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Torque

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, torque is the rotational version of a push or a pull. It measures how effectively a force can make an object spin or rotate around a pivot point.

Why this matters

Have you ever tried to open a heavy door at a museum or an old library? If you push right next to the hinges, the door barely budges, no matter how hard you shove. But if you push on the side farthest from the hinges, near the handle, it swings open with much less effort.

That difference you feel is torque. It’s not just about how hard you push (the force), but also where you push and in what direction. The same force can have a dramatically different effect depending on its application. Understanding torque is the key to understanding why wrenches have long handles, why seesaws balance, and how everything from a spinning top to a planet in orbit works. In this lesson, we'll break down exactly what torque is and how to calculate it.

Diagram

Calculating Torque on a Lever A diagram showing two equivalent methods for calculating torque. A horizontal bar is pivoted at its left end. A force F is applied to the right end at an angle theta. The diagram shows how to calculate torque either by finding the perpendicular component of the force (F-perp) or by finding the lever arm (r-perp). Axis of Rotation (O) r Line of Action F θ F⊥ = Fsin(θ) Method 1: Effective Force τ = r ⋅ F⊥ r⊥ = rsin(θ) Method 2: Lever Arm τ = F ⋅ r⊥ F∥ (creates no torque)
This diagram shows a horizontal bar pivoted at its left end, representing a lever. A force is applied to the right end at an angle, and the diagram illustrates the two equivalent ways to calculate the resulting torque: using the perpendicular component of the force or using the lever arm.

Concept map

flowchart TD
    A[Start: Analyze a Rigid System] --> B{Identify Axis of Rotation (Pivot)};
    B --> C[Identify all forces acting on the system];
    C --> D{For each force: Is it applied AT the pivot?};
    D -- Yes --> E[Torque = 0];
    D -- No --> F[Identify r, F, and θ];
    F --> G[Calculate Torque: τ = rFsin(θ)];
    G --> H{Is rotation CCW or CW?};
    H -- CCW --> I[Torque is Positive (+)];
    H -- CW --> J[Torque is Negative (-)];
    I --> K[Sum all torques: Στ = τ₁ + τ₂ + ...];
    J --> K;
    E --> C;
    K --> L[End: Net Torque Found];

Core explanation

In our previous units, we talked a lot about forces. Forces cause objects to accelerate linearly—to move from one point to another. But what causes things to rotate? A spinning basketball, a door swinging on its hinges, or a lug nut being loosened by a wrench? That's where torque comes in.

Torque is the rotational equivalent of force.

Just like a net force causes a linear acceleration, a net torque causes a rotational (or angular) acceleration.

The Three Ingredients of Torque

Imagine you're trying to loosen a stubborn lug nut on a car tire with a wrench. What makes you more effective?

  1. The amount of force (F): Pushing or pulling harder obviously helps. More force can produce more torque.
  2. The distance from the pivot (r): If you apply the force at the very end of the wrench's handle, you have more leverage than if you push on it close to the nut. The distance from the axis of rotation (the center of the nut) to the point where you apply the force is crucial. This distance is often called the radius or position vector.
  3. The angle of the force (θ): For the best result, you pull perpendicular (at 90°) to the wrench handle. If you push or pull along the length of the wrench (parallel to it), nothing happens! The nut won't turn. The angle between the force vector and the position vector matters immensely.

These three ingredients give us the fundamental equation for the magnitude of torque (represented by the Greek letter tau, τ):

τ = r * F * sin(θ)

Where:

  • τ is the torque, measured in Newton-meters (N·m).
  • r is the distance from the axis of rotation to the point where the force is applied.
  • F is the magnitude of the applied force.
  • θ is the angle between the force vector F and the position vector r.

Two Ways to Think About the Torque Equation

Method 1: The Effective Force (F_⊥)

Think of the equation as τ = r * (Fsinθ).

The term Fsin(θ) represents the component of the force that is perpendicular to the lever arm r. Let's call this F_⊥ (F-perp).

A visual showing a force F being broken down into components parallel and perpendicular to a lever arm r.

Imagine pushing on a door. The part of your push that is perpendicular to the door makes it swing. The part of your push that is parallel to the door (aimed toward the hinges) just jams the door into its frame; it doesn't help with rotation.

So, F_⊥ = Fsin(θ) is the only part of the force that contributes to the torque. The parallel component, F_∥, does nothing for rotation.

*`τ = r F_⊥`**

Method 2: The Effective Distance, or "Lever Arm" (r_⊥)

Now, let's group the terms differently: τ = (rsinθ) * F.

The term rsin(θ) has a special name: the lever arm, sometimes written as r_⊥.

