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AP QUIZ #9 ADVANCED FORCES B3-RT54: PERSON IN A MOVING ELEVATOR—SCALE READING A person who weighs 600 N is standing on a scale in an elevator. The elevator is identical in all cases. The velocity and acceleration of the elevators at the instant shown are given. A B C D v = 3 m/s v = 2 m/s v = 3 m/s v = 1 m/s a = 2 m/s2 a = 2 m/s2 a=0 a = 2 m/s2 Rank the scale reading. OR 1 Greatest 2 3 4 Least All the same All zero Cannot determine Explain your reasoning. Answer: A = D > C > B. The scale is providing the normal force on the person, and the scale reading indicates the magnitude of this force. For cases A and D that normal force is larger than the person’s weight because the scale also has to accelerate the person upward. For case C the normal force is equal in magnitude to the person’s weight, and for case B it is less than the person’s weight since the person is accelerating downward. B3-RT55: Person in an Elevator Moving Downward—Scale Reading A person who weighs 600 N is standing on a scale in an elevator. The elevator is identical in all cases. The velocity and acceleration of the elevators at the instant shown are given. A B C D v = 3 m/s v = 3 m/s v = 3 m/s v=0 a = 2 m/s2 a = 2 m/s2 a=0 a = 9.8 m/s2 Rank the scale reading. OR 1 Greatest 2 3 4 Least All the same All zero Cannot determine Explain your reasoning. Answer: A > C > B > D. The scale is providing the normal force on the person, and the scale reading indicates the magnitude of this force. For case A that normal force is larger than the person’s weight because the scale also has to accelerate the person upward. For case C the normal force is equal, in magnitude, to the person’s weight, for case B it is less than the person’s weight since the person is accelerating downward, and in case D the scale reading will be zero since the person is in free fall. B3-CT58: PERSON IN AN ELEVATOR—SCALE READING A person who weighs 500 N is standing on a scale in an elevator. The elevator is identical in both cases. In both cases the elevator is moving at a constant speed, upward in case A and downward in case B. A B v = 3 m/s v = 3 m/s Will the scale reading be (a) greater in case A, (b) greater in case B, or (c) the same in both cases? Explain. Answer: The scale reading will be the same in both cases. By Newton’s Second Law the net force on the person must be zero in both cases, since the elevator is not accelerating. So the normal force on the person by the scale (which is the reading on the scale) must be equal to the weight of the person in both cases. B3-RT68: Hanging Blocks—Tension Two blocks are connected by strings, and are pulled upward by a second string attached to the upper block. The lower block is the same in all cases, but the mass of the upper block varies. The acceleration and velocity for each system at the instant shown is given. A B C D E F 15 g 15 g 20 g 20 g 10 g 10 g v = 1 m/s v = 1 m/s v = 1 m/s v = 1 m/s v=0 a = 0 a = 0 a = 0 a = 2 m/s2 a = 2 m/s2 v = 1 m/s a = 2 m/s2 Rank the tension in the string between the blocks. OR 1 Greatest 2 3 4 5 6 Least All the same All zero Cannot determine Explain your reasoning. Answer: A > B = D = E > C = F. These tensions can be compared by considering a free-body diagram of the lower block. The tension minus the weight of the lower block must equal the mass of the lower block times its acceleration in the upward direction. Since the lower blocks are all identical, the tension must be greatest for blocks that are accelerating upward, less for blocks with no acceleration, and least for blocks that are accelerating downward. The mass of the upper block does not matter, and the velocity of the blocks at the instant shown does not matter. B3-SCT69: HANGING STONE CONNECTED TO BOX—FREE-BODY DIAGRAMS A massless rope connects a box on a horizontal surface and a hanging stone as shown below. The rope passes over a massless, frictionless pulley. The box is given a quick tap so that it slides to the right along the horizontal surface. The figure below shows the block after it has been pushed while it is still moving to the right. The mass of the hanging stone is larger than the mass of the box. There is friction between the box and the horizontal surface. Free-body diagrams that a student has drawn to scale for the box and for the hanging stone are shown. v = 2 m/s Fon stone by rope 240 g Fon box by rope Fon stone by earth Fon box by surface Fon box by surface Fon box by earth Four students discussing these