Torque Calculations to Know for AP Physics 1 (2025)

1. What You Need to Know

Torque is the rotational “push” that causes an object to start rotating, speed up its rotation, slow down, or change its rotational direction. On AP Physics 1, torque calculations show up in:

• Static equilibrium (beams, ladders, sign/hanging mass problems)
• Rotational dynamics (finding \alpha using \sum \tau = I\alpha)
• Rolling motion (torque from friction causing rolling)
• Pulleys with rotational inertia (tensions create net torque)

Core definition (the one you must know cold)

Torque about a chosen pivot/axis is the cross product:

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

• \vec{r}: vector from the pivot to the point where the force acts
• Magnitude:

\tau = rF\sin\theta

where \theta is the angle between \vec{r} and \vec{F}.

The “lever arm” version (fastest for AP problems)

\tau = F d

• d is the perpendicular distance (moment arm) from the pivot to the force’s line of action.

Critical reminder: If the force’s line of action passes through the pivot, then d=0 and the torque is zero even if the force is huge.

Why sign matters (CCW vs CW)

In 2D AP problems, you’ll assign a sign:

• Counterclockwise (CCW) torque: positive
• Clockwise (CW) torque: negative

You must be consistent within a problem.

2. Step-by-Step Breakdown

A) How to compute torque from a force (most common skill)

1. Choose the pivot/axis you’re taking torque about.

2. Draw \vec{r} from pivot to where the force acts.

3. Find the component of force perpendicular to \vec{r}:

   • F_\perp = F\sin\theta

4. Compute magnitude:

   \tau = rF\sin\theta = rF_\perp

   or use \tau = Fd if you can identify the moment arm d quickly.

5. Assign a sign (CCW +, CW -) by imagining the rotation the force alone would cause.

Mini-check: If the force is applied at the pivot or directly toward/away from the pivot (so \theta = 0 or 180^\circ), then \tau=0.

B) How to solve static equilibrium torque problems (beams/ladders)

For an object not rotating and not accelerating:

\sum F_x = 0,\quad \sum F_y = 0,\quad \sum \tau = 0

Procedure:

1. Draw a clean FBD (include weights at centers of mass!).
2. Pick a pivot that eliminates the most unknown forces (forces at the pivot contribute zero torque about that pivot).
3. Write \sum \tau = 0 using signed torques.
4. Use \sum F_x = 0 and \sum F_y = 0 to finish unknowns.

Decision point: If there are multiple unknown support forces, choosing the pivot at one support often kills those unknown torques instantly.

C) How to do rotational dynamics (object actually angularly accelerating)

When the object rotates about a fixed axis:

\sum \tau = I\alpha

Procedure:

1. Choose the axis (often through a pivot or the center).
2. Compute each external torque about that axis.
3. Sum them with signs to get \sum \tau.
4. Use the appropriate I for that axis.
5. Solve for \alpha.

Mini-example (annotated):
A force F is applied perpendicular to a rod of length L pivoted at one end.

• Torque: \tau = FL
• If rod is uniform: I = \frac{1}{3}ML^2
• Angular acceleration:

\alpha = \frac{\tau}{I} = \frac{FL}{\frac{1}{3}ML^2} = \frac{3F}{ML}

3. Key Formulas, Rules & Facts

Torque essentials

How to find the moment arm d quickly

• Draw the line of action of the force.
• Drop a perpendicular from the pivot to that line.
• That perpendicular distance is d.

Special cases:

• If the force is perpendicular to \vec{r}, then d=r.
• If the force is parallel to \vec{r}, then d=0.

Common torque contributors in typical AP setups

Moments of inertia you actually use with torque problems

(Only included because you’ll often pair with \sum \tau = I\alpha.)

If needed (sometimes given/allowed in AP1 contexts):

I_{\text{pivot}} = I_{\text{cm}} + Md^2

where d is distance from CM to pivot (parallel-axis theorem).

Units and interpretation

• Unit: \text{N}\cdot\text{m}
• Torque is not energy (even though \text{N}\cdot\text{m} matches joule); energy is scalar work, torque is a rotational effect.

4. Examples & Applications

Example 1: Door torque (classic angle trap)

You push on a door at a point r=0.80\,\text{m} from the hinge with force F=20\,\text{N} at 30^\circ to the door (i.e., to \vec{r}).

Setup:

• Use \tau = rF\sin\theta with \theta=30^\circ.

\tau = (0.80)(20)\sin 30^\circ = 16\cdot 0.5 = 8\,\text{N}\cdot\text{m}

Key insight: Only the perpendicular component of your push produces torque.

