Classical-Mechanics · Unit 21 · Video 1 · Interactive Practice

Force Times Moment Arm: Defining Torque About a Point and Choosing Its Sign

IKey Formulas

Formula	Name	What it says
$\vec{\tau}_S = \vec{r}_{S,P} \times \vec{F}_P$	Torque about $S$	Position vector from $S$ to $P$ first, force second
$\tau_S \equiv \|\vec{\tau}_S\| = rF\sin\theta,\ \ 0 \le \theta \le \pi$	Magnitude	In $\text{N}\cdot\text{m}$ ; zero when $\theta = 0$ or $\theta = \pi$
$\tau_S = r_\perp F = rF_\perp,\quad r_\perp = r\sin\theta,\ \ F_\perp = F\sin\theta$	Moment arm	$r_\perp$ is the perpendicular distance from $S$ to the line of action
$\vec{\tau}_S = +\,r_\perp F\,\hat{n}_1 = -\,r_\perp F\,\hat{n}_2$	Planar sign	For a force turning counterclockwise about $S$ as you view the plane: $+$ with counterclockwise positive ( $\hat{n}_1$ , toward you), $-$ with clockwise positive ( $\hat{n}_2 = -\hat{n}_1$ )

Key Insight: A torque is always about a point: about a different $S$ the same force can have a different moment arm, and even the opposite sign. In a planar problem the sign carries meaning only once the positive normal, and the side the plane is viewed from, is declared.

IIForce Times Moment Arm

One torque, two readings: $\tau_S = r_\perp F = rF_\perp$ , with $r_\perp$ measured from $S$ to the line of action.

r

(m) 0.80

F_x

along the door (N) 12

F_y

across the door (N) 16

IIIA Torque Is Always About a Point

One fixed force, but its torque depends on $S$ : zero on its line of action, opposite signs on either side.

S_x

(m) −1.0

S_y

(m) −2.0

IVOne Vector, Two Signs

Clockwise and counterclockwise depend on the side you view from; the vector $\vec{r}_{S,P} \times \vec{F}_P$ does not.

Viewer elevation 55° Declared positive (seen from above)

VQuiz Questions

Problem 1 · A Wrench at an Angle

Given: A mechanic pulls on a wrench with a force of magnitude $F = 80\ \text{N}$ , applied at $r = |\vec{r}_{S,P}| = 0.25\ \text{m}$ from the centre $S$ of the bolt. The angle between $\vec{r}_{S,P}$ and $\vec{F}_P$ is $\theta = 30\degree$ — find the magnitude $\tau_S$ of the torque about $S$ .

✅ Correct!

\tau_S = rF\sin\theta = (0.25\ \text{m})(80\ \text{N})(0.500) = 10\ \text{N}\cdot\text{m}

— the same as

r_\perp F = (0.125\ \text{m})(80\ \text{N})

❌ Not quite.

rF = 20\ \text{N}\cdot\text{m}

would be the torque only if the force were perpendicular to

\vec{r}_{S,P}

. At

30\degree

only

F_\perp = F\sin 30\degree = 40\ \text{N}

turns the bolt.

❌ Close, but that is the cosine.

rF\cos\theta

is built from the force component along

\vec{r}_{S,P}

, whose line runs straight through

S

, so it cannot turn the bolt. The magnitude uses

\sin\theta

❌ Not quite. The moment arm is

r_\perp = r\sin\theta = 0.125\ \text{m}

, never

r/\sin\theta

: the perpendicular from

S

to the line of action can be no longer than

r

itself.

Show solution

The magnitude of the torque about $S$ is the product of the two magnitudes and the sine of the angle between the vectors:

\tau_S = rF\sin\theta = (0.25\ \text{m})(80\ \text{N})\sin 30\degree = (20\ \text{N}\cdot\text{m})(0.500) = 10\ \text{N}\cdot\text{m}

Both readings of the same product agree. Grouping the sine with the distance, $r_\perp = r\sin\theta = 0.125\ \text{m}$ and $r_\perp F = (0.125\ \text{m})(80\ \text{N}) = 10\ \text{N}\cdot\text{m}$ . Grouping it with the force, $F_\perp = F\sin\theta = 40\ \text{N}$ and $rF_\perp = (0.25\ \text{m})(40\ \text{N}) = 10\ \text{N}\cdot\text{m}$ .

