Phys 201A. Homework 8 Solutions

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1 Phys 01A Homewor 8 Solutions 15. (b) The static frictional force that blocs A and B eert on each other has a magnitude f. The force that B eerts on A is directed to the right (the positive direction), while the force that A eerts on B is directed to the left. Blocs B and C also eert static frictional forces on each other, but these forces have a magnitude f, because the normal force pressing B and C together is twice the normal force pressing A and B together. The force that C eerts on B is directed to the right, while the force that B eerts on C is directed to the left. In summary, then, bloc A eperiences a single frictional force +f, which is the net frictional force; bloc B eperiences two frictional forces, f and +f, the net frictional force being f +f = +f; bloc C eperiences a single frictional force +f, which is the net frictional force. It follows that f s, A = f s, B = f s, C /. 16) The magnitude of the inetic frictional force is proportional to the magnitude of the normal force. The normal force is smallest in B, because the vertical component of F compensates for part of the bloc s weight. In contrast, the normal force is greatest in C, because the vertical component of F adds to the weight of the bloc. 3. (b) Since the boes move at a constant velocity, they have no acceleration and are, therefore, in equilibrium. According to ewton s second law, the net force acting on each bo must be zero. Thus, ewton s second law applied to each bo gives two equations in two unnowns, the magnitude of the tension in the rope between the boes and the inetic frictional force that acts on each bo. ote that the frictional forces acting on the boes are identical, because the boes are identical. Solving these two equations shows that the tension is one-half of the applied force..81) REASOIG AD SOLUTIO If the + ais is taen to be parallel to and up the ramp, then F = ma gives T f mg sin 30.0 = ma where f = µ. Hence, Also, F y = ma y gives T = ma + µ + mg sin 30.0 (1) mg cos = 0 since no acceleration occurs in this direction. Then Substitution of Equation () into Equation (1) yields = mg cos ()

2 T = ma + µ mg cos mg sin 30.0 T = (05 g)(0.800 m/s ) + (0.900)(05 g)(9.80 m/s )cos (05 g)(9.80 m/s )sin 30.0 = ) REASOIG The bo comes to a halt because the inetic frictional force and the component of its weight parallel to the incline oppose the motion and cause the bo to slow down. The distance that the bo travels up the incline can be can be found by solving Equation.9 ( v v 0 a ) for. Before we use this approach, however, we must first determine the acceleration of the bo as it travels along the incline. SOLUTIO The figure above shows the free-body diagram for the bo. It shows the resolved components of the forces that act on the bo. If we tae the direction up the incline as the positive direction, then, ewton's second law gives F mg sin f ma or mg sin F ma where we have used Equation 4.8, f. In the y direction we have F F mg cos 0 or F mg cos y since there is no acceleration in the y direction. Therefore, the equation for the motion in the direction becomes mg sin mg cos ma or a g (sin cos ) According to Equation.9, with this value for the acceleration and the fact that v = 0 m/s, the distance that the bo slides up the incline is v 0 v0 (1.50 m/s) 0.65 m a g(sin cos ) (9.80 m/s )[sin 15.0 (0.180)cos 15.0 ] 99) SSM REASOIG In order to start the crate moving, an eternal agent must supply MAX a force that is at least as large as the maimum value fs s, where s is the coefficient of static friction (see Equation 4.7). Once the crate is moving, the magnitude of the frictional force is very nearly constant at the value f, where is the coefficient of inetic friction (see Equation 4.8). In both cases described in the problem

statement, there are only two vertical forces that act on the crate; they are the upward normal force, and the downward pull of gravity (the weight) mg.