free-body diagrams make the following contentions: Ali: Brianna: Carlos: Dante: “I think there is a problem with the free-body diagram for the hanging stone. The two forces should have the same magnitude.” “But the stone is moving upward – there should be a larger force in that direction.” “No, the diagram for the hanging stone is okay, but there is a problem with the diagram for the box. The frictional force is in the wrong direction.” “No, all three of you are wrong. Both free-body diagrams are correct because both show the way the objects would be accelerating.” With which, if any, of these students do you agree? Ali _______ Brianna _______ Carlos ______ Dante ______ None of them _______ Explain your reasoning. For the instant shown, Carlos is correct. Since the box is sliding to the right, the friction on the box by the surface will be to the left. So all horizontal forces on the box would be to the left and the block would slow down (as we expect). If, as Ali claimed, the forces on the stone had the same magnitude, the stone would have no net force and would not accelerate, and if (as Brianna claimed) the force on the stone by the rope was greater than the weight of the stone then the stone would speed up. B3-RT70: Hanging Stone Connected to Box on Rough Surface—Acceleration In each case shown below, a box is sliding along a horizontal surface. There is friction between the box and the horizontal surface. The box is tied to a hanging stone by a massless rope running over a massless, frictionless pulley. All these cases are identical except for the different initial velocities of the boxes. vo = 8 m/s A B 100 g 180 g vo = 3 m/s C 100 g 180 g vo = 5 m/s D 100 g 180 g vo = 5 m/s 100 g 180 g Rank the magnitudes of the accelerations of these boxes at the instant shown. Box moving to the right Box moving to the left Explain your reasoning. Answer: A = D > B = C. Fon box by surface Fon box by surface Since the surfaces are identical and the blocks all Fon box by rope Fon box by rope have the same mass, the magnitude of the friction Fon box by surface force is the same in all cases. However, when the box is moving to the left the frictional force is to the Fon box by surface Fon box by earth Fon box by earth right (opposite the direction of the tension force on the box) and when the box is moving to the right the frictional force is to the left, in the same direction as the tension. The friction force doesn’t depend on the velocity. The acceleration is the same for all boxes that are moving in the same direction, since the free-body diagrams for these boxes will be the same. B3-RT71: MOVING STRING PASSING OVER A PULLEY—TENSION AT POINTS A student pulls on a massless string that passes over a frictionless pulley to a suspended mass. He is pulling the string horizontally so that, at the the mass is moving upward at a constant speed. A and is attached instant shown, B C D Rank the tension at the labeled points A, B, C, and D. OR 1 2 3 4 All All Least the same zero de Greatest Explain your reasoning. Answer: All the same. In the approximation that the string is massless, any small section of the string must have no net force acting on it (since it is not accelerating). So the tension on one end of that small section must have the same magnitude as the tension on the other end. By extension, the tension must be the same everywhere along the horizontal section of string and everywhere along the vertical section of string. The tension in the horizontal section of string exerts a counterclockwise torque on the pulley, and the tension in the vertical section of string exerts a clockwise torque on the pulley. Since the pulley has no angular acceleration, the net torque on it is zero. The tension in the horizontal portion of the string between the pulley and the hand must be equal to the tension in the vertical portion of the string. (The perpendicular distance between the line of action of the tension force and the pulley pivot is the radius of the pulley, and is the same for both sections of string.) B3-CT73: Ball Suspended from Ceiling by Two Strings—Tension A 0.5-kg ball is suspended from a ceiling by two strings. The ball is at rest. a) Is the tension in string 1 greater than, less than, or the same as the tension in string 2? Explain. String 1 TB1 60° 30° Answer: Greater since the forces on the ball must add to zero. Keeping the directions of the vectors as shown constrains the relative sizes of TB1 and TB2 as shown in the vector summation here. Note that TB1 and