Problem 2 · Distance Versus Moment Arm

Given: A $40\ \text{N}$ force acts at a point $P$ on a gate, $r = 0.50\ \text{m}$ from the hinge $S$ . The force is applied at an angle, so its line of action passes $0.30\ \text{m}$ from $S$ — find the magnitude $\tau_S$ of the torque about the hinge.

✅ Correct! The moment arm is the perpendicular distance from

S

to the line of action:

\tau_S = r_\perp F = (0.30\ \text{m})(40\ \text{N}) = 12\ \text{N}\cdot\text{m}

❌ Not quite.

0.50\ \text{m}

is the distance from

S

P

, not the moment arm;

rF

would require the force to be perpendicular to

\vec{r}_{S,P}

. The perpendicular distance from

S

to the line of action is

0.30\ \text{m}

❌ Not quite. Here

\sin\theta = 0.30/0.50 = 0.60

, so

|\cos\theta| = 0.80

, and

rF|\cos\theta| = 16\ \text{N}\cdot\text{m}

is built from the force component along

\vec{r}_{S,P}

— a component whose line passes through

S

and produces no torque.

❌ Close, but the sine is counted twice. The

0.30\ \text{m}

already equals

r\sin\theta

; multiplying

r_\perp F

\sin\theta

again gives

7.2\ \text{N}\cdot\text{m}

Show solution

The perpendicular distance from $S$ to the line of action is, by definition, the moment arm: $r_\perp = 0.30\ \text{m}$ . Torque is the force times its moment arm:

\tau_S = r_\perp F = (0.30\ \text{m})(40\ \text{N}) = 12\ \text{N}\cdot\text{m}

Check with the other readings. Since $r_\perp = r\sin\theta$ , the angle satisfies $\sin\theta = 0.30/0.50 = 0.60$ , so

rF\sin\theta = (0.50\ \text{m})(40\ \text{N})(0.60) = 12\ \text{N}\cdot\text{m}, \qquad rF_\perp = (0.50\ \text{m})(24\ \text{N}) = 12\ \text{N}\cdot\text{m}

with $F_\perp = F\sin\theta = 24\ \text{N}$ . The distance $r = 0.50\ \text{m}$ enters only together with $\sin\theta$ .

Problem 3 · One Force, Two Reference Points

Given: In the $xy$ -plane of the page, a $30\ \text{N}$ force pointing in the $+x$ direction acts at $P = (1.0\ \text{m},\ 2.0\ \text{m})$ . Counterclockwise is positive, so $\hat{n}_1$ points out of the page — find the torque about $S = (0,\ 0)$ and about $S' = (4.0\ \text{m},\ 3.0\ \text{m})$ .

Torque about S = (0, 0)

Torque about S′ = (4.0 m, 3.0 m)

✅ Correct! About

S

the moment arm is

2.0\ \text{m}

and the turn is clockwise,

-60\ \text{N}\cdot\text{m}\ \hat{n}_1

; about

S'

it is

1.0\ \text{m}

and counterclockwise,

+30\ \text{N}\cdot\text{m}\ \hat{n}_1

. One force, two different torques.

❌ Check the sense. The line of action runs

2.0\ \text{m}

above

S

and the force points to the right, so it turns the page clockwise about

S

: the component along

\hat{n}_1

is negative.

❌ Not quite.

|\vec{r}_{S,P}| = \sqrt{(1.0)^2 + (2.0)^2} = 2.24\ \text{m}

is the distance to

P

, not the moment arm. For a horizontal line of action the moment arm is the vertical distance from the reference point to that line.

❌ Not quite. A horizontal offset is measured along the line of action, and sliding a force along its own line changes nothing. The moment arm is the perpendicular (here vertical) distance from the reference point to the line

y = 2.0\ \text{m}

❌ Not quite. That is the torque about

S

. About

S'

the position vector, the moment arm and even the sense of rotation change: there is no torque of a force on its own, only a torque about a chosen point.

