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Understanding
Torque

Master the rotational equivalent of force: learn how forces cause objects to spin, and how lever arm, force, and angle determine the twisting effect.

AP Physics 1

What Is Torque?

Torque is the rotational analog of force. While a net force causes linear acceleration, a net torque causes angular acceleration. Every time you open a door, turn a wrench, or spin a merry-go-round, you are applying torque.

The Three Ingredients

Force \( F \)
The magnitude of the applied force: a bigger push means more torque.
Lever Arm \( r \)
The distance from the pivot to the point where the force is applied.
Angle \( \theta \)
The angle between the force vector and the lever arm: maximum torque occurs at \( 90^\circ \).

Visualizing Torque

Pivot
\( r \)
\( F \)
\( \theta \)
 rotation

The Torque Equation

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

\(\tau\)
Torque (Greek letter tau): measured in newton-meters (N·m).
\(r\)
Lever arm: the straight-line distance from the axis of rotation to the point of force application (in meters).
\(F\)
Applied force: the magnitude of the force (in newtons).
\(\theta\)
Angle between the force vector \( \vec{F} \) and the position vector \( \vec{r} \).

Vector Form: Cross Product

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

Torque is fundamentally a vector quantity. Its direction is determined by the right-hand rule: curl the fingers of your right hand from \( \vec{r} \) toward \( \vec{F} \), and your thumb points in the direction of \( \vec{\tau} \).

Magnitude
\( |\vec{\tau}| = rF\sin\theta \)
Direction
Perpendicular to the plane of \( \vec{r} \) and \( \vec{F} \) (right-hand rule).

Torque vs. Force

Force Torque
Symbol \( \vec{F} \) \( \vec{\tau} \)
Units Newtons (N) Newton-meters (N·m)
Effect Linear acceleration Angular acceleration
Newton's 2nd Law \( \vec{F}_{\text{net}} = m\vec{a} \) \( \vec{\tau}_{\text{net}} = I\vec{\alpha} \)
Depends on pivot? No Yes

Sign Convention for Torque

\(+\) (CCW) Counterclockwise = Positive \(-\) (CW) Clockwise = Negative

How to Calculate Torque

1
Identify the pivot (axis of rotation). All distances are measured from here.
2
Draw the lever arm \( r \): the straight line from the pivot to the point of force application.
3
Identify the angle \( \theta \) between \( \vec{r} \) and \( \vec{F} \). If the force is perpendicular, \( \theta = 90^\circ \) and \( \sin\theta = 1 \).
4
Compute \( \tau = rF\sin\theta \).
5
Assign the sign: positive for CCW, negative for CW.

Example 1: Perpendicular Force

Pivot
\( r = 0.5 \text{ m} \)
\( F = 20 \text{ N} \)
\( 90^\circ \)

Example 2: Force at an Angle

Pivot
\( r = 2.0 \text{ m} \)
\( F = 50 \text{ N} \)
\( 30^\circ \)

Example 3: Rotational Equilibrium

Fulcrum
\( W_1 = 30 \text{ N} \)
\( r_1 = 1.5 \text{ m} \)
\( W_2 = ? \)
\( r_2 = 2.0 \text{ m} \)

Illustrating the relationship

This separate introductory illustration uses the stated force and distance values.

The same 20 N perpendicular force at lever arms 0.20 m and 0.80 m produces 4 N·m and 16 N·m.
The same 20 N perpendicular force at lever arms 0.20 m and 0.80 m produces 4 N·m and 16 N·m.

A separate worked example

The following source example uses its own apparatus and numbers; its companion figure appears with the solution.

A uniform horizontal beam of length L = 4.0 m and mass M = 20 kg is attached to a wall by a hinge at its left end. A cable attached to the right end of the beam makes an angle of 30° above the horizontal and holds the beam in static equilibrium. A block of mass m = 10 kg hangs from the beam at a point 3.0 m from the hinge. Find the tension T in the cable.

