Torque is the rotational analog of force: it measures how hard a force twists an object around a pivot. It equals r*F*sin(theta), where r is the distance from the pivot to where the force is applied and theta is the angle between r and the force.

Why is there a sin(theta) in the torque formula?

Only the component of force perpendicular to r can cause rotation. That perpendicular component is F*sin(theta): it is zero when the force points along r (theta = 0) and maximal when the force is perpendicular to r (theta = 90 degrees).

Torque (τ = rF sin θ) Explained — Derivation, Examples & Quizzes

Q: Why does pushing a door near the hinge barely move it?

Torque is r*F*sin(theta): near the hinge r is tiny, so the torque is tiny even with a big force. The handle maximizes r, which is why handles sit at the far edge of the door.

Q: What are the units of torque?

Newton-meters (N*m): force (newtons) times distance (meters). It is dimensionally the same as a joule, but torque is never called energy — torque twists, energy is the capacity to do work.

Where it comes from

Force alone does not tell you whether something spins. Push a door and three things decide what happens:

Before reading on: a heavy door — do you push at the handle or near the hinge? Straight into the door face, or along it toward the hinge? Predict which choices spin it best, then check yourself below.

handle, 0.8 m out

τ = 0.8 × 40 × sin 90° = 32 N·m

Push 40 N at the handle, perpendicular: the door swings easily.

near hinge, 0.2 m out

τ = 0.2 × 40 × sin 90° = 8 N·m

Same 40 N near the hinge: one quarter the twist. Distance r matters.

along the door

τ = 0.8 × 40 × sin 0° = 0 N·m

Pushing along the door toward the hinge: zero twist, however hard you shove. Angle θ matters.

So torque needs three ingredients: how hard (F), how far out (r), and how squarely (θ). The sin θ is the formula’s way of saying only the perpendicular part of the push can twist: sin 90° = 1 rewards a square push, sin 0° = 0 kills a push along the rod.

Physicists write this compactly as a cross product, τ = r × F: its magnitude is rF sin θ, and its direction follows the right-hand rule (fingers from r to F, thumb points along τ — out of the page for counterclockwise twists).

Derivation

Torque is defined as the twist a force produces about a pivot. The formula falls out of one observation: a force applied at an angle can be split into a part that twists and a part that merely pulls. Watch which part survives.

F

=

F_∥ + F_⊥

Step 1 — split the force. Resolve F into components parallel and perpendicular to r. The parallel part pulls straight along the rod, through the pivot’s line — it can stretch or compress the rod, but it cannot rotate it.

F_⊥

=

F sin θ

Step 2 — the twisting part. From the right triangle of components, the perpendicular piece is F sin θ, with θ the angle between r and F tail-to-tail.

τ

=

r × (twisting component) = r F sin θ

Step 3 — assemble. Twist = (distance from pivot) × (component that twists). The parallel part contributed nothing, so it drops out. ∎

d

=

r sin θ ⇒ τ = Fd

Same idea, regrouped. d is the lever arm — the perpendicular distance from the pivot to the force’s line of action. Torque = force × lever arm. Two groupings, one formula.

Why isn’t it just rF? Because “rF” claims every push twists equally. The door test kills that: pushing along the door (F ≠ 0, r ≠ 0) gives no rotation at all. The sin θ is load-bearing — it is the whole difference between force and torque.

How to use it

The procedure, every time:

Mark the pivot. Torque is always about something — a hinge, an axle, a balance point. No pivot, no torque.
Draw r: an arrow from the pivot to the point where the force is applied.
Find θ tail-to-tail. Slide the force arrow so its tail sits on r’s tail; θ is the angle between them. (0° ≤ θ ≤ 180°.)
Compute τ = rF sin θ, and give it a sign: counterclockwise positive is the standard convention (out of the page by the right-hand rule).
Sum for the net torque: Στ = τ₁ + τ₂ + … — signs included. Opposing twists fight each other.

The angle trap

Problems sometimes give the angle to the perpendicular instead of to the rod. If a force hits a wrench at 30° from perpendicular, then θ (to the wrench) is 60° — or just use τ = rF cos(30°), same number. Always ask: “angle to what?” before choosing sin or cos.

τ net = Σ r i F i sin θ i torques add like signed numbers — clockwise fights counterclockwise Say it: “the net torque is the signed sum of r F sine theta over every force”

Common mistake: measuring θ from the wrong reference — e.g. using the 30° a force makes with the vertical as if it were the angle to the rod. Redraw r and F tail-to-tail; the angle between the two arrows is θ, full stop.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: 25 N at 0.9 m, square on

Pivot and r. Pivot at the hinge; r = 0.9 m to the push point.
Angle. The push is perpendicular to the rod: θ = 90°, sin θ = 1.
Compute. τ = 0.9 × 25 × 1 = 22.5 N·m.
Sanity check. Units N·m ✓; a firm push near a meter out giving tens of N·m feels right.

