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MCAT Torque, Static Equilibrium and Lever Practice Questions

By StudyStore · Published 2026-11-04 · 8 original questions · about 12 min

Eight free MCAT-style physics practice questions on torque, lever arms, rotational equilibrium, balancing beams, mechanical advantage and support forces, with worked explanations.

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Quick concept primer

Torque measures the turning effect of a force: torque = r F sin(theta), where r is the distance from the pivot to the point of application and theta is the angle between the force and the lever arm. Only the component of force perpendicular to the lever arm produces torque, so a force pointed directly at or away from the pivot produces none. The SI unit is the newton-meter.

An object is in static equilibrium when the net force is zero and the net torque about any point is zero. Choose the pivot at an unknown force to eliminate it from the torque equation. A uniform beam's weight acts at its center. Levers trade force for distance: mechanical advantage = effort arm / load arm, and ideal work in equals work out.

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A 20 N force is applied perpendicular to a wrench handle at a distance of 0.50 m from the bolt. What torque does it produce?

  1. 10 N m
  2. 40 N m
  3. 20 N m
  4. 0.025 N m
Show answer and explanation

Answer: A. 10 N m

With the force perpendicular to the handle, torque = rF = 0.50 m x 20 N = 10 N m. Multiplying gives the right result, while dividing (as in 40 or 0.025) does not. If the force were not perpendicular, sin(theta) would reduce the torque.

Question 2

A 40 N force acts on a door handle 0.50 m from the hinge, at 30 degrees to the door surface (the lever arm). What is the torque? (sin 30 = 0.50)

  1. 20 N m
  2. 10 N m
  3. 17 N m
  4. 40 N m
Show answer and explanation

Answer: B. 10 N m

Torque = r F sin(theta) = 0.50 x 40 x 0.50 = 10 N m. Only the perpendicular component of the force, 20 N, contributes to turning the door. Using the full 40 N would overestimate the torque at 20 N m.

Question 3

A uniform seesaw is balanced at its center. A 40 kg child sits 2.0 m from the pivot. How far from the pivot must a 50 kg child sit on the other side to balance it?

  1. 1.6 m
  2. 2.5 m
  3. 2.0 m
  4. 1.0 m
Show answer and explanation

Answer: A. 1.6 m

Balance requires equal torques: 40 x g x 2.0 = 50 x g x d, so d = 80/50 = 1.6 m. The heavier child sits closer to the pivot, and g cancels from both sides.

Question 4

Which conditions are both required for an extended object to be in static equilibrium?

  1. Net force zero and net torque zero
  2. Net force zero only
  3. Net torque zero only
  4. Constant nonzero acceleration and zero torque
Show answer and explanation

Answer: A. Net force zero and net torque zero

Zero net force prevents translation and zero net torque prevents rotation. Two equal and opposite forces applied at different points can have zero net force but still spin an object, so the torque condition is needed separately.

Question 5

A uniform 6.0 m plank weighing 300 N is supported at each end. A 300 N load sits 2.0 m from the left end. What is the force exerted by the right support?

  1. 300 N
  2. 350 N
  3. 150 N
  4. 250 N
Show answer and explanation

Answer: D. 250 N

Take torques about the left end: R_right x 6.0 = 300 x 2.0 + 300 x 3.0 (plank weight at its center) = 600 + 900 = 1500, so R_right = 250 N. The total downward force is 600 N, so the left support provides 350 N.

Question 6

A lever has an effort arm of 1.5 m and a load arm of 0.50 m. Ignoring friction and the weight of the lever, what is the largest load that can be balanced by an effort of 30 N?

  1. 10 N
  2. 45 N
  3. 90 N
  4. 15 N
Show answer and explanation

Answer: C. 90 N

Balance torques: effort x 1.5 = load x 0.50, so load = 30 x 1.5 / 0.50 = 90 N. The mechanical advantage is 3. The effort must move three times as far as the load, so work is conserved.

Question 7

A force is applied to a bolt with a wrench in a direction pointing directly along the wrench handle toward the bolt. What torque does it produce?

  1. The maximum possible torque
  2. Zero torque
  3. A torque equal to the force times the handle length
  4. A torque equal to half the maximum
Show answer and explanation

Answer: B. Zero torque

When the force is parallel to the lever arm, theta is 0 degrees and sin(theta) = 0, so there is no torque. Torque is maximum when the force is perpendicular to the lever arm, which is why wrenches are pushed at right angles to the handle.

Question 8

A student uses a longer wrench to loosen a stuck bolt but applies the same perpendicular force. What happens to the torque?

  1. It decreases because the wrench is heavier
  2. It stays the same because force is unchanged
  3. It becomes zero
  4. It increases because the lever arm increases
Show answer and explanation

Answer: D. It increases because the lever arm increases

Torque is proportional to lever arm length when the force is constant and perpendicular. A longer wrench gives more torque for the same effort, which is why long handles and cheater pipes help with stuck bolts.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.

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