Torque — Formula and Examples

Torque is the turning effect of a force about a pivot or axis. In introductory physics, the torque magnitude from one force is

$$\tau = rF\sin\theta$$

Here, $r$ is the distance from the pivot to where the force is applied, $F$ is the force magnitude, and $\theta$ is the angle between the radius line and the force. The key idea is that only the perpendicular part of the force creates rotation. If the force points directly toward or away from the pivot, then $\theta = 0$ and the torque is $0$.

What Torque Means In Plain Language

Torque is the rotational version of push strength. A larger torque means a stronger tendency to make an object turn.

Torque gets larger when:

1. a larger force
2. a larger distance from the pivot
3. a force applied more nearly perpendicular to the radius

That is why a door opens more easily when you push near the handle than near the hinge. The same force produces more turning effect when the lever arm is longer.

Torque Formula: What Each Part Does

You can read

$$\tau = rF\sin\theta$$

as three separate ideas:

• $r$: how far from the pivot the force is applied
• $F$: how large the force is
• $\sin\theta$: how much of that force is perpendicular to the radius

Another useful form is

$$\tau = rF_{\perp}$$

where $F_{\perp}$ is the component of the force perpendicular to the radius. This is often the fastest way to reason about a problem.

In SI units, torque is measured in newton-meters, written as $\mathrm{N \cdot m}$. That has the same dimensions as energy, but it is not the same physical quantity. Torque describes turning effect, not stored or transferred energy.

Worked Example: Torque On A Door

Suppose you push on a door with a force of $25\ \mathrm{N}$ at a point $0.80\ \mathrm{m}$ from the hinge. The hinge is the pivot.

If you push perpendicular to the door, then $\theta = 90^\circ$ and $\sin 90^\circ = 1$. The torque magnitude is

$$\tau = rF\sin\theta = (0.80)(25)(1) = 20\ \mathrm{N \cdot m}$$

So the door experiences $20\ \mathrm{N \cdot m}$ of torque.

Now keep the same force and the same distance, but push at an angle of $30^\circ$ to the radius. Then

$$\tau = (0.80)(25)\sin 30^\circ = (0.80)(25)(0.5) = 10\ \mathrm{N \cdot m}$$

The force is unchanged, but the torque is smaller because less of the force is perpendicular. This is the main idea many students miss: the full force does not always contribute to turning.

When Torque Is Zero

Torque is zero in either of these cases:

1. the force is applied at the pivot, so $r = 0$
2. the force acts along the radius, so $\theta = 0$ or $180^\circ$

Both cases give no lever arm for rotation, even if the force itself is large.

Clockwise Vs. Counterclockwise Torque

In many introductory problems, torque is given a sign based on rotation direction. A common convention is:

1. counterclockwise torque is positive
2. clockwise torque is negative

This sign choice is a convention, not a separate physical law. Use the convention your course or problem specifies, but keep it consistent.

Common Torque Formula Mistakes

Using $F$ Instead Of The Perpendicular Component

If the force is angled, you usually cannot use just $rF$. You need the perpendicular part, which is why the $\sin\theta$ factor matters.

Measuring Distance From The Wrong Point

The distance must be measured from the pivot or axis. If the pivot is the hinge of a door, measure from the hinge to the point where the force is applied.

Forgetting That A Force Through The Pivot Gives Zero Torque

If the line of action passes through the pivot, the lever arm is zero, so the torque is zero even if the force itself is large.

Mixing Up Torque And Force

Force can cause translation, while torque causes rotation. A large force does not guarantee a large torque if it is applied very close to the pivot or along the radius.

Where Torque Is Used

Torque appears whenever rotation matters. Common cases include:

1. opening doors
2. using wrenches and screwdrivers
3. balancing seesaws and beams
4. analyzing motors, wheels, and pulleys
5. solving rotational dynamics and static equilibrium problems

In static equilibrium, the net torque about a chosen pivot must be zero. In rotational dynamics, net torque is what changes rotational motion.

Try A Similar Problem

A wrench is $0.30\ \mathrm{m}$ long, and you apply a force of $40\ \mathrm{N}$ perpendicular to it. Compute the torque, then compare it with the torque from the same force applied at $45^\circ$. That quick comparison makes the role of angle much easier to see.