Example 2: Uniform beam with a hanging mass (static equilibrium)

A horizontal uniform beam of length L is hinged at the left end. A mass m hangs from the right end. Beam mass is M.

Pick pivot at hinge (kills hinge force torques).
Torques about hinge:

• Hanging mass: clockwise torque -(mg)(L)
• Beam weight acts at center: clockwise torque -(Mg)(L/2)
• Suppose a cable at the right end provides upward tension component T_y giving CCW torque +(T_y)(L)

Equilibrium:

\sum \tau = (T_y)L - (mg)L - (Mg)\frac{L}{2} = 0

Solve:

T_y = mg + \frac{Mg}{2}

Key insight: Put your pivot at an unknown-force point to eliminate it.

Example 3: Ladder against a frictionless wall (big AP favorite)

A uniform ladder length L rests on rough ground and leans against a **frictionless** wall. Ladder weight Mg acts at its center.

Forces:

• At ground: normal N_g (up), friction f_g (horizontal)
• At wall (frictionless): normal N_w (horizontal)
• Weight: Mg at midpoint

Choose pivot at ground contact to eliminate N_g and f_g torques.
Let ladder make angle \theta with ground.

• Torque from wall normal N_w: CCW with moment arm L\sin\theta
• Torque from weight Mg at midpoint: CW with moment arm \frac{L}{2}\cos\theta

Equilibrium:

N_w(L\sin\theta) - Mg\left(\frac{L}{2}\cos\theta\right)=0

So:

N_w = \frac{Mg}{2}\cot\theta

Key insight: Moment arms come from perpendicular distance; for horizontal force at the top, the perpendicular distance is vertical height L\sin\theta.

Example 4: Pulley with rotational inertia (tension difference creates torque)

Two masses m_1 and m_2 hang on either side of a pulley of radius R and moment of inertia I. Tensions differ: T_1 and T_2.

Pulley torque equation about its axle:

\sum \tau = (T_2 - T_1)R = I\alpha

No-slip constraint between rope and pulley:

a = \alpha R

Key insight: If the pulley has inertia, then T_1 \neq T_2 (that difference is what produces net torque).

5. Common Mistakes & Traps

1. Using \tau=rF without the sine
   What goes wrong: You forget the angle and overestimate torque.
   Fix: Default to \tau=rF\sin\theta unless you’re sure the force is perpendicular.

2. Using the wrong angle \theta
   What goes wrong: You plug in the angle between the force and the horizontal, not between \vec{r} and \vec{F}.
   Fix: Always identify \vec{r} first; \theta is the angle from \vec{r} to \vec{F}.

3. Not using the perpendicular distance (moment arm) correctly
   What goes wrong: You use the distance to the point of application rather than the perpendicular distance to the force line of action.
   Fix: Either use rF\sin\theta or explicitly draw the line of action and find d.

4. Forgetting the beam’s own weight (or placing it at the wrong spot)
   What goes wrong: You treat a uniform beam’s weight as acting at an end.
   Fix: Uniform object weight acts at its center of mass (midpoint for uniform rod).

5. Mixing torques about different pivots
   What goes wrong: You compute one torque about the hinge and another about the center.
   Fix: In a single \sum \tau equation, every torque must be about the same pivot.

6. Sign errors (CW/CCW)
   What goes wrong: You assign signs based on force direction instead of rotation direction.
   Fix: Imagine the object free to rotate about the pivot: does that force make it turn CW or CCW?

7. Assuming equilibrium means only \sum \tau=0
   What goes wrong: You forget \sum F_x=0 and \sum F_y=0, so you can’t solve all unknowns.
   Fix: For statics in 2D, you typically need three equations: \sum F_x=0, \sum F_y=0, \sum \tau=0.

8. Calling torque “joules”
   What goes wrong: You treat \text{N}\cdot\text{m} as energy.
   Fix: Torque unit is \text{N}\cdot\text{m}; energy is joules. Same base unit, different physical quantity.

6. Memory Aids & Quick Tricks

7. Quick Review Checklist

• You can compute torque using either \tau=rF\sin\theta **or** \tau=Fd (with d perpendicular).
• You always define torque about a specific pivot/axis.
• You assign torque signs by the rotation it causes (CCW +, CW -).
• For equilibrium: \sum F_x=0, \sum F_y=0, and \sum \tau=0.
• For rotational dynamics: \sum \tau = I\alpha (about the same axis as I).
• Forces through the pivot create zero torque.
• Weight forces act at the center of mass (especially for uniform beams).
• In pulley inertia problems, T_1\neq T_2 and \ (T_2-T_1)R=I\alpha.

You’ve got this—set the pivot smart, keep the perpendicular distance idea front-and-center, and torque problems become algebra.