5.3 - Torque

Introduction: The Rotational Equivalent of Force

Torque is the rotational version of force. Just as a force causes linear acceleration, a torque causes angular acceleration.

When a force acts on a rigid system at some distance from the axis of rotation, and it isn’t aimed straight toward or away from the axis, it creates a torque.

The key idea: torque measures the “twisting effect” of a force, meaning its ability to cause rotation.

Lever Arm (Moment Arm)

The lever arm, also called the moment arm, is the perpendicular distance from the axis of rotation to the line of action of the force.

Symbol:

r_\perp

(sometimes written as

d

). Units: meters (m).

A way to picture it: imagine using the force vector as a compass to draw a circle. The axis of rotation is where you plant the compass point, and the lever arm is the radius of that circle.

Torque Diagrams (Rigid Body Diagrams)

Unlike free body diagrams, where the forces are drawn at the center of mass, torque diagrams put the forces at their actual points of application relative to the axis of rotation.

Why? Because where the force is applied matters. The same force creates different torques depending on where it acts.

These are sometimes called rigid body diagrams, because they show the extended body rather than just a point mass.

On the AP exam, these are called force diagrams. A force diagram works like a free body diagram: each force gets an arrow showing its direction and its size compared with the other forces (a force twice as big gets an arrow about twice as long). What’s new is that each arrow starts where that force actually acts, so the diagram shows the location of every force relative to the axis of rotation. You’ll often draw both: a free body diagram to analyze the linear motion and a force diagram to analyze the torques.

The Mathematical Definition of Torque

Torque is defined with the cross product of the position vector and the force vector:

\vec{\tau} = \vec{r} \times \vec{F}

where: -

\vec{\tau}

= torque (the Greek letter tau), in N·m -

\vec{r}

= position vector from the axis to the point where the force is applied (m) -

\vec{F}

= force vector (N) -

\times

= the cross product

The cross product multiplies two vectors and gives you a vector result.

In general, the cross product of any two vectors

\vec{A}

and

\vec{B}

is a vector with magnitude

|\vec{A} \times \vec{B}| = AB\sin\theta

where

\theta

is the angle between the two vectors when they’re drawn tail to tail. - Direction: perpendicular to both

\vec{A}

and

\vec{B}

, so it’s normal to the plane the two vectors lie in. The right-hand rule (below) tells you which of the two perpendicular directions it is. - Order matters:

\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})

, the same size but the opposite direction. Always write

\vec{r} \times \vec{F}

, with

\vec{r}

first. - Parallel vectors: if

\theta = 0^\circ

180^\circ

, the cross product is zero.

Magnitude of Torque

The magnitude of the torque is:

\tau = rF\sin\theta

where: -

\tau

= magnitude of the torque (N·m) -

r

= distance from the axis to the point where the force is applied (m) -

F

= magnitude of the force (N) -

\theta

= angle between

\vec{r}

and

\vec{F}

Alternative forms:

\tau = r_\perp F = rF_\perp

where: -

r_\perp = r\sin\theta

is the lever arm, the perpendicular distance from the axis to the force’s line of action. -

F_\perp = F\sin\theta

is the component of the force perpendicular to

\vec{r}

Maximum and Minimum Torque

Maximum torque happens when the force is perpendicular to the position vector

\vec{r}

(

\theta = 90°

\tau_{max} = rF\sin(90°) = rF

Minimum (zero) torque happens when the force is parallel to the position vector

\vec{r}

(

\theta = 0°

180°

\tau_{min} = rF\sin(0°) = 0

The read: pushing directly toward or away from the axis creates no rotation at all, just compression or tension.

Why Doorknobs Are Where They Are

A practical application: doorknobs sit as far as possible from the hinges (the axis of rotation) because of

\tau = rF\sin\theta

For the same force

F

: - A larger

r

(far from the hinge) gives a larger torque, so the door is easier to open. - A smaller

r

(near the hinge) gives a smaller torque, so the door is harder to open.

The design principle: to get more torque out of less force, maximize the lever arm.

More examples: - Wrenches have long handles to boost torque. - Wheelbarrows put the handles far from the wheel. - Seesaws work best with the riders far from the pivot.

Direction of Torque (The Right-Hand Rule)

Since torque comes from a cross product, it’s a vector with a direction.

The Right-Hand Curl Rule

The steps: 1. Point your fingers in the direction of

\vec{r}

, from the axis toward the force. 2. Curl your fingers toward the direction of

\vec{F}

. 3. Your thumb points in the direction of

\vec{\tau}

Because

\vec{\tau}

is perpendicular to both

\vec{r}

and

\vec{F}

, when those two vectors lie in the plane of the page, the torque points straight out of the page or straight into it: - Out of the page: the force tends to turn the system counterclockwise. - Into the page: the force tends to turn the system clockwise. That’s why, for rotation about one axis, you can treat torque as a single number with a sign, which is what the sign convention below does.

