Torque τ = r × F — definition, physical meaning: Worked examples

This page is the "no surprises left" workshop for Torque τ = r × F. We will hit every kind of situation torque can throw at you: both directions of spin, the force aimed straight at the pivot, a force with zero lever arm, the maximum-twist case, a real-world word problem, and an exam-style trap. Nothing here uses a symbol we have not already earned in the parent note — but as a one-line refresher:

Recall The only three formulas we need

Magnitude: $τ = r F sin θ$ (full distance × sideways force). Component (2D, in the $x y$ -plane): $τ_{z} = x F_{y} - y F_{x}$ (positive = anticlockwise = out of the page toward you). Vector: $τ = r \times F$ . Here $r$ always starts at the pivot and points to where the force touches.

The scenario matrix

Every torque problem you will ever meet lands in one of these cells. The table lists the cell, what makes it special, and which worked example below covers it.

Cell	What is special about it	Covered by
A. Perpendicular push	$θ = 9 0^{\circ}$ , maximum twist	Ex 1
B. Angled push	$0^{\circ} < θ < 9 0^{\circ}$ , only $F sin θ$ counts	Ex 2
C. Force along the radius	$θ = 0^{\circ}$ or $18 0^{\circ}$ → zero torque	Ex 3
D. Positive (anticlockwise) sign	$τ_{z} > 0$ , out of page	Ex 4
E. Negative (clockwise) sign	$τ_{z} < 0$ , into page	Ex 5
F. Zero lever arm	force line passes through the pivot	Ex 6
G. Real-world word problem	translate words → $r$ , $F$ , $θ$	Ex 7
H. Exam twist (net torque of two forces)	add signed torques, watch signs	Ex 8
I. Limiting behaviour	how $τ$ varies as $θ$ sweeps	Ex 9

Example 1 — Perpendicular push (Cell A)

Worked example Statement

A mechanic pushes with $F = 40 N$ at the end of a $0.25 m$ spanner, perpendicular to the spanner. Find the torque about the bolt.

Forecast: perpendicular means "all of the force is sideways" — do you expect this to be the biggest, smallest, or middling twist for this $r$ and $F$ ?

Step 1 — Identify the angle: $θ = 9 0^{\circ}$ . Why this step? Torque depends on the angle between $r$ and $F$ ; we must pin it down before touching the sine.

Step 2 — Read the sine: $sin 9 0^{\circ} = 1$ . Why this step? At $9 0^{\circ}$ the entire force is the "spinning" component $F sin θ = F$ , so no force is wasted pushing along the arm.

Step 3 — Multiply: $τ = r F sin θ = 0.25 \times 40 \times 1 = 10 N\cdotm$ . Why this step? Plug into $τ = r F sin θ$ .

Verify: Units $m \times N = N\cdotm$ ✓. This is the maximum possible twist for this $r$ and $F$ (sine caps at 1), matching the forecast.

Figure — Torque τ = r × F — definition, physical meaning

Example 2 — Angled push (Cell B)

Worked example Statement

Same spanner ( $r = 0.25 m$ ), same effort ( $F = 40 N$ ), but you push at $θ = 3 0^{\circ}$ to the spanner. Find the torque.

Forecast: you use the same muscle force. Will the twist be more or less than $10 N\cdotm$ ? By roughly what factor?

Step 1 — Split the force: the part along the spanner is $F cos 3 0^{\circ}$ , the sideways part is $F sin 3 0^{\circ}$ . Why this step? Only the sideways (perpendicular-to- $r$ ) part can circulate the point around the pivot; the along-part just tugs toward/away from the bolt.

Step 2 — Sideways force $= 40 \times sin 3 0^{\circ} = 40 \times 0.5 = 20 N$ . Why this step? This is the only force that produces spin.

Step 3 — Torque $= r \times F_{⊥} = 0.25 \times 20 = 5 N\cdotm$ . Why this step? Torque = full distance × sideways force.

