5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously.

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1 5. REASONING AND SOLUTION An object will not necessarily accelerate when two or more forces are applied to the object simultaneously. The applied forces may cancel so the net force is zero; in such a case, the object will not accelerate. The resultant of all the forces that act on the object must be nonzero in order for the object to accelerate.

2 6. REASONING AND SOLUTION Since the father and the daughter are standing on ice skates, there is virtually no friction between their bodies and the ground. We can assume, therefore, that the only horizontal force that acts on the daughter is due to the father, and similarly, the only horizontal force that acts on the father is due to the daughter. a. According to Newton's third law, when they push off against each other, the force exerted on the father by the daughter must be equal in magnitude and opposite in direction to the force exerted on the daughter by the father. In other words, both the father and the daughter experience pushing forces of equal magnitude. b. According to Newton's second law, F = ma. Therefore, a = F / m. The magnitude of the net force on the father is the same as the magnitude of the net force on the daughter, so we can conclude that, since the daughter has the smaller mass, she will acquire the larger acceleration.

3 10. REASONING AND SOLUTION The mass of an object is a quantitative measure of its inertia. The mass of an object is an intrinsic property of the object and is independent of the location of the object. The weight of an object is the gravitational force exerted on the object by the earth. The gravitational force depends on the distance between the object and the center of the earth. Therefore, when an object is moved from sea level to the top of a mountain, its weight will change, while the mass of the object remains constant.

4 11. REASONING AND SOLUTION The weight of the ball always acts downward. The force of air resistance will always act in the direction that is opposite to the direction of motion of the ball. The net force on the ball is the resultant of the weight and the force of air resistance. a. As the ball moves upward, the force of air resistance acts downward. Since air resistance and the weight of the ball act in the same direction in this case, the net force on the ball will be greater in magnitude than the weight of the ball. b. As the ball falls downward, the force of air resistance is upward. Since air resistance and the weight of the ball act in opposite directions, the net force that acts on the ball will be smaller in magnitude than the weight of the ball. Note that in both cases, the net force points downward since the object accelerates downward in both cases.

5 16. REASONING AND SOLUTION The apparent weight will differ from the true weight only in an accelerating elevator. When the scale in an elevator reads the true weight, the only conclusion that can be made is that the elevator has zero acceleration. Therefore, one cannot conclude whether the elevator is moving with any constant velocity upward, any constant velocity downward, or whether the elevator is at rest, since each of these conditions involves zero acceleration.

6 10. REASONING AND SOLUTION F E =Fcosθ = (70 N) cos 38 = 570 N F N =Fsin θ = (70 N) sin 38 = 440 N

7 11. REASONING Newton s second law gives the acceleration as a =(F)/m. Since we seek only the horizontal acceleration, it is the x component of this equation that we will use; a x =(F x )/m. For completeness, however, the free-body diagram will include the vertical forces also (the normal force F N and the weight W). SOLUTION The free body diagram is shown at the right, where F 1 F = 59.0 N = 33.0 N = y F N When F 1 is replaced by its x and y components, we obtain the free body diagram in the second drawing. F W F 1 +x Choosing right to be the positive direction, we have ΣF F cosθ F x 1 a = = x m m +y F N a x ( 59.0 N) cos 70.0 ( 33.0 N) = = 1.83 m/s 7.00 kg F F 1 cos +x The minus sign indicates that the horizontal acceleration points to the left. W F 1 sin

8 16. REASONING For both the tug and the asteroid, Equation.8 x = v 0 t + 1 at applies with v 0 = 0 m/s, since both are initially at rest. In applying this equation, we must be careful and use the proper acceleration for each object. Newton s second law indicates that the acceleration is given by a = ΣF/m. In this expression, we note that the magnitudes of the net forces acting on the tug and the asteroid are the same, according to Newton s action-reaction law. The masses of the tug and the asteroid are different, however. Thus, the distance traveled for either object is given by, where we use for ΣF only the magnitude of the pulling force x = v 0 t + 1 at = 1 ΣF t m SOLUTION Let L be the initial distance between the tug and the asteroid. When the two objects meet, the distances that each has traveled must add up to equal L. Therefore, L = x T + x A = 1 a T t + 1 a A t L = 1 ΣF t + 1 ΣF t = 1 ΣF m T m T Solving for the time t gives m A m A t t = L ΣF m T m A = ( 450 m) 1 ( 490 N) 3500 kg kg = 64 s

9 18. REASONING Newton s law of universal gravitation indicates that the gravitational force that each uniform sphere exerts on the other has a magnitude that is inversely proportional to the square of the distance between the centers of the spheres. Therefore, the maximum gravitational force between two uniform spheres occurs when the centers of the spheres are as close together as possible, and this occurs when the surfaces of the spheres are touching. Then, the distance between the centers of the spheres is the sum of the two radii. SOLUTION When the bowling ball and the billiard ball are touching, the distance between their centers is r = r Bowling + r Billiard. Using this expression in Newton s law of universal gravitation gives Gm m Gm m F = = r r r Bowling Billiard Bowling Billiard ( + Bowling Billiard ) 11 ( N m / kg )( 7. kg)( 0.38 kg) ( 0.11 m m) = = N

10 0. REASONING AND SOLUTION The forces that act on the rock are shown at the right. Newton's second law (with the direction of motion as positive) is R ΣF = mg R = ma Solving for the acceleration a gives ( 45 kg)( 9.80 m/s ) ( 50 N) mg R a = = = m 45 kg 4. m/s mg

11 8. REASONING AND SOLUTION Thefigureattheright shows the three spheres with sphere 3 being the sphere of unknown mass. Sphere 3 feels a force F 31 due to the presence of sphere 1, and a force F 3 due to the presence of sphere. The net force on sphere 3 is the resultant of F 31 and F 3. Note that since the spheres form an equilateral triangle, each interior angle is 60. Therefore, both F 31 and F 3 make a 30 angle with the vertical line as shown. 3 F F 1.0m Furthermore, F 31 and F 3 have the same magnitude given by F = GMm 3 r where M is the mass of either sphere 1 or and m 3 is the mass of sphere 3. The components of the two forces are shown in the following drawings: F Fcosθ Fcosθ 30.0 F 3 Fsin θ Fsin θ Clearly, the horizontal components of the two forces add to zero. Thus, the net force on sphere 3 is the resultant of the vertical components of F 31 and F 3 : F 3 = Fcos θ = GMm 3 r cos θ. The acceleration of sphere 3 is given by Newton's second law: a 3 = F 3 = GM ( m 3 r cosθ = = m/s ( ) ( 1.0m) cos N m /kg ).80kg

12 3. REASONING AND SOLUTION Since both motions are characterized by constant acceleration, it follows that 1 y a J t J J = y 1 E a E t E where the subscripts designate those quantities that pertain to Jupiter and Earth. Since both objects fall the same distance, the above ratio is equal to unity. Solving for the ratio of the times yields t J t E = a E a J = GM E / R E GM J / R J = R J R E M E = ( 11. ) M J = 0.68

Week 4 Homework/Recitation: 9/21/2017 Chapter4: Problems 3, 5, 11, 16, 24, 38, 52, 77, 78, 98. is shown in the drawing. F 2

Week 4 Homework/Recitation: 9/1/017 Chapter4: Problems 3, 5, 11, 16, 4, 38, 5, 77, 78, 98. 3. Two horizontal forces, F 1 and F, are acting on a box, but only F 1 is shown in the drawing. F can point either