L5F11 - P123 F11 Lecture 5 30 September 2010 Newton’s...

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Unformatted text preview: P123 F11 Lecture 5 30 September 2010 Newton’s Laws of Motion TOPICS FROM PREVIOUS LECTURE (motion in 2-D) • Components of Equation of Motion vx = vox + axt vy = voy + ayt x = xo + voxt + 1/2axt2 y = yo + voyt + 1/2ayt2 etc., … • Projectile Motion Superposition of two motions: Constant vx and free-fall vy Then solve for : y = f(x) or trajectory: y vo voy h vox R x 2 vo sin 2 o h 2g 2 vo sin 2 o R g • Relative Velocity: (e.g., cars on highway, airplane in wind, boat on river, …) Fixed Frame: A Moving Frame: B Moving Object: P vP / A vP / B vB / A 1 i‐Clicker y vo voy h vox R x ax =0 ; ay = -g 2 NEWTON’S LAWS OF MOTION So Far: Follow motion in 1‐D and 2‐D based solely on definitions of v and a in terms of position as a function of time. (KINEMATICS) NOW: Determine motion based on the underlying physical cause of that motion. (DYNAMICS) “Laws of Motion” formulated by Newton over 300 years ago Forces and interactions Common experience: Force is a “push” or a “pull” pull push Force is a vector Different types of forces: • Contact Force (kick soccer ball, push book, tow car, … ) • Long range forces, act through “empty” space (electric force, magnetic force, gravitational force, … ) Unit of force in SI units: Newton [N] We will focus on contact forces and gravity (near the earth’s surface) 3 Superposition of forces Forces are vectors add to give resultant. R F1 R Ry F2 Rx R F F1 F2 F3 ... NET FORCE: Rx Fx F1x F2 x F3 x ... R y Fy F1 y F2 y F3 y ... 2 R Rx2 R y Rz2 EQUILIBRIUM: Net force = 0 F3 F3 y F3 x F2 Fx 0; F1 Fy 0; F 0 ?! Fz 0 4 Newton’s first law Ancient Idea: The natural state of an object is at rest. If an object moves, it’s because a force acts on it. NEWTON: The nature of a body is to resist change in motion. Newton’s First Law When no net force acts on an object, its acceleration is zero If F 0 then a 0 An isolated object with no net force acting on it is either at rest, or is moving with a constant velocity. INERTIA: The tendency of a body to resist change in . v INERTIAL FRAMES OF REFERENCE An inertial frame of reference is one in which Newton’s 1st Law is valid (The frame is not accelerating!!) zB zA Inertial frame xA vB / A yB yA xB Also inertial frame if vB / A constant. • A reference frame that moves at constant velocity relative to a distant star is the best approx. of an inertial frame. • Earth is approximately an inertial frame • Accelerating elevator is NOT an inertial frame. 5 Newton’s Second Law From Newton’s 1st Law, a 0 when F 0 Background: F 0 the net force causes the object to But when accelerate in the same direction as the net force. The a magnitude of is directly proportional to F Mass and Force: • Force changes motion of an object • Mass (inertia) resists change in motion m F or F ma a Mass = weight scalar vector BUT…. Weight is proportional to mass. • The mass of a body is the same everywhere in the universe. • The weight of a body is not!! 6 Statement Of Newton’s Second Law Background: The acceleration of an object is directly proportional to the net force acting on that object, and inversely proportional to its mass. F or F ma a m F ma Fx ma x ; Fy ma y ; Fz ma z Unit of force: Newton [N] 1 N = (1 kg) X (1 m/s2) = 1kg.m/s2 ( in cgs units: 1 dyne = 1 g.cm/s2) (in British units: 1 lb = 1 slug X 1 ft/s2) (1 lb = 4.448 N) EXAMPLE: A block (m = 5.0 kg) is at rest at t = 0. Then, a constant force Fx acts on it. At t = 3.0 s, the block has moved 2.3 m. Find Fx. m Fx Fx ma x ax Have: m, need ax. But, also have t and (x - xo). 2 ( 2 .3 m ) 4 .6 ax m/s 2 0.51 m/s 2 ( x xo ) vo t 1 a x t 2 2( x xo ) a x 2 2 (3.0 s) 9 t2 2 2 So: Fx (5.0 kg )(0.51 m/s ) 2.6 kg m/s 2.6 N 7 i‐Clicker T mg Elevator moves, but at constant velocity a = 0 F = ma, therefore in particular F 0 Fy 0 Therefore, T – mg = 0 ; T = mg. 8 Mass And Weight Fg Weight is the attractive force, , exerted by the earth’s gravity on an object. Fg = weight of object = w ag Because and then in free fall (i.e., no force F ma other than gravity acting on object) we have w mg So we can compare masses by measuring weight at earth’s surface. Note: • • • Weight is a force that acts on a body at all times g varies with location, we will use g = 9.8 m/s2. Avoid incorrect usage (i.e. an object’s weight is 2 kg) Measuring forces: Convenient to use elongation of a spring. (later, we will discuss Hooke’s law which states that the elongation of a spring is directly proportional to the force exerted on it.) 9 Newton’s Third Law Whenever two isolated objects interact, the forces that the objects exert on each other are always equal in magnitude and opposite in direction. Restated: If object A exerts a force on B (an “action”) then B exerts a force on A (a “reaction”). These two forces: • Have the same magnitude • Are opposite in direction • Act on different bodies!!!!! Suppose two balls collide FB on A A “action‐reaction pair” FA on B B FB on A FA on B Forces exist in pairs, acting on different objects (e.g., bat hits ball; hammer hits nail; foot kicks butt … ) 10 Consider a box on a table on the earth n (normal force: force of table on box) Fg w n n ' Reaction to (force of box on table) Fg Reaction to weight of box Simpler to understand if consider forces on box alone: n Box is at rest, so: Fx 0 F 0 Fy 0 x: no forces y: n – mg = 0 n = mg w n and w are NOT action/reaction pair. They act on same object. n n' and ARE action/reaction pair Free Body Diagrams: an aid to solving problems • F ma • • • • applied to one object with mass m (or a group moving as one). Use only forces applied to this body Draw diagram for each body separately Choose convenient coordinate system Apply Newton’s Laws in component form. 11 EXAMPLE: Apparent contradiction – a horse pulling a cart. If the force of horse on cart equals the force of cart on horse how does the cart move???? FH on C FC on H Isolate cart and draw free body diagram n n FH on C w y w FH on C At the instant the cart moves, it accelerates: 1st 2nd F 0 law: so is NOT constant (acceleration) v law: F ma F F y x 3rd nw0 ma x FHC a x FHC m law: Note that , are NOT reaction pair!!! nw 12 x i‐Clicker action/reaction pair FT on C MILK nC FC on T FH on T nT mCg mTg FT on C FC on T 13 EXAMPLE: A skier of mass 65.0 kg is pulled by a tow rope up a snow‐covered ski slope at a constant speed. The rope and the slope are inclined by 26o. Draw a free body diagram and calculate the tension in the tow rope. T T y y n x x 26o 26o w wsin(26o) moving, but no acceleration F 0 consider components: Fy 0 n w cos(26 ) Fx 0 T w sin( 26 ) T w sin( 26 ) mg sin( 26 ) T = (65.0 kg)*(9.80 m/s2)*sin(26o) T = 279 N 14 ...
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This note was uploaded on 01/03/2012 for the course PHYSICS 750:123 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.

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