3 statement, there are only two vertical forces that act on the crate; they are the upward normal force, and the downward pull of gravity (the weight) mg. Furthermore, the crate has no vertical acceleration in either case. Therefore, if we tae upward as the positive direction, ewton's second law in the vertical direction gives F mg 0, and we see that, in both cases, the magnitude of the normal force is F mg. SOLUTIO a. Therefore, the applied force needed to start the crate moving is MAX fs s mg (0.760)(60.0 g)(9.80 m/s ) 447 b. When the crate moves in a straight line at constant speed, its velocity does not change, and it has zero acceleration. Thus, ewton's second law in the horizontal direction becomes P f = 0, where P is the required pushing force. Thus, the applied force required to eep the crate sliding across the doc at a constant speed is P f mg (0.410)(60.0 g)(9.80 m/s ) REASOIG AD SOLUTIO If the + ais is taen in the direction of motion, ΣF = 0 gives where Then F f mg sin = 0 f = µ F µ mg sin = 0 (1) Also, ΣF y = 0 gives so mg cos = 0 = mg cos () Substituting Equation () into Equation (1) and solving for F yields F = mg( sin + µ cos ) F = (55.0 g)(9.80 m/s )[sin (0.10)cos 5.0 ] = 86 39) SSM REASOIG The boo is ept from falling as long as the total static frictional force balances the weight of the boo. The forces that act on the boo are shown in the following free-body diagram, where P is the pressing force applied by each hand.

In this diagram, note that there are two pressing forces, one from each hand. Each hand also applies a static frictional force, and, therefore, two static frictional forces are shown.

4 In this diagram, note that there are two pressing forces, one from each hand. Each hand also applies a static frictional force, and, therefore, two static frictional forces are shown. The maimum static frictional force is related in the usual way to a normal force, but in this problem the normal force is provided by the pressing force, so that = P. SOLUTIO Since the frictional forces balance the weight, we have Solving for P, we find that MAX s s s f F P W P W s 39 45) SSM REASOIG AD SOLUTIO Four forces act on the sled. They are the pulling force P, the force of inetic friction f, the weight mg of the sled, and the normal force eerted on the sled by the surface on which it slides. The following figures show free-body diagrams for the sled. In the diagram on the right, the forces have been resolved into their and y components.

5 Since the sled is pulled at constant velocity, its acceleration is zero, and ewton's second law in the direction of motion is (with right chosen as the positive direction) F Pcos f ma 0 From Equation 4.8, we now that f, so that the above epression becomes In the vertical direction, P cos F 0 (1) F Psin F mg ma 0 () y y Solving Equation () for the normal force, and substituting into Equation (1), we obtain Pcos mg P sin 0 Solving for, the coefficient of inetic friction, we find P cos (80.0 ) cos 30.0 mg Psin (0.0 g) (9.80 m/s ) (80.0 ) sin ) REASOIG The magnitude of the inetic frictional force is given by Equation 4.8 as the coefficient of inetic friction times the magnitude of the normal force. Since the slide into second base is horizontal, the normal force is vertical. It can be evaluated by noting that there is no acceleration in the vertical direction and, therefore, the normal force must balance the weight. To find the player s initial velocity v 0, we will use inematics. The time interval for the slide into second base is given as t = 1.6 s. Since the player comes to rest at the end of the slide, his final velocity is v = 0 m/s. The player s acceleration a can be obtained from ewton s second law, since the net force is the inetic frictional force, which is nown from part (a), and the mass is given. Since t, v, and a are nown and we see v 0, the appropriate inematics equation is Equation.4 (v = v 0 + at). SOLUTIO a. Since the normal force balances the weight mg, we now that = mg. Using this fact and Equation 4.8, we find that the magnitude of the inetic frictional force is f F mg g 9.8 m/s 390

6 b. Solving Equation.4 (v = v 0 + at) for v 0 gives v 0 = v at. Taing the direction of the player s slide to be the positive direction, we use ewton s second law and Equation 4.8 for the inetic frictional force to write the acceleration a as follows: a F m m mg g The acceleration is negative, because it points opposite to the player s velocity, since the player slows down during the slide. Thus, we find for the initial velocity that v0 v g t 0 m/s m/s 1.6 s 7.7 m/s

Solution of HW4. and m 2

Solution of HW4. and m 2 Solution of HW4 9. REASONING AND SOLUION he magnitude of the gravitational force between any two of the particles is given by Newton's law of universal gravitation: F = Gm 1 m / r where m 1 and m are the