1. A bucket of water (total mass = 20.0 kg) hangs straight down from a rope. (As

Question

1. A bucket of water (total mass = 20.0 kg) hangs straight down from a rope. (Assume that the rope has negligible mass.)

a. (1 pt.) Use the bucket shown (at right) to construct a freebody diagram of the bucket. (Specifically, draw ALL force vectors acting on the bucket and LABEL their names.)

b. (3 pts.) While the bucket is being raised at a constant speed of 1.00 m/s, what is the magnitude of the tension force in the rope? Show your work completely.

c. (4 pts.) Suppose the bucket starts with an upward speed of 0.500 m/s. It undergoes a constant acceleration so that the bucket its speed is 2.20 m/s at a vertical displacement of 1.40 m above its starting height. What is the magnitude of the tension force in the rope during this acceleration? Show your work completely.

d. (2 pt.) Much later… Suppose that, at a particular moment, the magnitude of the tension in the rope is 175 N. What are the magnitude and direction of the bucket’s acceleration at that moment? Show your work completely.

Two boxes of mass mA and mB are stacked on a table, as shown at right.

a. (2 pts.) For now, the entire configuration is at rest. Use the figures of box A and box B provided BELOW to construct a free-body diagram for each box. Namely, draw ALL force vectors acting on each box, and LABEL all of their names:

(4 pts. total; –1 pt. per error) You do NOT need to show your work for the following fill-in-the-blank questions. Express all of your final answers… • ONLY in terms of mA, mB, g, and (when appropriate) v1 or a1 • in SIMPLEST algebraic form While all boxes are sitting continuously at rest…

b. What is the magnitude of the net (total) force acting on box B? ________________

c. What is the magnitude of the table’s normal force? ________________ While the table is in an elevator descending at constant speed v1… (Note: v1 has a positive value.)

d. What is the magnitude of the net (total) force acting on box B? ________________

e. What is the magnitude of the table’s normal force? ________________ While the table is in an elevator accelerating upward at constant acceleration a1…

f. What is the magnitude of the net (total) force acting on box B? ________________

g. What is the magnitude of the table’s normal force? ________________ The elevator cable snaps, and the elevator (and all of its contents) fall freely. In this case…

h. What is the magnitude of the net (total) force acting on box B? ________________

i. What is the magnitude of the table’s normal force? ________________ mA mB table mB mA

4. You (mass = 50.0 kg) are standing on a “bathroom scale” while riding in an elevator. A friend is standing nearby in the elevator (but not standing on the scale), and in some of the cases below, the friend applies an upward or downward force on you. NOTE: A “bathroom scale” always reads the magnitude of the normal force of the floor on you. For the following questions that ask for a “scale reading,” please provide normal force, not mass. You do NOT need to show your work for the following fill-in-the-blank questions. Be sure to provide proper units and significant figures!

a. (1 pt.) The elevator is sitting continuously at rest, and your friend is not touching you. What is the reading on the scale? __________________

b. (1 pt.) While the elevator is sitting continuously at rest, your nearby friend pushes vertically downward on your shoulders with 100. N of force. What is the reading on the scale? __________________

c. (1 pt.) While the elevator is descending at constant speed of 1.0 m/s, your nearby friend pulls vertically upward on your body with 100. N of force. What is the reading on the scale? __________________

d. (1 pt.) While the elevator is accelerating downward at constant acceleration of 1.0 m/s2 , and your friend is not touching you, what is the reading on the scale? __________________

e. (2 pts.) While the elevator is accelerating upward at constant acceleration of 1.0 m/s2 , your nearby friend pushes vertically downward on your shoulders with 100. N of force. What is the reading on the scale? __________________

f. (2 pts.) At a time when the scale reading is 425 N, and your friend is not touching you, what are the magnitude and direction of the elevator’s acceleration? __________________

BONUS: (+1 pt.) The elevator cable snaps, and the elevator (and all of its contents) fall freely. If your friend is not touching you during this time, what is the reading on the scale? _________________ (Thought question [NOT for credit]: How is it possible that some scales provide readouts in kilograms (mass) instead of force (newtons or pounds)? What calculation or conversion has the scale manufacturer performed, and what assumptions have they made?)

Explanation / Answer

1) a) The forces acting on the bucket are:
* tensile force, pulling upward
* force of gravity, pulling downward

b) As there is no acceleration, the tensile force in the rope is the same as the force of gravity on the bucket.
T = m×g =  (20 kg) × (9.8 m/s²) = 196 N

c) First find the acceleration of the bucket.
v² = u² + 2×a×s
(2.2 m/s)² = (0.5m/s)² + 2×a×(1.4 m)
a = 1.64 m/s²

The forces acting on the bucket are:
* tensile force, pulling upward
* force of gravity, pulling downward
As the bucket accelerates upward, you know that the tensile force must be stronger than the force of gravity, and the net force on the bucket is:
Fn = T - m×g
By Newton's second law, you know that
(net force on object) = (mass of object) × (acceleration of object)

so
T - m×g = m×a
T = m×a + m×g
T = m×(a + g)
T= (20 kg)×((1.64 m/s²) + (9.8 m/s²))  = 228.8 N

d) The forces acting on the bucket are:
* tensile force, pulling upward with 175 N
* force of gravity, pulling downward with 196 N
As the force of gravity is stronger than the tensile force, you know that the bucket will accelerate downward, and the net force on the bucket is:
Fn = (196 N) - (175 N) = 21 N
Again, by Newton's second law, you can find the acceleration.
(21 N) = (20 kg) × a
a = 1.05 m/s² (downward)

4) mass = 50 kg, g = 9.8 m/s^2

a) the reading on the scale = 50 X 9.8 = 490N
b) the reading on the scale = 50 X 9.8 +100 = 590N  

c) the reading on the scale = 50 X (9.8 - 1) - 100 = 50 X 8.8 -100 = 340N

d)  the reading on the scale = 50 X (9.8 - 1) = 440N

e)  the reading on the scale = 50 X (9.8 + 1) + 100 = 640N

f) Force = 425N
50 X (9.8 - a) = 425
Implies = 490 - 50a = 425
a = 1.3 m/s^2 the elevator is moving downwards.

Bonus: When elevator is in fall free mode then effective acceleration = 0 and hence reading on the scale = 0N

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