Sign Convention (Simplified)

For problems in a plane (2D rotation): - Counterclockwise rotation gives a positive torque (+). - Clockwise rotation gives a negative torque (−).

A memory aid: “righty tighty, lefty loosey.” Turning a screw clockwise (right) tightens it, and turning it counterclockwise (left) loosens it. The phrase doesn’t map perfectly onto our sign convention, but it helps you keep the rotation directions straight.

Example Problem: Drawbridge

Problem: It takes a torque of 5000 N·m to raise a drawbridge. The cable pulling the bridge makes an angle

\theta = 30°

with the bridge, and it’s attached 6.0 m from the pivot (the hinge).

What cable tension is needed to produce this torque?

Would it be easier to raise the drawbridge if the angle $\theta$ were larger or smaller?

Part (a): Finding the Tension

Step 1: Write the torque equation

\tau = rF\sin\theta

Here, the force making the torque is the tension

T

\tau = rT\sin\theta

Step 2: Solve for the tension

T = \frac{\tau}{r\sin\theta}

Step 3: Substitute the values

Given:

\tau = 5000

N·m,

r = 6.0

\theta = 30°

T = \frac{5000}{(6.0)\sin(30°)}

T = \frac{5000}{(6.0)(0.5)}

T = \frac{5000}{3.0}

T \approx 1667 \text{ N}

Answer: the tension has to be about 1700 N (1667 N to be precise).

Note: that’s the magnitude of the full tension force, not just a component. The

\sin\theta

term accounts for the fact that only the perpendicular part of the tension contributes to the torque.

Part (b): The Effect of Changing the Angle

The question: would it be easier (need less tension) if

\theta

were larger or smaller?

Look at the equation:

T = \frac{\tau}{r\sin\theta}

T

and

\theta

have an inverse relationship through

\sin\theta

: - As

\sin\theta

increases,

T

decreases, so it gets easier. - As

\sin\theta

decreases,

T

increases, so it gets harder.

Since

\sin\theta

grows as

\theta

goes from 0° to 90°, increasing

\theta

decreases the tension, which makes it easier to raise.

Answer: it would be easier if the angle

\theta

were larger.

Verification with Numbers

Try

\theta = 45°

T = \frac{5000}{(6.0)\sin(45°)}

T = \frac{5000}{(6.0)(0.707)}

T = \frac{5000}{4.24}

T \approx 1180 \text{ N}

Comparing: - At

\theta = 30°

T \approx 1667

N. - At

\theta = 45°

T \approx 1180

Sure enough, increasing

\theta

dropped the required tension. ✓

The read: when the cable pulls more perpendicular to the bridge (a larger

\theta

), more of the tension goes into producing rotation, so you need less total tension.

The optimal angle is

\theta = 90°

, with the cable perpendicular to the bridge, where

\sin(90°) = 1

and the tension is at its minimum.

Example Problem: Net Torque from a Force Diagram

Problem: A uniform 2.00 kg rod, 1.20 m long, turns on a frictionless pin through its center, O. The force diagram shows every force on the rod at the instant it’s horizontal:

F_1 = 15.0

N straight up at the right end,

F_2 = 40.0

N at the left end pointing up and to the left at 30.0° above the rod,

F_3 = 25.0

N at the right end pulling straight along the rod, the rod’s weight

W = 20.0

N (using

g = 10

m/s²), and the force from the pin. Find the torque from each force about O and the net torque. Take counterclockwise (out of the page) as positive.

Solution

Step 1: Forces at the axis. The weight acts at the center of mass, which is at O, and the pin pushes on the rod at O. Both have

r = 0

, so both make zero torque, no matter how big they are.

Step 2: The force along the rod.

F_3

pulls straight along the rod, so its line of action passes through O. The angle between

\vec{r}

and

\vec{F}_3

is 0°, so

\tau_3 = rF_3\sin 0^\circ = 0

Step 3: The perpendicular force.

F_1

is perpendicular to the rod, 0.600 m from O, and it tends to turn the rod counterclockwise:

\tau_1 = rF_1\sin 90^\circ = (0.600\text{ m})(15.0\text{ N})(1) = +9.00\text{ N}\cdot\text{m}

Step 4: The angled force.

F_2

acts 0.600 m from O at 30.0° to the rod. Pulling up on the left end, it tends to turn the rod clockwise:

\tau_2 = -rF_2\sin 30.0^\circ = -(0.600\text{ m})(40.0\text{ N})(0.500) = -12.0\text{ N}\cdot\text{m}

Check with the lever arm: extend

F_2

’s line of action. Its perpendicular distance from O is

r_\perp = (0.600\text{ m})\sin 30.0^\circ = 0.300

m, and

(0.300\text{ m})(40.0\text{ N}) = 12.0

N·m. Same answer.

