Solution of HW4. and m 2

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1 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 masses of the particles and r is the distance between them. Since the particles have equal masses, we can arrange the particles so that each one experiences a net gravitational force that has the same magnitude if we arrange the particles so that the distance between any two of the particles is the same. herefore, the particles should be placed at the corners of an equilateral triangle with all three sides of equal length. 15. REASONING AND SOLUION If the elevator were at rest, or moving with a constant velocity, the scale would read the true weight of mg = 98 N. When the elevator is accelerating, the scale reading will differ from 98 N and will display the apparent weight,, which is given by Equation 4.6: = mg + ma where a, the acceleration of the elevator, is positive when the elevator accelerates upward and negative when the elevator accelerates downward. a. When the apparent weight is = 75 N, the apparent weight is less than the true weight (mg = 98 N) so a must be negative. he elevator is accelerating downward. b. When the apparent weight is = 10 N, the apparent weight is greater than the true weight (mg = 98 N) so a must be positive. he elevator is accelerating upward. 18. REASONING AND SOLUION Since the sled moves with constant velocity, the force of kinetic friction is present. he magnitude of this force is given by µ k, where µ k is the coefficient of kinetic friction and is the magnitude of the normal force that acts on the sled. Furthermore, the horizontal component of the applied force must be equal in magnitude to the force of kinetic friction, since there is no acceleration. When the person pulls on the sled, the vertical component of the pulling force tends to decrease the magnitude of the normal force relative to that when the sled is not being pulled or pushed. On the other hand, when the person pushes on the sled, the vertical component of the pushing force tends to increase the normal force relative to that when the sled is not being pulled or pushed. herefore, when the sled is pulled, the magnitude of the force of kinetic friction, and therefore the magnitude of the applied force, is less than when the sled is pushed. 6. REASONING AND SOLUION here are three forces that act on the ring as shown in the figure below. he weight of the block, which acts downward, and two forces of tension that act along the rope away from the ring. Since the ring is at rest, the net force on the ring is zero. he weight of the block is balanced by the vertical components of the tension in the rope. Clearly, the rope can never be made

2 horizontal, for then there would be no vertical components of the tension forces to balance the weight of the block. W 3. SSM REASONING AND SOLUION a. Combining Equations 4.4 and 4.5, we see that the acceleration due to gravity on the surface of Saturn can be calculated as follows: g Saturn = G M Saturn r Saturn ( ( ) kg) = N m /kg ( = 10.5 m/s m) b. he ratio of the person s weight on Saturn to that on earth is WSaturn mgsaturn gsaturn 10.5 m/s = = = = W mg g 9.80 m/s earth earth earth REASONING AND SOLUION a. he apparent weight of the person is = mg + ma = (95.0 kg)(9.80 m/s m/s ) = 1.10 X 10 3 N b. = (95.0 kg)(9.80 m/s ) = 931 N c. = (95.0 kg)(9.80 m/s 1.30 m/s ) = 808 N 41. REASONING In each of the three cases under consideration the kinetic frictional force is given by f k. However, the normal force varies from case to case. o determine the normal force, we use Equation 4.6 ( = mg + ma) and thereby take into account the acceleration of the elevator. he normal force is greatest when the elevator accelerates upward (a positive) and smallest when the elevator accelerates downward (a negative).

3 SOLUION a. When the elevator is stationary, its acceleration is a = 0 m/s. Using Equation 4.6, we can express the kinetic frictional force as f k ( mg + ma) m( g + a) [ ( )] = 1. N = ( )( 6.00 kg) ( 9.80 m/s )+ 0 m/s b. When the elevator accelerates upward, a = +1.0 m/s. hen, f k ( mg + ma) m( g + a) [ ( )] = 3.8 N = ( )( 6.00 kg) ( 9.80 m/s )+ 1.0 m/s c. When the elevator accelerates downward, a = 1.0 m/s. hen, f k ( mg + ma) m( g + a) [ ( )] = 18.6 N = ( )( 6.00 kg) ( 9.80 m/s )+ 1.0 m/s 51. REASONING he book is kept from falling as long as the total static frictional force balances the weight of the book. he forces that act on the book are shown in the following free-body diagram, where is the pressing force applied by each hand. f s f s W 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. he maximum 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 =. SOLUION Since the frictional forces balance the weight, we have Solving for, we find that f s = ( µ s )= ( µ s )= W

