Example. F and W. Normal. F = 60cos 60 N = 30N. Block accelerates to the right. θ 1 F 1 F 2

Transcription

1 Physic 3 Lecture 7 Newton s 3 d Law: When a body exerts a force on another, the second body exerts an equal oppositely directed force on the first body. Frictional forces: kinetic friction: fk = μk N static friction f N s < μ s Examples.

2 Example Two forces, F and F, act on the kg block shown in the drawing. The magnitudes of the forces are F =60 N and F = 5 N. What is the horizontal acceleration (magnitude and direction) of the block? 60 0 F Normal W Normal force cancels the y components of F F,y + Normaly W = 0,x ( o ) F = 60cos 60 N = 30N F and W F = F F = 30N 5N = 5N net,x F F m 60 N 5 N 5 kg θ o 60 θ a? a = F /m= 5N/5kg = m / s net Block accelerates to the right.

3 conceptual question Consider a person standing in an elevator that t is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) )larger than b) identical to c) smaller than the downward weight W of the person.

4 Example: with balance scale marked in Newtons, not the usual scale, but a quite reasonable one. A 00 kg man stands on a bathroom scale in an elevator. Starting ti from rest, the elevator ascends, attaining its maximum speed of. m/s in 0.80 s. It travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for.5 s and comes to rest. What does the scale register (a) before the elevator starts to move? (b) during the first 0.8 s? (c) while the elevator is traveling at constant speed? (d) during the negative acceleration? Scale reading =Normal force N? N mg = ma : scale = N (in Newtons) m v f Δt 00 kg. m/s 0.8 s a) : a=0; N = mg : scale = N = 980 Newtons b) : N mg = ma : scale = N a =Δv / Δ t = (.m / s) / 0.8s =.5m / s N = m(g + a);scale = m(g + a) scale = (00kg)(9.8m / s +.5m / s ) = 30Newtons c) : N = mg : scale = N = 980Newtons v i. m/s N v f 0 d) : a = (vf v i) / Δ t =.m / s / (.5s) = 0.8m / s Δt.5 s N = m(g + a);scale = m(g + a) = 900Newtons

5 No Acceleration: Static Equilibrium All objects are at rest and remain so. The net force on any object must vanish. I.e. on an object: F = 0 i i Example: Three ropes are arranged so as to support a 4 kg mass as shown below. Determine the tension in each rope. T A ( 0 Fi,x = 0= T + T cos 60 ) T i T 60 o T 3 T 3 4 kg 4 kg Solve by considering forces at point A mg i 0= T+ T /; T = T / ( 0 ) F = 0 = T sin 60 T i,y 3 ( 0 ) T = T / sin 60 =.6T 3 3 T3 = mg = 39N;T 39.N;T = 45.3N 453N T =.6N

6 Atwood s machine Consider the Atwood machine to the right. The massless string passes over a massless and frictionless pulley. It is under tension T, which we define to be the magnitude of the tension force. By this definition it is a positive number. Choosing up to be positive, what is the net force on mass? a) T-m g b) T+m g c) m g-t d) none of the above m m

7 Choosing up to be positive, what is the net force on mass? a) m g-t b) T+m g c) T-m g d) none of the above Atwood s machine m m

8 From Newton s nd law: T m g T = = m a m a + m g Atwood s machine Also T m g = m a T = m a + m g m If m exceeds m, m goes down and m goes up. If m exceeds m, m goes down and m goes up. In either Putting it together: case m a + m g = m a + m g Δy = Δy = ma + mg v = v ma + ma = mg mg a = a ( m ) m g net force a = pulling m m + m up total mass and m down. m

9 Reading Quiz 4. An action/reaction pair of forces A. point in the same direction. B. act on the same object. C. are always long-range forces. D. act on two different objects. Slide 4-3

10 Newton s Third Law When a body bd exerts a force on another, the second dbd body exerts an equal oppositely directed force on the first body. Note: the two forces act on different bodies Force on body due to body : F Force on body due to body : F 3 d law: body body F =- F

11 Example Two skaters, an 8 kg man and a 48kg woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes the man with a force of 45 N due east. Determine the accelerations (magnitude and direction) of the man and the woman. East x components (West is positive) a a man woman 45N = = 8kg 45N = = 48kg 0.55m / s east 0.94m / s west

12 quiz Two skaters, an 00 kg man and a 50 kg woman, are standing on ice. Neglect any friction between the skate blades and the ice. By pushing the man, the woman is accelerated at m/s in the direction of due west. What is the corresponding acceleration of the man? a) 4 m/s due east b) m/s due east c) m/s due east d).5 m/s due east F = F m a = m a woman_on_man man_on_woman man man woman woman East choosing west to be negative (just to be contra ry) mwoman 50 aman,x = a woman,x = ( m/s ) = m/s east m 00 man Note: the direction would still be east even if you chose west to be positive

13 F = 60 N Example N m =0 kg N m = 5 kg A block with mass 5 kg and a second block with mass 0 kg are supported by a frictionless surface. A force of 60 N is applied to the 0 kg mass. What is the force of the 5 kg block on the 0 kg block? x components: d body : N,x = ma x; N,x = N,x from Newtons 3 law body : F + N = ma ; F = N + ma x,x x x,x x ( ) F = N + ma = ma + ma = m + m a ax = F x /( m + m) N = N = ma = mf / m + m x,x x x x x ( ),x,x x x N = 5kg60N / (5kg) = 0N N,x,x = 0N to the left If objects move together, the acceleration is governed by the total mass

14 Friction Friction impedes the motion of one object along the surfaces of another. It occurs because the surfaces of the two objects temporarily stick together via microwelds. The frictional force can be larger if the two surfaces are at rest with respect to each other. Experimentally we have two cases: N kinetic friction: f v k fk = μk N N static friction f F s f < μ N s s The coefficient of static friction generally exceeds that for kinetic friction: μ > s μ k Frictional forces always oppose the motion of one surface with respect to the other.

15 Example with static friction Consider the figure below, with M =05 kg and M =44. kg. What is the minimum static coefficient of friction necessary to keep the block from slipping. T = M g If M doesn t move T= f s μ N = μ Mg s s Putting it together Mg μs Mg M s M μ If this isn t true, M will slip

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