Forces and Free-Body Diagrams AP Physics 1 FRQ – Exam Style Questions etc.
Forces and Free-Body Diagrams AP Physics 1 FRQ
Unit: 2. Force and Translational Dynamics
Weightage : 10-15%
Exam Style Practice Questions, Forces and Free-Body Diagrams AP Physics 1 FRQ
Question
A uniform meterstick, which weighs 1.5 N, is supported by two spring scales. One scale is attached 20 cm from the left-hand edge; the other scale is attached 30 cm from the right-hand edge, as shown in the preceding diagram.
(a) Which scale indicates a greater force reading? Justify your answer qualitatively, with no equations or calculations.
(b) Calculate the reading in each scale.
(c) Now the right-hand scale is moved closer to the center of the meterstick but is still hanging to the right of center. Explain your answers to the following in words with reference to your calculations in (b).
(i) Will the reading in the left-hand scale increase, decrease, or remain the same?
(ii) Will the reading in the right-hand scale increase, decrease, or remain the same?
(d) Now the scales are returned to their original locations, as in the diagram. Where on the meterstick could a 0.2-N weight be hung so as to increase the reading in the right-hand spring scale by the largest possible amount? Justify your answer.
▶️Answer/Explanation
Ans:
Part (a)
Consider the center of the meterstick as the fulcrum; then the weight of the meterstick provides no torque. The oppositely directed torques applied by each scale must be equal, because the meterstick is in equilibrium. Torque is force times distance from the fulcrum; since the righthand scale’s torque calculation includes a smaller distance from the fulcrum, the right-hand scale must apply more force in order to multiply to the same torque.
Part (b)
For this calculation, consider the left-hand scale as the fulcrum—that way, the left-hand scale provides no torque, and we only have to solve for one unknown variable. Set counterclockwise torques equal to clockwise torques, with \(T_{2}\) the reading in the right-hand scale.
The weight of the meterstick provides the clockwise torque; the right-hand scale provides the counterclockwise torque. \(**(T_{2})(50cm)=(1.5N)(30cm)\)
Solve for \(T_{2}\) to get \(T_{2}\) = 0.9 N
Next, the sum of the scale readings has to be the 1.5 N weight of the meterstick: 0. 9 N + \(T_{1}\) = 1.5 N
Giving \(T_{1}\) = 0.6 N
Part (c)
(i) Look at the starred calculation in Part (b). By moving the right-hand scale closer to the center, the scale will be less than 50 cm from the left-hand scale; but the meterstick’s center will still be 30 cm from the fulcrum. So when we solve for \(T_{2}\), we’re dividing (1.5 N)(30 cm) by a smaller value, giving a bigger \(T_{2}\) reading. But the question asks for the reading in the left-hand scale, which adds to \(T_{2}\) to the same 1.5 N. A bigger \(T_{2}\) adds to a smaller \(T_{1}\) to get 1.5 N. Answer: decrease.
(ii) See Part (i): \(T_{2}\), the reading in the right-hand scale, will increase.
Part (d)
Again, start from the equilibrium of torques using the left-hand scale as the fulcrum: \((T_{2})\)(50 cm) = (1.5 N)(30 cm)
Hanging a 0.2-N weight would provide a clockwise torque that would add to the torque applied by the meterstick’s weight on the right of this equation. Algebraically, \(T_{2}\) is increased by adding to the numerator of the right side of this equation. We want to add the biggest possible torque. Torque is force times distance from the fulcrum. We want, then, the largest possible distance from the fulcrum, which would be the right-hand edge of the meterstick, 80 cm from the left-hand scale.
Question
An elevator ride consists of the following portions.
a. Initially, the elevator is on the ground floor and quickly speeds up in the upward direction on its way to the top floor.
b. Most of the trip upward is at a constant velocity. c. Suddenly, near the top floor, the elevator comes to a stop just in time for the top floor.
d. Now the button for the first floor is pressed, and the elevator speeds downward suddenly.
e. Most of the trip downward is spent at constant velocity.
f. Suddenly, at the very bottom, the elevator comes to a stop.
(a) Draw a qualitative free-body diagram (showing the relative vector sizes) for each portion of the elevator ride as experienced by a passenger. Next to each, describe how the person would feel (normal, heavier than usual, lighter than usual, etc.).
(b) Explain the difference among the passenger’s mass, gravitational weight, and apparent weight throughout the trip. When (if ever) are these the same, and when (if ever) are they different? Which quantities change during the trip, and which ones do not?
(c) Describe what would happen if, on the way up (during step (b)), the cables were to break. Ignore any frictional effect between the elevator itself and the elevator shaft. Be sure to include a discussion of the relative motion of the elevator car and the passenger. Your answer should include a discussion or sketch of the height above ground, the velocity, and the acceleration experienced by the passenger from the time of the break until the end of the motion.
(d) Sketch the expected behavior of the person’s height as a function of time for situation (c) above. Also sketch the person’s \(V_{y}\) as a function of time as well.
▶️Answer/Explanation
Ans:
(a) 
(b) A person’s mass is a measure of his or her inertia. This value does not change due to acceleration or changes in location. Gravitational weight is a force due to the interaction of the person’s mass and the planet on which he or she is standing (including the distance between their centers). Although technically this value is slightly smaller as you get higher above sea level, the differences within a building on Earth are negligible. Apparent weight is the contact forces your body experiences, which give you your subjective experience of “weight.” In this case, the normal force and the changing values of the normal force explain the changes the person would experience on the elevator ride.
(c) The elevator and passenger would experience free fall. The only force would be the downward mg , and the normal force would be zero. Hence, the person would feel weightless. Since the car was on the way up when the cables broke, both passenger and elevator would maintain the same relative velocity to each other as both continued upward, slowed down, and then reversed direction and continued to speed up while falling. The entire time, the passenger would feel weightless.
(d) 