4 54. REASONING he free-body diagram in the drawing at the right shows the forces that act on the clown (weight = W). In this drawing, note that denotes the pulling force. Since the rope passes around three pulleys, forces of magnitude are applied both to the clown s hands and his feet. he normal force due to the floor is, and the maximum static frictional f s force is f s. At the instant just before the clown s feet move, the net vertical and net horizontal forces are zero, according to Newton s second law, since there is no acceleration at this instant. W SOLUION According to Newton s second law, with upward and to the right chosen as the positive directions, we have + W = 0 Vertical forces and f s = 0 Horizontal forces From the horizontal-force equation we find = f s. But f s = µ s. From the vertical-force equation, the normal force is = W. With these substitutions, it follows that = f s = µ s = µ s ( W ) Solving for gives = µ s W = ( 0.53) ( 890 N) = 310 N 1+ µ s REASONING AND SOLUION Let the tension in wire 1 be 1 and the tension in wire be. he sum of the vertical forces acting on the point where the wires join must be zero. 1 sin sin 55.0 mg = 0 (1) Similarly, the horizontal forces must add to zero so 1 cos cos 55.0 = 0 () Solving Equation () for 1, substituting into Equation (1), and rearranging yields = 317 N

5 Using this result in Equation () gives 1 = 49 N 67. REASONING AND SOLUION Newton's second law applied to block 1 (4 N) gives Object 1 Object = m 1 a 1 Similarly, for block (185 N) N 1 m g = m a W 1 W If the string is not to break or go slack, both blocks must have accelerations of the same magnitude. hen a 1 = a and a = a. he above equations become = m 1 a (1) m g = m a () a. Substituting Equation (1) into Equation () and solving for a yields a = m g m 1 + m =.99 m/s b. Using this value in Equation (1) gives 79. SSM REASONING he box comes to a halt because the kinetic frictional force and the component of its weight parallel to the incline oppose the motion and cause the box to slow down. he distance that the box travels up the incline can be can be found by solving Equation.9 (v = v0 + ax ) for x. Before we use this approach, however, we must first determine the acceleration of the box as it travels along the incline. SOLUION he figure above shows the free-body diagram for the box. It shows the resolved components of the forces that act on the box. If we take the direction up the incline as the positive x direction, then, Newton's second law gives F x = mgsinθ f k = ma x or mgsinθ µ k = ma x where we have used Equation 4.8, f k. In the y direction we have

6 F y = mgcosθ = 0 or = mgcosθ since there is no acceleration in the y direction. herefore, the equation for the motion in the x direction becomes mgsinθ µ k mgcosθ = ma x or a x = g(sinθ + µ k cosθ ) According to Equation.9, with this value for the acceleration and the fact that v = 0 m/s, the distance that the box slides up the incline is x = v 0 a = v 0 g(sinθ + µ k cosθ ) = (1.50 m/s) (9.80 m/s )[sin (0.180)cos 15.0 ] = 0.65 m 76. REASONING AND SOLUION If the +x axis is taken to be parallel to and up the ramp, then F x = ma x gives f k mg sin 30.0 = ma x where f k. Hence, (1) Also, F y = ma y gives = ma x + µ k + mg sin 30.0 mg cos 30.0 = 0 since no acceleration occurs in this direction. hen () = mg cos 30.0 Substitution of Equation () into Equation (1) yields = ma x + µ k mg cos mg sin 30.0 = (05 kg)(0.800 m/s ) + (0.900)(05 kg)(9.80 m/s )cos (05 kg)(9.80 m/s )sin 30.0 = 730 N

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