Torque Calculator: rF sin theta and Lever Arm

Torque equals rF sin theta: the distance from the pivot times the force times the sine of the angle between them. The calculator above returns the torque in newton meters plus the lever arm r sin theta. Counterclockwise torque is positive, and zero net torque means rotational equilibrium.

Distance from pivot r (m)Measured from the axis of rotation to where the force is applied.Force F (N)Angle theta (degrees)Angle between the arm and the force. 90 means pushing straight across the arm.

torque (tau)

10 N m

sin(90) = 1: a perpendicular push uses the full distance r, the most effective angle.

Steps

1.lever arm: r_perp = r sin(theta) = 0.4 m x sin(90) = 0.4 m
2.tau = r F sin(theta) = 0.4 m x 25 N x sin(90) = 10 N m
3.same result through the lever arm: tau = F r_perp = 25 N x 0.4 m = 10 N m

AP Physics: Unit 5 (topics 5.3 Torque, 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form). Torque is Topic 5.3 of AP Physics 1 Unit 5 (Torque and Rotational Dynamics), which carries 10-15% of the multiple-choice section. AP Physics C: Mechanics covers the same tau = rF sin theta relation in its own Unit 5.

What the calculator above computes

The calculator above computes torque from the AP equation sheet formula $\tau = rF\sin\theta$ : enter the distance $r$ from the pivot (the axis of rotation) to the point where the force is applied, the force magnitude $F$ , and the angle $\theta$ between the position vector and the force. It returns the torque $\tau$ in newton meters (N·m) and the lever arm $r_{\perp} = r\sin\theta$ in meters.

Torque measures how effectively a force twists an object about a pivot. A bigger force, a longer distance from the pivot, or an angle closer to $90^\circ$ all increase it. The number itself is only half the skill, though. For the full free-response method behind these numbers, work through the how to calculate torque guide.

Every term in tau = rF sin theta

Each symbol in $\tau = r_{\perp}F = rF\sin\theta$ comes straight from the AP Physics 1 equation sheet:

Symbol	Meaning	SI unit
$\tau$	torque about the chosen pivot	N·m
$r$	distance from the pivot to the point where the force acts	m
$F$	magnitude of the applied force	N
$\theta$	angle between the position vector and the force	degrees

The angle trips up more students than anything else. $\theta$ sits between the line from the pivot to the application point and the direction of the force. A force applied perpendicular to a wrench handle has $\theta = 90^\circ$ , so $\sin\theta = 1$ and the torque hits its maximum value of $rF$ . A force aimed straight at the pivot has $\theta = 0^\circ$ and produces no torque at all.

The lever arm: r sin theta

The lever arm (also called the moment arm) is the perpendicular distance from the pivot to the line of action of the force, and it equals $r_{\perp} = r\sin\theta$ . The calculator above reports it alongside the torque because the two forms of the equation are interchangeable: $\tau = rF\sin\theta$ and $\tau = r_{\perp}F$ always give the same number.

The lever arm view explains everyday intuition. Pushing a door at its handle gives the force a long lever arm, so a small push creates a large torque. Pushing near the hinges shrinks $r$ , and pushing along the door toward the hinge line shrinks $\sin\theta$ to zero. Either way the lever arm collapses and the door barely rotates. A quick diagram test: extend the line of the force. If that line passes through the pivot, the lever arm and the torque are both zero.

Torque sign: counterclockwise counts as positive

Torque has direction, and in AP Physics 1 you track it with a sign. The standard convention: torques that would spin the object counterclockwise are positive, and torques that would spin it clockwise are negative. The calculator above gives the magnitude; you assign the sign by asking which way the force would rotate the object about your chosen pivot.

The convention itself is arbitrary. You could call clockwise positive and every answer would still come out consistent, as long as you never switch mid-problem. On free-response questions, state your convention in one line (taking counterclockwise as positive, for example) before summing torques. Graders look for a consistent sign scheme, and writing it down protects you from the classic error of adding two torques that should subtract.

Rotational equilibrium: net torque equals zero

An object is in rotational equilibrium when the net torque about the pivot equals zero: $\Sigma\tau = 0$ . Combined with $\Sigma F = 0$ for translational equilibrium, this idea (Topic 5.5 in Unit 5) is the setup behind every balanced-beam, hanging-sign, and ladder problem.

The working recipe: pick a pivot, list every force, compute each torque with $rF\sin\theta$ , attach signs, and set the sum to zero. A smart pivot choice does half the work for you. Put the pivot at the point where an unknown force acts, and that force drops out of the torque equation because its $r$ is zero.

When the net torque is not zero, the object gains angular acceleration, and Unit 5 shifts from statics to motion. The angular equations that describe what happens next live in the rotational kinematics guide, where $\omega$ and $\alpha$ play the roles that velocity and acceleration play in linear motion.