Show answer and explanation

Step 2 — Identify Forces and Pivot: Three forces produce torque about the hinge: (1) the weight of the beam, Mg, acting downward at the center of mass (L/2 = 2.0 m from the hinge); (2) the weight of the block, mg, acting downward at 3.0 m from the hinge; (3) the tension T in the cable, acting at 30° above horizontal at the end of the beam (4.0 m from the hinge). The hinge force is eliminated by choosing the hinge as the pivot, since its lever arm is zero. Step 3 — Write the Torque Equation: Setting counterclockwise as positive and applying Στ = 0 about the hinge: T × L × sin 30° − Mg × (L/2) − mg × (3.0 m) = 0 Step 4 — Substitute Known Values: Inserting the given values with g = 9.8 m/s²: T × 4.0 × sin 30° − (20)(9.8)(2.0) − (10)(9.8)(3.0) = 0 → T × 2.0 − 392 − 294 = 0 Step 5 — Solve for T: Combining and solving for T: 2.0T = 686 → T = 343 N ≈ 340 N Step 6 — Check Reasonableness: The total weight supported is (20 + 10)(9.8) = 294 N. The tension (343 N) exceeds this because the cable acts at only 30°—its vertical component is T sin 30° = 171.5 N, meaning the hinge must supply the remaining vertical support. The cable tension must be large to compensate for the small angle, which is physically consistent.

AP Physics 1 • beam 20 kg • load 10 kg; illustrative apparatus not to scale.
AP Physics 1 • beam 20 kg • load 10 kg; illustrative apparatus not to scale.

Muscle force is amplified by a short anatomical moment arm

A horizontal forearm holding a dumbbell accompanies an exact lever model with0.050m muscle arm,0.16m forearm center of mass and0.35m hand load.A short muscle moment arm requires a large muscle force to balance the hand load. In the vertical-pull model, both the torque balance about the elbow and the vertical force balance must hold.

Forearm torque and force balance

Forearm torque and force balance. A short muscle moment arm requires a large muscle force to balance the hand load. In the vertical-pull model, both the torque balance about the elbow and the vertical force balance must hold.A short muscle moment arm requires a large muscle force to balance the hand load. In the vertical-pull model, both the torque balance about the elbow and the vertical force balance must hold.

Newton's 2nd Law for Rotation

$$ \sum \tau = I\alpha $$

\( \sum \tau \)
Net torque on the object about the chosen axis (N·m).
\( I \)
Moment of inertia: the rotational analog of mass (\( \mathrm{kg\,m}^2 \)).
\( \alpha \)
Angular acceleration (\( \text{rad/s}^2 \)): the rate of change of angular velocity.

The Moment Arm Shortcut

Instead of computing \( rF\sin\theta \), you can find the perpendicular distance from the pivot to the line of action of the force. This perpendicular distance is the moment arm \( r_\perp \).

Pivot
\( F \)
\( r_\perp \)
\( \theta \)

Common Mistakes

✕ Using cos instead of sin
The formula uses \( \sin\theta \), not \( \cos\theta \). Maximum torque is at \( 90^\circ \), not \( 0^\circ \).
✕ Wrong pivot distance
Measure \( r \) from the axis of rotation, not from the center of mass or some other point.
✕ Forgetting the sign
Torque has a direction. Always assign \( + \) for CCW and \( - \) for CW, then sum correctly.

Summary of Key Relationships

Concept Linear Rotational
Cause of acceleration Force \( F \) Torque \( \tau \)
Inertia Mass \( m \) Moment of inertia \( I \)
Acceleration \( a \) (\(\text{m/s}^2\)) \( \alpha \) (\(\text{rad/s}^2\))
Newton's 2nd Law \( F_{\text{net}} = ma \) \( \tau_{\text{net}} = I\alpha \)
Equilibrium \( \sum F = 0 \) \( \sum \tau = 0 \)
Units \(\text{N}\) \(\text{N} \cdot \text{m}\)

Question 1 of 5

A force of 40 N is applied perpendicularly at the end of a 0.3 m wrench. What is the magnitude of the torque?

Question 2 of 5

A 100 N force is applied at an angle of 30° to a 2.0 m lever arm. What is the torque?

Question 3 of 5

A force is applied directly through the pivot point of a rotating beam. What is the torque produced?

Question 4 of 5

A seesaw is balanced with a 60 N child sitting 1.5 m from the fulcrum. A second child sits 2.0 m from the fulcrum on the other side. What is the weight of the second child?

Question 5 of 5

A net torque of 24 N·m acts on a wheel with moment of inertia \( I = 4.0 \) kg·m². What is the angular acceleration?