Verify: $τ = r F sin θ = 0.25 \times 40 \times 0.5 = 5 N\cdotm$ ✓. It is exactly half of Example 1's $10 N\cdotm$ because $sin 3 0^{\circ} = 0.5$ — same effort, half the twist.

Example 3 — Force along the radius (Cell C)

Example 4 — Positive (anticlockwise) torque (Cell D)

Example 5 — Negative (clockwise) torque (Cell E)

Example 6 — Zero lever arm (line of action through pivot) (Cell F)

Example 7 — Real-world word problem (Cell G)

Worked example Statement

A child sits on a seesaw $1.2 m$ from the central pivot and weighs $300 N$ . Gravity pulls straight down while the seesaw beam is horizontal. Find the torque the child produces about the pivot.

Forecast: the weight acts vertically, the arm is horizontal. Are they perpendicular? What does that make $sin θ$ ?

Step 1 — Set up: $r = 1.2 m$ (pivot → child), $F = 300 N$ (weight, downward). Why this step? $r$ must start at the pivot; the "force" here is the child's weight.

Step 2 — Angle between horizontal arm and vertical weight: $θ = 9 0^{\circ}$ , so $sin θ = 1$ . Why this step? Horizontal ⟂ vertical, so no force is wasted along the beam.

Step 3 — $τ = r F sin θ = 1.2 \times 300 \times 1 = 360 N\cdotm$ .

Verify: Units N·m ✓. If a second child on the other side must balance this, they need to supply the same $360 N\cdotm$ of opposite-sense torque — the everyday balancing rule (see Equilibrium of Rigid Bodies). ✓

Example 8 — Exam twist: net torque of two forces (Cell H)

Worked example Statement

On a horizontal rod pivoted at its centre, a force pushes down at $x = + 2 m$ with $F_{1} = 10 N$ , and another pushes down at $x = - 3 m$ with $F_{2} = 10 N$ . Both forces are vertical (down). Find the net torque about the pivot and the resulting sense of spin.

Forecast: both forces push down, but on opposite sides. Do their torques add up or fight each other?

Step 1 — Torque of force 1: $r_{1} = (2, 0)$ , $F_{1} = (0, - 10)$ , so $τ_{1} = x F_{y} - y F_{x} = (2) (- 10) - 0 = - 20 N\cdotm$ (clockwise). Why this step? A downward push on the right end rotates the rod clockwise.

Step 2 — Torque of force 2: $r_{2} = (- 3, 0)$ , $F_{2} = (0, - 10)$ , so $τ_{2} = (- 3) (- 10) - 0 = + 30 N\cdotm$ (anticlockwise). Why this step? A downward push on the left end rotates the rod the other way — the sign flips because $x$ is negative.

Step 3 — Net torque: $τ_{net} = τ_{1} + τ_{2} = - 20 + 30 = + 10 N\cdotm$ . Why this step? Torques about the same axis are signed scalars in 2D, so we just add them.

Verify: $τ_{net} = + 10 N\cdotm$ → net anticlockwise. The trap is thinking "both forces point the same way so torques add in magnitude" — but opposite sides give opposite signs. The two torques fight, and the longer arm ( $3 > 2$ ) wins. ✓

Example 9 — Limiting behaviour as the angle sweeps (Cell I)

Wrap-up

Recall Which cell is this problem in?

The single fastest way to avoid errors: before computing, name the cell. Force perpendicular to arm? ::: Cell A — full $r F$ . Force at an angle? ::: Cell B — use $r F sin θ$ . Force along the arm (or line through pivot)? ::: Cells C/F — torque is zero. Which sign in 2D? ::: $τ_{z} = x F_{y} - y F_{x}$ ; $+$ anticlockwise, $-$ clockwise (Cells D/E). Several forces? ::: Cell H — add signed torques about the same axis.

Related deeper dives: use these torques in Newton's Second Law for Rotation, connect the cross product itself in Cross Product (Vector Algebra), and balance them in Equilibrium of Rigid Bodies.