter 13 22 Equations of Motion: Normal and Tangential Components - Section 13.5 Todays Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. In-Class Activities: Reading quiz Applications Equation of motion in n-t coordinates Concept quiz Group problem solving Attention quiz Reading Quiz 1. The normal component of the equation of motion is written as Fn = man, where Fn is referred to as the _______. A) impulse C) tangential force B) centripetal force D) inertia force 2. The positive n direction of the normal and tangential coordinates is ____________. A) B) C) D) normal to the tangential component always directed toward the center of curvature normal to the bi-normal component All of the above Engr222 Spring 2004 Chapter 13 23 Applications Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding at high speeds. If the cars maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle () required to prevent the car from sliding? Applications - continued Satellites are held in orbit around the earth by using the earths gravitational pull as the centripetal force the force acting to change the direction of the satellites velocity. Knowing the radius of orbit of the satellite, how can we determine the required speed of the satellite to maintain this orbit? Engr222 Spring 2004 Chapter 13 24 Normal and Tangential Components When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. The normal direction (n) always points toward the paths center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle. Equations of Motion Since the equation of motion is a vector equation , F = ma, it may be written in terms of the n & t coordinates as Ftut + Fnun = mat + man Here Ft & Fn are the sums of the force components acting in the t & n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: Ft = mat and Fn = man . Since there is no motion in the binormal (b) direction, we can also write Fb = 0. Engr222 Spring 2004 Chapter 13 25 Normal and Tangential Accelerations The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of Ft, the particles speed will either be increasing or decreasing. The normal acceleration, an = v2/, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the paths center of curvature. Thus, Fn will always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from = dy [1 + ( )2]3/2 dx d2 y dx2 Solving Problems with n-t Coordinates Use n-t coordinates when a particle is moving along a known, curved path. Establish the n-t coordinate system on the particle. Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts inward (the positive ndirection). The tangential acceleration (at) may act in either the positive or negative t direction. Apply the equations of motion in scalar form and solve. It may be necessary to employ the kinematic relations: at = dv/dt = v dv/ds an = v2/ Engr222 Spring 2004 Chapter 13 26 Given:At the instant = 60, the boys center of mass G is momentarily at rest. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords. Find: The boys speed and the tension in each of the two supporting cords of the swing when = 90. Plan: 1) Since the problem involves a curved path and finding the force perpendicular to the path, use n-t coordinates. Draw the boys free-body and kinetic diagrams. 2) Apply the equation of motion in the n-t directions. 3) Use kinematics to relate the boys acceleration to his speed. Example Solution: 1) The n-t coordinate system can be established on the boy at some arbitrary angle . Approximating the boy and seat together as a particle, the free-body and kinetic diagrams can be drawn. Free-body diagram Kinetic diagram n n 2T Example - continued = man mat W t T = tension in each cord W = weight of the boy t Engr222 Spring 2004 Chapter 13 27 2) Apply the equations of motion in the n-t directions. (a) Fn = man => 2T W sin = man Using an = v2/ = v2/10, w = 60 lb, and m = w/g = (60/32.2), we get: 2T 60 sin = (60/32.2)(v2/10) (1) (b) Ft = mat => W cos = mat => 60 cos = (60/32.2) at Solving for at: at = 32.2 cos (2) Note that there are 2 equations and 3 unknowns (T, v, at). One more equation is needed. Example - continued Example - continued 3) Apply kinematics to relate at and v. v dv = at ds where ds = d = 10 d => v dv = 32.2 cos ds = 32.2 cos (10 d ) v 90 => v dv = 322 cos d 2 => v = 322 sin 2 0 60 90 => v = 9.29 ft/s 60 This v is the speed of the boy at = 90. This value can be substituted into equation (1) to solve for T. 2T 60 sin(90) = (60/32.2)(9.29)2/10 T = 38.0 lb (the tension in each cord) Engr222 Spring 2004 Chapter 13 28 Concept Quiz 1. A 10 kg sack slides down a smooth surface. If the normal force on the surface at the flat spot, A, is 98.1 N () , the radius of curvature is ____. A) 0.2 m C) 1.0 m B) 0.4 m v=2m/s D) None of the above A 2. A 20 lb block is moving along a smooth surface. If the normal force on the surface at A is 10 lb, the velocity is ________. A) 7.6 ft/s C) 10.6 ft/s B) 9.6 ft/s D) 12.6 ft/s A =7 ft Group Problem Solving Given: A 200 kg snowmobile with rider is traveling down the hill. When it is at point A, it is traveling at 4 m/s and increasing its speed at 2 m/s2. Find: The resultant normal force and resultant frictional force exerted on the tracks at point A. Plan: 1) Treat the snowmobile as a particle. Draw the freebody and kinetic diagrams. 2) Apply the equations of motion in the n-t directions. 3) Use calculus to determine the slope and radius of curvature of the path at point A. Engr222 Spring 2004 Chapter 13 29 Group Problem Solving - continued Solution: 1) The n-t coordinate system can be established on the snowmobile at point A. Treat the snowmobile and rider as a particle and draw the free-body and kinetic diagrams: W F N n = t man n mat t W = mg = weight of snowmobile and passenger N = resultant normal force on tracks F = resultant friction force on tracks Group Problem Solving - continued 2) Apply the equations of motion in the n-t directions: Fn = man => W cos N = man Using W = mg and an = v2/ = (4)2/ => (200)(9.81) cos N = (200)(16/) => N = 1962 cos 3200/ Ft = mat => W sin F = mat Using W = mg and at = 2 m/s2 (given) => (200)(9.81) sin F = (200)(2) => F = 1962 sin 400 (2) (1) Engr222 Spring 2004 Chapter 13 30 Group Problem Solving - continued 3) Determine by differentiating y = f(x) at x = 10 m: y = -5(10-3)x3 => dy/dx = (-15)(10-3)x2 => d2y/dx2 = -30(10-3)x dy [1 + ( )2]3/2 [1 + (-1.5)2]3/2 dx = = d2 y -0.3 x = 10 m dx2 Determine from the slope of the curve at A: dy tan = dy/dx = tan-1 x = 10 m dx From Eq.(1): N = 1962 cos(56.31) 3200/19.53 = 924 N From Eq.(2): F = 1962 sin(56.31) 400 = 1232 N (dy/dx) = tan-1 (-1.5) = 56.31 Attention Quiz 1. The tangential acceleration of an object A) represents the rate of change of the velocity vectors direction. B) represents the rate of change in the magnitude of the velocity. C) is a function of the radius of curvature. D) Both B and C. 2. The ball has a mass of 20 kg and a speed of v = 30 m/s at the instant it is at its l...