Torque τ = r × F — definition, physical meaning: Question bank

Before we start, three words you must own so every reveal reads cleanly:

Pivot / axis — the fixed line the body would rotate about; $r$ is always drawn from here.
Line of action — the infinite straight line you get by extending the force arrow both ways.
==Moment arm $d_{⊥}$ == — the shortest (perpendicular) distance from the pivot to that line of action; equals $r sin θ$ .

True or false — justify

A force with larger magnitude always produces larger torque.

False. Torque is

r F sin θ

, so a huge force aimed straight at the pivot (

θ = 0

) gives zero torque while a gentle sideways push far out gives plenty.

If two forces have the same magnitude and same moment arm, they produce the same-magnitude torque.

True.

τ = d_{⊥} F

depends only on the moment arm and the force size, not on where along the line of action the force is drawn.

Torque and energy can be added together because both are measured in N·m.

False. They share dimensions but torque is a vector and energy a scalar; you never add a twisting axis to a number of joules.

A force applied exactly at the pivot produces zero torque no matter its direction.

True. There

r = 0

, so

r \times F = 0

for every

F

— nothing twists a point you're pushing right at.

Sliding a force along its own line of action changes the torque about a fixed pivot.

False. The moment arm

d_{⊥}

(perpendicular distance from pivot to that line) is unchanged, so

τ = d_{⊥} F

stays fixed — this is why the line of action, not the contact point, is what matters.

Reversing the force direction (

F \to - F

) flips the sense of rotation.

True.

r \times (- F) = - (r \times F)

, so the torque vector flips, turning anticlockwise into clockwise.

Doubling only the distance

r

(same force, same angle) doubles the torque.

True.

τ = r F sin θ

is linear in

r

, so twice the lever length gives twice the twist — the door-handle rule.

Torque is always perpendicular to the force that creates it.

True.

τ = r \times F

is perpendicular to both

r

and

F

by the definition of the cross product, so it can never point along the force.

r

and

F

both lie in the

x y

-plane, the torque must point along

\pm \hat{k}

True. The cross product of two in-plane vectors is normal to that plane, so a planar problem always yields a torque along the out-of-plane axis.

A radial (inward or outward) pull can never spin an object about the pivot.

True. A radial force is parallel/antiparallel to

r

(

θ = 0

18 0^{\circ}

), and

sin θ = 0

, so it only stretches or compresses toward the axis, never circulates.

Spot the error

"Push the wrench at

3 0^{\circ}

to its length, so use

τ = r F cos 3 0^{\circ}

Error: the twisting part is the component perpendicular to

r

, which is

F sin θ

, not

F cos θ

. The

cos

instinct comes from projecting onto an axis; torque wants the sideways leftover.

r

runs from where the force is applied to the pivot."

Error: direction is backwards.

r

starts at the pivot and points to the application point; reversing it flips the torque's sign.

"Torque is zero, so the object is in equilibrium."

Error: zero net torque means no angular acceleration, but the object can still spin at constant rate, and it may also have a net force accelerating its center of mass — see Equilibrium of Rigid Bodies.

"The force is bigger, and the object still didn't rotate, so torque must have a flaw."

Error: no flaw — if the extra force is aimed along

r

, its

sin θ = 0

contributes zero torque. Always check the moment arm before blaming the formula.

τ = r F sin θ

, and I measured

θ

as the angle to the horizontal."

Error:

θ

is the angle between $r$ and $F$ , not between the force and some fixed axis. Using the wrong reference angle gives the wrong

sin θ

"Two forces give torques

+ 8

and

- 8

N·m, so the total twisting is

16

N·m."

Error: signs encode opposite senses, so they cancel to a net torque of

0

; you add signed values, not magnitudes.

Why questions

Why does torque use the cross product instead of an ordinary product of

r

and

F

Because the twist depends on the angle between position and force and produces a directed rotation axis; the cross product

r F sin θ

built on Cross Product (Vector Algebra) captures exactly the perpendicular part and the axis via the right-hand rule.

Why is torque maximal when the force is perpendicular to

r

9 0^{\circ}

all of the force is "sideways" (tangential), so

sin θ = 1

and every newton contributes to circulation; any tilt wastes part of the force radially.

Why can the same torque be read as "

r \times F_{⊥}

" and "

d_{⊥} \times F

Both are just re-bundlings of

r F sin θ

: group the

sin θ

with

F

to get the sideways force, or with

r

to get the moment arm — multiplication doesn't care where the factor sits.

Why does the same physical push open a door easily at the handle but barely at the hinge?

The moment arm

d_{⊥}

(distance from hinge line to your push) is large at the handle and tiny near the hinge, and

τ = d_{⊥} F

scales directly with it.

Why is torque the rotational partner of force in Newton's law?

Just as force sets linear acceleration, net torque sets angular acceleration through

τ_{n e t} = I α

— see Newton's Second Law for Rotation and Moment of Inertia.

Why does torque relate to angular momentum rather than to plain momentum?

Torque is the time rate of change of Angular Momentum (

τ = d L / d t

), the rotational echo of "force is the rate of change of momentum."

Edge cases

What is the torque if the force acts exactly along the line through the pivot?

Zero, because

F ∥ r

gives

θ = 0

18 0^{\circ}

and

sin θ = 0

— the force only pushes toward/away from the axis.

What is the torque about a pivot placed on the force's own line of action?

Zero: the moment arm

d_{⊥}

0

since the line passes through the pivot, so

τ = d_{⊥} F = 0

regardless of force size.

What torque does a zero-magnitude force produce, whatever its position?

Zero, since

τ = r F sin θ

has

F = 0

as a factor — no force, no twist, at any distance or angle.

r

and

F

point in the exact same direction, what is the torque?

Zero — the parallelogram they span collapses to a flat line of zero area, so

∣ r \times F ∣ = 0

θ

sweeps from

0^{\circ}

past

9 0^{\circ}

18 0^{\circ}

at fixed

r

and

F

, how does

τ

behave?

It rises from

0

to a maximum at

9 0^{\circ}

then falls back to

0

18 0^{\circ}

, tracing the shape of

sin θ

— symmetric about the perpendicular.

Does the torque about a pivot depend on the choice of pivot?

Yes — moving the pivot changes

r

(and

d_{⊥}

), so the same force generally gives different torques about different axes; only a couple (equal, opposite, offset forces) gives a torque independent of pivot.

If the whole force acts at the center of mass through the pivot, can gravity ever create a torque about that pivot?

No — if the pivot coincides with the center of gravity, the weight's line of action passes through the pivot,

d_{⊥} = 0

, and the gravitational torque vanishes; see Center of Mass & Gravity.