General Physics I Spring Applying Newton s Laws

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1 General Physics I Spring 2011 pplying Newton s Laws 1

2 Friction When you push horizontally on a heavy box at rest on a horizontal floor with a steadily increasing force, the box will remain at rest initially, i.e., remain in equilibrium. Since the box is in equilibrium, the net force in the horizontal direction is zero, i.e., there must be a force opposing your push. This force is the force of static friction. Static friction is friction that acts when surfaces in contact are at rest relative to each other. 2

3 Static Friction s you continue to increase the applied force, the magnitude of the static friction (f s ) rises to match the magnitude of the applied force and the box remains at rest. Eventually, the box will begin to move relative to the floor when the applied force exceeds the magnitude of the maximum static friction force (f s,max ) for the two surfaces. Experimentally, it has been found that f s,max is approximately proportional to the magnitude of the normal force acting on each surface: f s,max = µ s n, where µ s is the coefficient of static friction for the two surfaces. 3

4 Kinetic Friction When there is relative motion between the surfaces, the friction force becomes kinetic friction (f k ). Kinetic friction is in the opposite direction to the motion of one surface relative to the other. It has also been found experimentally that the magnitude of the kinetic friction force is approximately proportional to the magnitude of the normal force: f k = µ k n, where µ k is the coefficient of kinetic friction. For the same two surfaces, usually µ k < µ s. Note that both surfaces experience the friction force, according to Newton s third law. 4

5 Kinetic Friction Kinetic friction is independent of speed and area of contact. 5

6 Static and Kinetic Friction Graph of friction force vs. time as the box is pushed. 6

7 Friction and Driving Under normal circumstances, the force that propels your car forward is static friction between the tires are the road. The bottom surface of a tire does not slip relative to the road. Note that friction is in the same direction as the motion of the car. That is why it is important to note that friction opposes relative motion between surfaces. (The tire pushes back on the road; friction opposes this attempted backward relative motion and so acts in the forward direction.) When you are braking, it is also better that static friction slows you down. If the brakes lock, the tires slip and the friction will be kinetic friction, which is usually less than the maximum static friction force. nti-lock brakes prevent this from happening. When the roads are icy, the coefficients of static and kinetic friction are much smaller. The tires slip much more easily and the friction forces are much reduced, making driving more difficult. 7

8 Workbook: Chapter 5, Question 21 8

9 Interacting Objects s we saw before, when objects interact, they experience forces that come in action/reaction pairs due to Newton s third law. Each force that constitutes a pair acts on a different object. These forces have to be treated carefully when solving problems involving interacting objects. Note that the forces of a pair have the same magnitude and this is a useful piece of information in solving problems involving interacting objects. The figure and freebody diagrams illustrate these ideas. F on F = F on w on n F hand n w F on 9

10 Interacting Objects: ccelerations In many cases, the accelerations of two interacting bodies will be related. For example, for two blocks being pushed while they are in contact, their accelerations are equal. Otherwise, the blocks would separate. pplying Newton s 2 nd law to each block yields: lock F = F F = m a (1) lock Fx = F = m a (2) on Since F = F, we can substitute for on on F in Eq. (1) to get F m a= m a or, F on hand x hand on = m m + a. hand F on w n F hand n w F on x x 10

11 Interacting Objects: ccelerations For two objects connected by a rope or string that doesn t stretch or shrink, the magnitudes of the accelerations are the same. In the example to the right, though the directions of the accelerations are different, their magnitudes are the same. Thus, a = a. In terms of components, a = a. x y (Note that the acceleration of is in the negative y direction and so a is actually negative.) y 11

12 In this course, we will assume that ropes or strings are massless and pulleys are massless and frictionless. The tension in a massless rope is the same everywhere in the rope, including at both ends (which are connected to other objects). The rope exerts a force on a connected object equal to the tension. (The object of course exerts a reaction force on the rope having the same magnitude as the tension.) Ropes and Pulleys 12

13 When a rope or string passes over a pulley, the magnitude of the tension in the rope is the same on both sides of the pulley. y n T Ropes and Pulleys x Frictionless surface y T w T = T w 13

14 Workbook: Chapter 5, Question 26, 29 14

15 Textbook: Chapter 5, Problem 74 y T T m = 100 kg w w First, we have to find the acceleration of the 100-kg block. Use: y = y ( ) 1 2 i + vy i t+ 2 a ( t). f y Knowns: y = 1 m, y = 0 m, t = 6.0 s, ( v ) = 0. Unknown: a. f Solving for a gives: a y i y i y y 2 y yi ( vy) f i t = = 2( 1 m 0 0) = m/s 2. ( t) 2 (6.0 s) 2 15

16 Problem 74 (continued) pply Newton's 2nd law to the 100-kg block. F = T w = m a. y y So, T = w + m a = m g + m a = m ( g + a ). y y y 100 kg[9.8 m/s 2 ( m/s 2)] 974 N. T = + = pply Newton's 2nd law to block of unknown mass. F = T w = m a. So, y y T = w + m a = m g + m a = m ( g + a ). Thus, m y y y =. ( g + a ) y Now, T = T = 974 N. Further, a = a = m/s 2. y y Hence, T m = 974 N = 99 kg. + (9.8 m/s m/s 2) 16

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