Common mistakes with rF sin theta

Five errors show up constantly in Unit 5 work:

Using cosine instead of sine. Torque wants the component of the force perpendicular to $r$ . If a problem gives the angle between the force and the perpendicular direction, the angle you feed to sine is $90^\circ$ minus the given one.
Leaving lengths in centimeters. A 25 cm wrench is $r = 0.25$ m. Mixing cm and m throws the answer off by a factor of 100.
Forgetting where gravity acts. The weight of a uniform beam acts at its center of mass, so a 4.0 m uniform beam pivoted at one end feels gravity at $r = 2.0$ m.
Ignoring the pivot choice. Torque is always measured about a specific point. Change the pivot and every $r$ changes with it.
Dropping signs. Sum torques with signs attached, never as bare magnitudes.

The how to calculate torque guide walks through each of these with diagrams.

Torque on a wrench pulled at 60 degrees

You pull on the end of a 0.25 m wrench with a force of 80.0 N, at an angle of $60^\circ$ to the handle. Find the torque about the bolt and the lever arm.

List the givens: $r = 0.25$ m, $F = 80.0$ N, $\theta = 60^\circ$ , and $\sin 60^\circ = 0.866$ .
Find the lever arm: $r_{\perp} = r\sin\theta = (0.25\ \text{m})(0.866) = 0.217\ \text{m}$ .
Find the torque: $\tau = rF\sin\theta = (0.25\ \text{m})(80.0\ \text{N})(0.866) = 17.3\ \text{N·m}$ .
Check with the lever arm form: $\tau = r_{\perp}F = (0.217\ \text{m})(80.0\ \text{N}) = 17.36\ \text{N·m}$ , which rounds to $17.3\ \text{N·m}$ . The small difference comes from rounding the lever arm.

The torque is $17.3$ N·m and the lever arm is $0.217$ m.

Balancing a seesaw (rotational equilibrium)

A 30.0 kg child sits 2.0 m to the left of a seesaw pivot. How far from the pivot must a 40.0 kg child sit on the right side for the seesaw to balance?

Weight of the left child: $F_1 = m_1 g = (30.0\ \text{kg})(9.8\ \text{m/s}^2) = 294\ \text{N}$ . Both weights act straight down, perpendicular to the horizontal seesaw, so $\sin\theta = 1$ for each torque.
Left torque, which rotates the seesaw counterclockwise (positive): $\tau_1 = +(294\ \text{N})(2.0\ \text{m}) = +588\ \text{N·m}$ .
Weight of the right child: $F_2 = (40.0\ \text{kg})(9.8\ \text{m/s}^2) = 392\ \text{N}$ . Sitting a distance $r$ to the right, the clockwise torque is $\tau_2 = -(392\ \text{N})\,r$ .
Set the net torque to zero: $588\ \text{N·m} - (392\ \text{N})r = 0$ , so $r = 588/392 = 1.5\ \text{m}$ .

The 40.0 kg child must sit $1.5$ m from the pivot on the opposite side.

Pushing a door: 90 degrees versus 30 degrees

You push on a door with 50.0 N at a point $r = 0.90$ m from the hinges. Compare the torque when you push perpendicular to the door ( $\theta = 90^\circ$ ) with the torque when you push at $\theta = 30^\circ$ .

Perpendicular push: $\tau = (0.90\ \text{m})(50.0\ \text{N})(\sin 90^\circ) = (0.90)(50.0)(1.00) = 45.0\ \text{N·m}$ . The lever arm is the full $0.90$ m.
Angled push: $\sin 30^\circ = 0.500$ , so $\tau = (0.90\ \text{m})(50.0\ \text{N})(0.500) = 22.5\ \text{N·m}$ . The lever arm shrinks to $(0.90\ \text{m})(0.500) = 0.45\ \text{m}$ .
Compare: the same 50.0 N push produces exactly half the torque at $30^\circ$ , because the lever arm is cut in half.

$45.0$ N·m at $90^\circ$ versus $22.5$ N·m at $30^\circ$ : same force, half the rotational effect.

Frequently asked questions

Why is torque measured in newton meters instead of joules?

They share the same base units but describe different quantities. Energy is force times distance moved along the force's direction, while torque is force times perpendicular distance from a pivot, with no motion required. Write torque as N·m on the AP exam and never relabel it as joules.

Which angle goes into rF sin theta?

Use the angle between the position vector (from the pivot to the point where the force acts) and the force itself. If the force is perpendicular to the lever, theta is 90 degrees and sin theta equals 1, so the torque is just rF. If a problem gives the angle measured from the perpendicular instead, subtract it from 90 degrees first.

When is torque zero?

In two cases: the force acts at the pivot itself (r = 0), or the force's line of action passes through the pivot (theta = 0 or 180 degrees, so sin theta = 0). In both cases the lever arm is zero, so the force cannot rotate the object no matter how large it is.

Is clockwise torque positive or negative?

By the usual convention, counterclockwise torques are positive and clockwise torques are negative. The choice is arbitrary, so any consistent convention works. On free-response questions, state your convention before summing torques so the grader can follow your signs.