The moment arm (called the lever arm in the AP Physics 1 CED) is the perpendicular distance from the axis of rotation to the line of action of a force. Multiply it by the force's magnitude and you get the torque, which is why a force aimed straight through the axis produces zero torque.
The moment arm is the perpendicular distance from the axis of rotation to the line of action of a force. The AP Physics 1 CED calls this exact quantity the lever arm (Topic 5.3, Essential Knowledge under 5.3.A), so treat the two names as interchangeable. To find it, extend the force vector into an infinite line (its line of action), then measure the shortest distance from the axis to that line. That shortest distance is always perpendicular to the line of action.
Here's why it matters for torque. Torque only comes from the force component perpendicular to the position vector from the axis to where the force is applied. The moment arm packages that geometry into a single distance, so torque becomes τ = F × (moment arm), which is the same thing as τ = rF sin θ where r sin θ is the moment arm. Intuitively, the moment arm measures how much rotational leverage a force has. Push a door at the handle and the moment arm is the full door width. Push at the hinge and the moment arm is zero, so the door doesn't budge no matter how hard you shove.
Moment arm lives in Topic 5.3 (Torque) within Unit 5: Torque and Rotational Dynamics. It directly supports learning objective 5.3.A (identify the torques exerted on a rigid system), where the essential knowledge defines the lever arm and states that torque comes only from the perpendicular force component. It also feeds 5.3.B (describe torques using force diagrams), because a force diagram has to show where each force acts relative to the axis, and that location is what sets the moment arm. Every torque calculation, equilibrium problem, and rotational dynamics setup in Unit 5 starts with correctly identifying each force's moment arm. Get the moment arm wrong and every torque downstream is wrong too.
Keep studying AP® Physics 1 Unit 5
Lever Arm (Unit 5)
These are two names for the identical quantity. The CED says lever arm, many textbooks and teachers say moment arm. On the exam, either phrasing points to the perpendicular distance from the axis to the force's line of action.
Axis of Rotation (Unit 5)
The moment arm is always measured from the axis of rotation, so the same force can have different moment arms depending on which axis you choose. A door's weight creates torque about the top hinge precisely because its line of action (through the center of mass) misses that hinge by a horizontal distance of w/2.
Force Diagram (Unit 5)
Force diagrams are free-body diagrams upgraded with location. Because they show where each force acts on the rigid system relative to the axis, they're the tool you use to read off each force's moment arm before computing torques.
Rigid System (Unit 5)
Moment arm only makes sense for an extended, rigid object. In Units 1-4 you treated objects as points, so where a force acted didn't matter. Unit 5 is where location starts mattering, and the moment arm is the number that captures it.
Expect moment arm to show up inside torque comparisons and equilibrium setups rather than as a standalone definition question. A classic multiple-choice move gives you two equal forces on a wheel, one tangent to the rim and one applied at the center, and asks for the net torque. The trap is forgetting that the force at the center has a moment arm of zero, so it contributes nothing. Another common stem asks why a uniform door's weight creates a torque about its top hinge; the answer is that the weight's line of action passes through the center of mass, a horizontal distance from the hinge, giving it a nonzero moment arm. On free-response questions, you'll use moment arms when you write torque equations for rotational equilibrium or apply τ = rF sin θ, and when you justify which forces produce zero torque about a chosen pivot. The single most valuable skill is choosing a smart axis so that an unknown force's moment arm is zero and it drops out of your equation.
These match only when the force is perpendicular to the position vector. The moment arm is the perpendicular distance from the axis to the force's line of action, which equals r sin θ. If a force of magnitude F acts at distance r from the axis at angle θ to the position vector, the torque is rF sin θ, not rF. When a force points directly toward or away from the axis (θ = 0° or 180°), r can be huge but the moment arm, and the torque, is zero.
The moment arm is the perpendicular distance from the axis of rotation to a force's line of action, and the AP CED calls this same quantity the lever arm.
Torque equals force times moment arm, which is the same as τ = rF sin θ because the moment arm equals r sin θ.
A force whose line of action passes through the axis of rotation has a moment arm of zero and produces zero torque, no matter how large the force is.
The same force has different moment arms about different axes, so always state which axis you're using before calculating torque.
Force diagrams show where each force acts on a rigid system relative to the axis, which is exactly the information you need to find each moment arm.
In equilibrium FRQs, pick your axis so an unknown force has zero moment arm, which makes that force vanish from your torque equation.
It's the perpendicular distance from the axis of rotation to the line of action of a force. Multiplying it by the force's magnitude gives the torque, so it's the geometric heart of every Unit 5 torque calculation.
Yes, they're identical. The AP Physics 1 CED uses 'lever arm' in Topic 5.3, but 'moment arm' is the common textbook synonym, and either term can appear on practice materials.
No, not in general. The moment arm is the perpendicular distance to the force's line of action, which equals r sin θ. They're only equal when the force is perpendicular to the position vector (θ = 90°).
Yes. If the force's line of action passes through the axis of rotation, the moment arm is zero and the force produces no torque. That's why a force applied at the center of a wheel can't spin it, and why pushing a door at its hinges does nothing.
Extend the force vector into a full line (its line of action), then draw the shortest line from the axis to it. That perpendicular distance is the moment arm. For a uniform door of width w pivoting at a hinge, gravity acts at the center of mass, so its moment arm about the hinge is w/2.
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