Torque — Worked Solutions

These solutions belong to Torque. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

A force acts at distance r from a pivot but points directly toward the pivot. What is its torque about that pivot?

1. Extend the force’s line of action toward the pivot. Why: Lever arm is perpendicular distance from the axis to that line.

2. The line passes through the pivot, so ℓ = 0 and τ = Fℓ = 0. Why: A nonzero distance to the application point does not guarantee a turning effect.

Back to this question.

Question 2

A 20 N force acts on a 0.40 m wrench at 30° to its position vector. What torque magnitude results?

1. Find the perpendicular component: F_perp = 20sin30° = 10 N. Why: Only the component perpendicular to the wrench produces torque.

2. τ = rF_perp = 0.40(10) = 4.0 N·m. Why: The angle is between the force and the wrench position vector.

Back to this question.

Question 3

Two forces have equal magnitudes. Which makes the larger torque about a given axis?

1. For the same force magnitude, τ = Fℓ. Why: Use the same reference axis for both comparisons.

2. A greater perpendicular lever arm gives greater torque. Why: The center of mass is not automatically the selected torque axis.

Back to this question.

Question 4

A light horizontal bar pivots at its center. A 6.0 N downward force acts 0.30 m to the left, and a 4.0 N downward force acts 0.20 m to the right. Determine the net torque using counterclockwise as positive.

1. Choose counterclockwise as positive and keep both force positions in the diagram. Why: Equal force directions can cause opposite turning senses on opposite sides.

2. The left contribution is +6.0(0.30) = +1.8 N·m. Why: A downward force left of the pivot tends counterclockwise.

3. The right contribution is −4.0(0.20) = −0.80 N·m. Why: A downward force right of the pivot tends clockwise.

4. Add to get τ_net = +1.0 N·m, tending counterclockwise. Why: Net torque is the signed sum about the stated pivot.

Back to this question.

Question 5

A meterstick is held about the same pivot in two trials. Trial A uses a 6.0 N pull applied 0.40 m from the pivot at 90° to the position vector. Trial B uses a 12 N pull applied 0.30 m from the pivot at 30° to that vector. Both pulls tend to turn the stick the same way. A student says B must make the larger torque because its force is larger. Evaluate the claim. Keeping B’s force and angle unchanged, choose a new application distance that would match A’s torque, and identify what distance must actually be measured.

1. Calculate τ_A = 0.40(6.0)sin90° = 2.4 N·m and τ_B = 0.30(12)sin30° = 1.8 N·m. Why: Torque depends on force and the perpendicular geometry, so force magnitude alone cannot rank the turning effects.

2. The claim is false: A has the larger torque. To match A, solve 2.4 = r(12)(0.5), giving r = 0.40 m. Why: Keeping force and angle fixed makes torque proportional to application distance from the pivot.

3. Measure from the pivot to the application point for r, and verify the angle between that position vector and the force. Alternatively measure the perpendicular lever arm, which must be 0.20 m for the matched B trial. Why: The lever arm is not generally the pivot-to-point distance; the two geometric methods must describe the same line of action.

Back to this question.