Torque: Force × Lever Arm

Torque is the rotational analog of force. It measures how effectively a force can cause rotation about a pivot. Two things determine the torque:

1. The magnitude of the force.
2. The lever arm — the perpendicular distance from the pivot to the line of action of the force.

The formula:

$\tau = r\,F\,\sin\theta$

where $r$ is the distance from the pivot to where the force is applied, $F$ is the force magnitude, and $\theta$ is the angle between $\vec r$ and $\vec F$. This is also the magnitude of the cross product $\vec\tau = \vec r \times \vec F$.

Step-by-Step Solution

Given: A wrench is being used to loosen a stuck bolt. Force $F = 100\;\text{N}$ is applied at a distance $r = 0.4\;\text{m}$ from the bolt, at angle $\theta$ from the wrench handle.

Find: The torque on the bolt for $\theta = 90°$ (perpendicular) and $\theta = 30°$.

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Step 1 — Torque at $\theta = 90°$ (perpendicular force, the most efficient case).

$\tau = r\,F\,\sin 90° = (0.4)(100)(1) = 40.0\;\text{N}\cdot\text{m}$

This is the maximum torque you can get from this force at this distance. All of the force "tries to spin the bolt" — none is wasted pulling along the wrench's length.

Step 2 — Torque at $\theta = 30°$.

$\tau = r\,F\,\sin 30° = (0.4)(100)(0.5) = 20.0\;\text{N}\cdot\text{m}$

Half the maximum, even though we're applying the same force at the same distance. The reason: only the perpendicular component $F\sin\theta = 50\;\text{N}$ contributes to spinning the bolt; the parallel component $F\cos\theta \approx 86.6\;\text{N}$ is pulling along the wrench and producing no torque.

Step 3 — How to maximize torque for a given force.

Two strategies:

1. Push perpendicular to the lever (use $\theta = 90°$).
2. Use a longer wrench (increase $r$). Doubling $r$ doubles the torque, with no increase in force required.

This is why mechanics use long breaker bars to crack stuck bolts — a 1-meter bar can apply 4× the torque of a 25-cm wrench with the same effort.

Step 4 — Equilibrium of torques (balance).

A see-saw is in balance when the net torque about the pivot is zero. If a child of weight $W_1$ sits at distance $r_1$ on one side and a child of weight $W_2$ at distance $r_2$ on the other:

$W_1 r_1 = W_2 r_2$

A heavier child sits closer to the pivot, and a lighter child sits farther away. This is also why you can balance a long stick on your finger if the stick's center of mass is directly above the contact point — the torque about that point is zero.

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Answer: Torque depends on the perpendicular component of force times the lever arm:

$\boxed{\;\tau = r\,F\,\sin\theta\;}$

For $r = 0.4\;\text{m}$, $F = 100\;\text{N}$:

• $\theta = 90°$ → $\tau = 40\;\text{N}\cdot\text{m}$ (maximum)
• $\theta = 30°$ → $\tau = 20\;\text{N}\cdot\text{m}$
• $\theta = 0°$ (force along wrench) → $\tau = 0\;\text{N}\cdot\text{m}$ (no rotation possible)

To maximize torque: use a perpendicular force on a long lever arm.

Try It

• Adjust the distance $r$, the force $F$, and the angle $\theta$.
• Watch the wrench update with the force vector and the resulting torque magnitude.
• Try $\theta = 0°$ — the force is along the lever and the torque drops to zero.
• The HUD shows both the perpendicular component of the force and the resulting torque.