### Chapter 2, second assignment

Answers to selected problems for Chapter 2 are found here.

 Exercise 15. Two 100-N weights are attached to a spring scale as shown. Does the scale read 0 N, 100 N, or 200 N, or does it give some other reading? As a help, we will consider the case in which one end of the rope is attached to the wall instead of to a weight. If the scale were attached to the wall at one end, it would read 100 N. We can agree on that. This is because a 100-N weight, suspended from the scale (even though the force is "bent" by the pulley) should read 100 N. But... let's analyze the forces involved. Assuming the weight of the scale is negligible, The weight exerts a 100-N force on the scale. The scale and the weight together exert a 100-N force on the wall to which the whole thing is fastened. The wall exerts a 100-N force on the scale and the weight, by Newton's Third Law. Now let's consider the situation shown in the drawing. The two weights each exert forces of 100 N on the scale. We agreed that, in a case where the scale was fastened to the wall and had only one 100-N weight, that the scale would read 100 N. However, we saw that there was a 100-N force on the scale from each end, yet the scale read 100 N. Therefore in this case, with a 100-N force on the scale from each end, the scale will read 100 N. Exercise 26. If you exert a horizontal force of 200 N to slide a crate across a factory floor at a constant velocity, how much friction is exerted by the floor on the crate? Is the force of friction equal and oppositely directed to your 200-N push? Does the force of friction make up the reaction force to your push? Why not? If the crate is moving at a contant velocity, the frictional force must be equal to the pushing force. Therefore it is 200 N. This frictional force is indeed equal and oppositly directed to the 200-N pushing force. However, the frictional force is NOT the reaction force to your push. That is because it is not coming from the crate itself, but from the contact between the crate and the floor. Furthermore, the frictional force would remain 200 N no matter how hard you pushed on the crate. For example, if you increased your push to 300 N the frictional force would still be 200 N, and the crate would begin to accelerate. Exercise 42. Why does vertically falling rain make slanted streaks on the side windows of a moving automobile? If the streaks make an angle of 45°, what does this tell you about the relative speed of the car and the falling rain? The velocity vectors of the rain and the car add, making a vector at an angle from the vertical that can be seen as rain streaks on the side windows. Since the velocity vectors of the rain and the car are perpendicular to each other (vertical and horizontal respectively, a 90° angle), if the streaks make a 45° angle it means that the speeds of the rain and the car are the same. Explanation: The sum of the angles of a triangle is 180°. The rain and car vectors are two equal sides (an isosceles triangle) with a 90° angle between them. The base angles of an isosceles triangle are equal, and (since the apical angle is 90°) these particular two must add up to 90°. 90 ÷ 2 = 45°. Exercise 46. Suppose a stone is being accelerated upward by a string. Now suppose that the string breaks and that the stone slows in its upward motion. Draw a force vector diagram of the stone when it reaches the top of its path. See below for a discussion of the net forces. The force vector diagram of the stone is very simple, and is shown at right. Exercise 47. What is the net force on the stone in Ex. 46 when it is at the top of its path? What is its instantaneous velocity? Its acceleration? The net force is the force of gravity, downward. The instantaneous velocity is zero; the acceleration is downward, with magnitude g (~10 m/s2). Exercise 49. Here is a stone on an incline, at rest, interacting with both the surface of the incline and a stationary block. (a) Identify all forces that act on the stone and draw appropriate force vectors. (b) Show that the net force on the stone is zero. (Hint 1: There are two normal forces on the stone. Hint 2: Be sure the vectors you draw are for forces that act on the stone, not on the surfaces by the stone.) There are two normal forces on the stone, one from the ramp and one from the block, and they make an angle to each other of 90°. There is also a gravitational force on the stone; this is straight downward. The vectors are shown below, not to scale. We know that the net force on the stone is zero because the stone is not moving! If the vectors were drawn to scale, the two normal force vectors would add to minus the gravitational force vector. Problem 3. You push with a 20-N horizontal force on a 2-kg mass resting on a horizontal surface against a horizontal friction force of 12 N. What is the acceleration? The net horizontal force is 20 - 12 = 8 N. An 8-N force on a 2-kg mass should produce an acceleration of 8 N ÷ 2 kg = 4 m/s2.