The lever arm is the perpendicular distance from the axis of rotation to the "line of action" of the force. The line of action is an imaginary line extending infinitely in both directions along the force vector.

This might feel a bit abstract, but it's incredibly useful. To find the lever arm, you draw your force, extend a line through it, and then find the shortest distance from the pivot to that line. That shortest distance will always form a 90° angle with the line of action.

*`τ = r_⊥ F`**

Both methods, τ = r * F_⊥ and τ = r_⊥ * F, give you the exact same answer. Pick the one that makes the most sense for the geometry of the problem you're solving.

Direction and Sign Convention

Torque is a vector. It has a direction. In AP Physics 1, we simplify this to two directions: clockwise (CW) and counter-clockwise (CCW).

By convention:

  • Counter-clockwise (CCW) torques are positive (+)
  • Clockwise (CW) torques are negative (-)

Think of a standard clock. The hands move clockwise—that's the negative direction for us. To unscrew most things (like a bottle cap or a screw), you turn counter-clockwise—that's the positive direction.

Force Diagrams

To analyze the torques on an object, we use force diagrams. These are like the free-body diagrams you know and love, but with one critical difference: where you draw the force matters.

In a free-body diagram, you could represent the object as a single dot. In a force diagram for torque, you must draw the object's shape and place the force vectors at the exact points where they are applied. This is because the distance r is essential for calculating torque.

For example, on a seesaw, the pivot force is at the center, while the weights of the children (like Priya and Marcus) are applied at the points where they are sitting.

Worked examples

Example 1

Loosening a Lug Nut

Priya is using a 0.5-meter long wrench to loosen a lug nut on her car. She applies a 100 N force to the end of the wrench at an angle of 60° relative to the handle. What is the magnitude of the torque she applies?

1. Identify the knowns and the goal.

  • Distance r = 0.5 m
  • Force F = 100 N
  • Angle θ = 60°
  • Goal: Find the torque, τ.

2. Choose the correct equation. The fundamental equation for torque is τ = rFsin(θ).

3. Plug in the values and solve. τ = (0.5 m) * (100 N) * sin(60°)

Remember from trigonometry that sin(60°) ≈ 0.866.

τ = (0.5 m) * (100 N) * (0.866) τ = 50 * 0.866 τ ≈ 43.3 N·m

Example 2

The Balancing Seesaw

Carlos (mass 40 kg) and Sofia (mass 30 kg) are on a 4-meter long seesaw, which is supported by a pivot at its center. Carlos sits 1.5 meters to the left of the pivot. Where must Sofia sit to balance the seesaw? (Assume g = 10 m/s² for simplicity).

1. Draw a force diagram and define directions. The seesaw is a horizontal line. The pivot is at the center (x=0).

  • Carlos is at r_C = 1.5 m to the left. His weight F_C = m_C * g pushes down. This will cause a counter-clockwise (positive) torque.
  • Sofia is at an unknown distance r_S to the right. Her weight F_S = m_S * g pushes down. This will cause a clockwise (negative) torque.
  • For the seesaw to balance, the net torque must be zero: Στ = 0.

2. Calculate the torques.

  • Force from Carlos: F_C = 40 kg * 10 m/s² = 400 N.
  • Force from Sofia: F_S = 30 kg * 10 m/s² = 300 N.

Both children are sitting on the seesaw, so their weight acts perpendicular to the board (θ = 90° and sin(90°) = 1).

  • Torque from Carlos: τ_C = r_C * F_C = 1.5 m * 400 N = +600 N·m (Positive because it's CCW).
  • Torque from Sofia: τ_S = r_S * F_S = r_S * 300 N. This will be a negative torque.

3. Set the net torque to zero and solve for the unknown. Στ = τ_C + τ_S = 0 (+600 N·m) + (-r_S * 300 N) = 0

This is a common mistake spot: Don't forget the negative sign for the clockwise torque!

600 = 300 * r_S r_S = 600 / 300 r_S = 2.0 m

Sofia must sit 2.0 meters to the right of the pivot to balance the seesaw. Since the seesaw is 4 meters long, the end is 2 meters from the center, so she must sit right at the end.

Try it yourself

Practice Problem 1

A 12-foot-long horizontal beam weighing 80 lbs is attached to a wall by a hinge. Its weight acts at its center of mass (6 ft from the hinge). A cable is attached to the far end of the beam (12 ft from the hinge) and pulls upward and back toward the wall at an angle of 30° above the beam. What is the torque exerted by the beam's own weight about the hinge?

Practice Problem 2

Using the same setup as above, the tension in the cable is 100 lbs. What is the torque exerted by the cable about the hinge? Is this torque positive (CCW) or negative (CW)? What is the net torque on the beam?

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