TB2 are perpendicular. So, TB1 must be greater than TB2. String 2 TB2 WBE Ball TB1 WBE Suppose that the ceiling in the picture is the ceiling of an elevator, and that the TB2 elevator is moving down at a constant speed of 2 m/s. b) Is the tension in string 1 greater than, less than, or the same as the tension in string 1 in the previous question (a) where the ball was at rest? Explain. Answer: The forces are the same as for the same system at rest. The net force is still zero since the ball is moving at a constant speed. The weight does not change (the ball is still on the surface of the earth) and so the tensions will not change. B3-SCT74: HANGING MASS—TENSION IN THREE STRINGS A hanging mass is suspended midway between two walls. attached to the left wall is horizontal while the string attached wall makes an angle with the horizontal as shown. This angle is larger than the angle () in Case B. Four students make the claims about the tensions in the strings: Abbie: Bobby: Che: Damian: Case A The string to the right () in Case A following Case B “I think the tensions in any string in case A is going to be the same as the equivalent string in case B. The weight is the same, and the weight is still going to be divided up among the three ropes.” “I think the tensions in the horizontal and vertical strings are the same, because they are exactly the same in both cases. But in Case B the diagonal rope is shorter, so the tension is more concentrated there.” “The diagonal string still has to hold the weight up by itself, because the horizontal string can’t lift anything. So the diagonal string still has the same tension. But in Case B it’s pulling harder against the horizontal string because of the angle, so the tension in the horizontal string has to go up.” “But the diagonal string is fighting harder against the weight in Case A – it is pointing more nearly opposite the weight. So it has to have a greater tension in Case A. And since the tension in the diagonal string is greater, and the tension in the vertical string is the same, the tension in the horizontal string must be less in Case A. The tensions still have to balance out so that they are the same in both cases.” With which, if any, of these students do you agree? Abbie _____ Bobby _____ Che _____ Damian _____ None of them______ Explain your reasoning. Answer: None of these responses is correct. The weight is the same for both cases, and because the weight is at rest we can conclude from a freebody diagram of the weight that the tension in the vertical string must be the same in both cases. But since the knot connecting the strings is at rest in both cases vector sum of forces acting on the knot is zero in both cases. Comparing vector sums, (with subscripts K, H, V, and D for knot, horizontal, vertical, and diagonal) we can see that the horizontal tension must be greater in case B. The tension in the diagonal string will also be larger in case B TKH TKD TKV TKD TKH TKD TKV TKD TKV TKH TKV TKH B3-CT80: Blocks Moving at Constant Speed—Tension in Connecting String Two identical blocks, 1 and 2, are connected by a massless A 10 cm/s to block 2 string. In Case A, a student pulls on a string attached so that the blocks travel to the right across a desk at a constant string speed of 10 cm/s. In Case B, the student pulls on a 1 2 attached to block 1 so that the same blocks travel across the same desk to the left at a constant speed of 20 cm/s. Will the tension in the diagonal string connecting the two B blocks be greater in Case A, greater in Case B, or the same 20 cm/s in both cases? Explain your reasoning. 1 2 Greater in case A. In both cases the net force acting on the trailing block is zero, since that block is moving at a constant speed. In case A, the tension in the connecting string has a component acting downward, on the trailing block, and since the net vertical force is zero, the normal force must be greater than the weight. In case B, the tension in the connecting string has a component acting upward on the trailing block, and the normal force on the block must therefore be less than the weight. Since the friction on the trailing block by the desk is proportional to the normal force on that block, the friction force on the block by the desk is greater in Case A than it is in Case B. And since the frictional force on the trailing block is equal to the horizontal component of the tension in the string connecting the blocks in both cases, the tension in this string must be greater in Case A than in Case B.