Common mistake: writing τ = rF cos θ by habit from work (W = Fd cos θ). Work rewards pushing along the motion; torque rewards pushing across the lever. Different geometry, different trig.

Your turn — F = 60 N at r = 0.5 m, perpendicular. τ = ?

Answer: 30 N·m. τ = 0.5 × 60 × sin 90° = 30 × 1 = 30 N·m.

Example 2 — angled push: 100 N on a 0.4 m wrench at 60°

Identify θ. 60° between the wrench (r) and the force, tail-to-tail.
Twisting component. F_⊥ = 100 × sin 60° ≈ 100 × 0.866 = 86.6 N. (The other 50 N just pulls along the wrench.)
Compute. τ = 0.4 × 86.6 ≈ 34.6 N·m.
Compare. A square push would give 40 N·m — the 60° angle cost about 13% of the twist.

Common mistake: grabbing sin 60° ≈ 0.866 but multiplying by the wrong r (e.g. the full 1 m handle when the push lands halfway). r is pivot-to-push-point, not the tool’s total length.

Your turn — F = 80 N at r = 0.3 m, θ = 30°. τ = ?

Answer: 12 N·m. τ = 0.3 × 80 × sin 30° = 24 × 0.5 = 12 N·m. A shallow 30° push wastes half the force.

Before reading on: a downward push on the left end of a pivoted rod — clockwise or counterclockwise? Decide, then check the sign below.

Example 3 — net torque with signs: a pivoted 1 m rod

Setup. Rod pivoted at its center. Left end: 20 N downward, r = 0.5 m. Right end: 30 N downward, r = 0.5 m. Convention: counterclockwise positive.
Left torque. Down on the left end swings that end down → counterclockwise → positive: τ₁ = +20 × 0.5 × sin 90° = +10 N·m.
Right torque. Down on the right end swings that end down → clockwise → negative: τ₂ = −30 × 0.5 × 1 = −15 N·m.
Net. Στ = 10 − 15 = −5 N·m — the rod twists clockwise; the heavier right side wins.

Common mistake: adding magnitudes (10 + 15 = 25) and ignoring signs. Torques are signed — opposing twists subtract. Always decide each torque’s direction before adding.

Your turn — Same rod. Left end: 10 N down. Right end: 10 N up. Net τ = ?

Answer: +10 N·m (counterclockwise). Left: down on the left → CCW → +10 × 0.5 = +5. Right: up on the right end lifts it → also CCW → +5. Total +10 N·m — the twists team up.

Example 4 — how hard must you push? Solving for F

Goal. A bolt needs 45 N·m. Your wrench is 0.3 m long; you push perpendicular at the end.
Rearrange. τ = rF sin θ ⇒ F = τ / (r sin θ).
Compute. F = 45 / (0.3 × 1) = 150 N (about 34 lb of push).
The lesson. A 0.6 m breaker bar would need only 75 N — this is why long wrenches exist: r is the cheapest way to buy torque.

Common mistake: solving F = τ/r but forgetting sin θ when the push is angled. At 30° the same bolt would need 45/(0.3 × 0.5) = 300 N — double the effort for a sloppy angle.

Your turn — Need 20 N·m with a 0.25 m wrench, pushing at 90°. F = ?

Answer: 80 N. F = 20 / (0.25 × 1) = 80 N.

Memorization tips

Say it aloud: “torque equals r times F times sine theta.” The rhythm matches the formula.
The door test (your 5-second self-check): handle = big r, square push = sin 90° = 1. If your setup gives less twist than the door handle, recheck r and θ.
Only the perpendicular part twists: sin θ measures “how square” the push is. 90° → full credit; 0° → zero.
Lever-arm form: τ = Fd with d = r sin θ — the perpendicular distance from pivot to the force’s line. One picture, two formulas.
Signs first, arithmetic second: decide clockwise vs counterclockwise for each torque before adding. Unsigned sums are the #1 torque error.
N·m is not joules: same dimensions, different meaning. Torque twists; energy does work. Never relabel one as the other.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and torque is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is torque?

Torque is the rotational analog of force: it measures how hard a force twists an object around a pivot. It equals r·F·sin θ, where r is the distance from the pivot to where the force is applied and θ is the angle between r and the force.

Why does pushing a door near the hinge barely move it?

Torque is r·F·sin θ: near the hinge r is tiny, so the torque is tiny even with a big force. The handle maximizes r, which is why handles sit at the far edge of the door.

Why is there a sin(θ) in the torque formula?

Only the component of force perpendicular to r can cause rotation. That perpendicular component is F·sin θ: it is zero when the force points along r (θ = 0°) and maximal when the force is perpendicular to r (θ = 90°).

What are the units of torque?

Newton-meters (N·m): force (newtons) times distance (meters). It is dimensionally the same as a joule, but torque is never called energy — torque twists, energy is the capacity to do work.

How do I handle the sign of torque?

Pick a convention and stick with it: counterclockwise positive is standard (equivalently, out-of-the-page positive by the right-hand rule). Add torques with their signs to get the net torque.

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