Check with the cross product: put O at the origin with

+x

to the right and

+y

up. Then

\vec{r}_2 = (-0.600\text{ m})\hat{i}

and

\vec{F}_2 = (-34.6\hat{i} + 20.0\hat{j})

N. Since

\hat{i} \times \hat{i} = 0

and

\hat{i} \times \hat{j} = \hat{k}

(out of the page):

\vec{\tau}_2 = (-0.600\,\hat{i}) \times (-34.6\,\hat{i} + 20.0\,\hat{j}) = (-0.600)(20.0)\,\hat{k} = -12.0\,\hat{k}\text{ N}\cdot\text{m}

The

-\hat{k}

direction is into the page, which means clockwise.

Step 5: Add the torques.

\sum\tau = \tau_1 + \tau_2 + \tau_3 + \tau_W + \tau_{pin} = 9.00 - 12.0 + 0 + 0 + 0 = -3.0\text{ N}\cdot\text{m}

Answer: the net torque about O is 3.0 N·m clockwise (into the page).

The read: the biggest force,

F_2

, doesn’t make a torque of

(40.0)(0.600) = 24.0

N·m, because only its component perpendicular to the rod,

(40.0\text{ N})\sin 30.0^\circ = 20.0

N, twists the rod. And

F_3

, at 25.0 N, makes no torque at all. In 5.5/5.6 you’ll see that a net torque like this one gives the rod an angular acceleration.

Units of Torque

The SI unit is the newton-meter (N·m).

Dimensional analysis:

[\tau] = [r][F] = \text{m} \cdot \text{N} = \mathrm{N·m}

An important distinction: torque has units of N·m, and energy has units of joules (J), and mathematically 1 J = 1 N·m. But we don’t write torque in joules, because the two are conceptually different: - Torque is a twisting effect (N·m). - Energy is work done (J).

Always use N·m for torque, not J.

Summary of Key Formulas

Concept	Formula	Notes
Torque (vector)	$\vec{\tau} = \vec{r} \times \vec{F}$	The cross product definition
Torque (magnitude)	$\tau = rF\sin\theta$	The general formula
Maximum torque	$\tau_{max} = rF$	When $\theta = 90°$ (perpendicular)
Zero torque	$\tau = 0$	When $\theta = 0°$ or $180°$ (parallel)
Alternative form	$\tau = r_\perp F = rF_\perp$	Using perpendicular components

Common Mistakes

Not checking your calculator mode. When you compute $\sin\theta$ , make sure the calculator is in degree mode if $\theta$ is in degrees.

Confusing torque with energy. Torque has units of N·m but it is not energy. Don’t write torque in joules.

Measuring r incorrectly. $r$ is the distance from the axis of rotation to the point where the force is applied, not just any distance.

Assuming every force creates torque. Forces applied right at the axis ( $r = 0$ ) create zero torque. Forces parallel to $\vec{r}$ (pointing straight toward or away from the axis) also create zero torque.

Using the wrong angle. $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ , not necessarily the angle the force makes with the horizontal or vertical.

Forgetting that direction matters. Torque is a vector. Clockwise and counterclockwise torques have opposite signs, so account for direction when you add multiple torques.

Putting forces at the center of mass in torque problems. For torque, draw the forces at their actual points of application, not at the center of mass.

Thinking a larger angle always means more torque. Not true. Torque depends on $\sin\theta$ , which peaks at $\theta = 90°$ . Past 90°, $\sin\theta$ starts dropping.

Confusing the lever arm with the object’s length. The lever arm is the perpendicular distance from the axis to the force, which can be shorter than the physical distance along the object.

Missing the inverse relationship in part (b). In $T = \frac{\tau}{r\sin\theta}$ , increasing $\sin\theta$ (the denominator) decreases $T$ . A lot of students wrongly think a bigger angle means more tension.

Adding torques and forces together. Don’t mix them. Torques add with torques, and forces add with forces. They’re different quantities.

Getting the clockwise/counterclockwise sign wrong. Be consistent. Pick a convention (counterclockwise as positive is standard) and stick with it through the whole problem.

Forgetting the perpendicular component. Only the component of force perpendicular to $\vec{r}$ makes torque. The parallel component contributes nothing.

Using $\tau = rF$ when the force isn’t perpendicular. That simplified formula only works when $\theta = 90°$ . Otherwise use $\tau = rF\sin\theta$ .

Not understanding why doorknobs sit far from hinges. This is a classic conceptual question. A larger $r$ means more torque for the same applied force, which makes the door easier to open.

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