One string, one shared pull

Mathematics · Mechanics · ages 17-18

Name ______________________ Date ____________

A 6 kg hanging block has only weight and upward tension T acting on it. Take downward as positive and call its signed acceleration a. With g = 10 N/kg, which equation describes the block?
- T - 60 = 6a
- 60 + T = 6a
- 60 - T = 6a
Hanging masses of 7 kg and 3 kg are joined by a taut, light, inextensible string over a fixed smooth pulley. Ignore air resistance and use g = 10 N/kg. Find their acceleration magnitude in m/s squared.
Answer: ______________
Zoe says tension has the same magnitude on both sides of a fixed smooth pulley when the taut string is light. Is Zoe right in this ideal model?
Circle one: True False
Two masses hang from a taut, light string over a fixed smooth pulley. Ignore air resistance. Which forces act on either hanging mass?
- Its weight and downward tension from the string.
- Its weight and upward tension from the string.
- Its weight and the other mass's weight directly.
- Its weight and upward contact force from the pulley.
A 3 kg mass on a smooth horizontal table connects to a hanging 2 kg mass by a taut, light, inextensible string over a fixed smooth pulley. The table section of string is horizontal. Ignore air resistance and use g = 10 N/kg. Find the tension in N.
Answer: ______________
A hanging mass M and a 2 kg mass use a taut, light, inextensible string over a fixed smooth pulley. Ignore air resistance. With g = 10 N/kg, M accelerates down at 2 m/s squared. Find M in kg.
Answer: ______________
Equal 3 kg masses hang from a taut, light, inextensible string over a fixed smooth pulley. Ignore air resistance and use g = 10 N/kg. Sam says their acceleration is zero even if they are already moving. Is Sam right?
Circle one: True False
In the ideal pulley model with two unequal hanging masses, equal tension magnitudes mean the net force on each mass is zero. Is this true?
Circle one: True False
Hanging masses of 5 kg and 3 kg are joined by a taut, light, inextensible string over a fixed smooth pulley. Ignore air resistance and use g = 10 N/kg. Find their acceleration magnitude in m/s squared.
Answer: ______________
Hanging masses of 4 kg and 1 kg use a taut, light, inextensible string over a fixed smooth pulley. Ignore air resistance; g = 10 N/kg. Both positive axes point down. The 4 kg mass has acceleration a > 0. Which pair has consistent signs?
- 40 + T = 4a; 10 + T = -a
- T - 40 = 4a; T - 10 = -a
- 40 - T = -4a; 10 - T = a
- 40 - T = 4a; 10 - T = -a

Answer key

For grown-ups. Fold this page away before handing over the rest.

One string, one shared pull W1-mt_pZ6zV0RFdC-s1

60 - T = 6a · Weight is 60 N downward and tension acts upward. With downward positive, net force is 60 - T, so 60 - T = 6a.
4 · Adding 70 - T = 7a and T - 30 = 3a gives 40 = 10a. Therefore a = 4 m/s squared.
True · The string has negligible mass, and the pulley adds no friction along it. Tension therefore has the same magnitude on both sides.
Its weight and upward tension from the string. · Each hanging mass has its own downward weight and upward string tension. The mass does not touch the pulley.
12 · The equations are T = 3a and 20 - T = 2a. Adding gives 20 = 5a, so a = 4 m/s squared and T = 12 N.
3 · The 2 kg mass accelerates upward: T - 20 = 2(2), so T = 24 N. Then 10M - 24 = 2M gives M = 3 kg.
True · Equal weights give zero acceleration. The masses may be at rest or moving at constant speed while the string stays taut.
False · The same tension magnitude acts upward at each mass, but their weights differ. Uniform tension does not mean each mass's forces balance.
2.5 · Adding 50 - T = 5a and T - 30 = 3a gives 20 = 8a. Thus a = 2.5 m/s squared.
40 - T = 4a; 10 - T = -a · Both weights are positive and both tensions negative on downward axes. The 1 kg mass accelerates upward, so its signed acceleration is -a.