❌ Check the sense. The line of action passes

1.0\ \text{m}

below

S'

, and a rightward push below a pivot turns counterclockwise, so the component along

\hat{n}_1

is positive.

Show solution

Step 1 — the line of action. The force points along $+x$ through $P = (1.0\ \text{m},\ 2.0\ \text{m})$ , so its line of action is the horizontal line $y = 2.0\ \text{m}$ . Each moment arm is the vertical distance from the reference point to that line.

Step 2 — about $S = (0,\ 0)$ . The moment arm is $r_\perp = 2.0\ \text{m}$ , so $\tau_S = r_\perp F = (2.0\ \text{m})(30\ \text{N}) = 60\ \text{N}\cdot\text{m}$ . The line runs above $S$ and the force points right, so the page turns clockwise about $S$ :

\vec{\tau}_S = -60\ \text{N}\cdot\text{m}\ \hat{n}_1

Step 3 — about $S' = (4.0\ \text{m},\ 3.0\ \text{m})$ . The line runs $1.0\ \text{m}$ below $S'$ , so $\tau_{S'} = (1.0\ \text{m})(30\ \text{N}) = 30\ \text{N}\cdot\text{m}$ , and a rightward push below the pivot turns counterclockwise:

\vec{\tau}_{S'} = +30\ \text{N}\cdot\text{m}\ \hat{n}_1

Check with the vector product ( $\hat{k} = \hat{n}_1$ out of the page):

\vec{r}_{S,P} \times \vec{F}_P = (1.0\,\hat{\imath} + 2.0\,\hat{\jmath})\ \text{m} \times 30\,\hat{\imath}\ \text{N} = 60\ \text{N}\cdot\text{m}\,(\hat{\jmath} \times \hat{\imath}) = -60\ \text{N}\cdot\text{m}\ \hat{k}

\vec{r}_{S',P} \times \vec{F}_P = (-3.0\,\hat{\imath} - 1.0\,\hat{\jmath})\ \text{m} \times 30\,\hat{\imath}\ \text{N} = -30\ \text{N}\cdot\text{m}\,(\hat{\jmath} \times \hat{\imath}) = +30\ \text{N}\cdot\text{m}\ \hat{k}

The same force gives torques of different size and opposite sign about the two points.

Problem 4 · One Turntable, Two Viewers

Given: A horizontal glass turntable of diameter $1.00\ \text{m}$ turns about a vertical axle through its centre $S$ . A $12\ \text{N}$ force acts at a point $P$ on the rim, along the rim's tangent, so that a viewer looking down from above sees it turn the table counterclockwise. A second viewer lies beneath the turntable, looking up, and declares clockwise, as she sees it, to be positive — find the direction of $\vec{\tau}_S$ and the signed torque she records.

Direction of the torque about S

Signed torque recorded by the viewer below

✅ Excellent!

\tau_S = RF = (0.50\ \text{m})(12\ \text{N}) = 6.0\ \text{N}\cdot\text{m}

along the upward vertical. Seen from below the turn is clockwise, so she records

+6.0\ \text{N}\cdot\text{m}

, while a viewer above who called clockwise positive would record

-6.0\ \text{N}\cdot\text{m}

for the same vector.

❌ Not quite. A torque never lies in the plane of

\vec{r}_{S,P}

and

\vec{F}_P

; it points along the normal to that plane, here along the vertical axle.

❌ Check the right-hand rule. Curl the fingers of your right hand counterclockwise as seen from above: the thumb points up, toward the viewer above.

❌ Not quite. The moment arm is the radius,

0.50\ \text{m}

: the tangent line at the rim is perpendicular to the radius, so its distance from

S

R

, not the diameter.

❌ Close, but that is the reading of a viewer above who calls clockwise positive. From below, the same turn looks clockwise, and clockwise, as she sees it, is what she calls positive.

❌ Not quite. Check both the moment arm (the radius, since a tangent at the rim is perpendicular to the radius) and which way the table turns as seen from below.

Show solution

Step 1 — magnitude. The force is tangent to the rim, so it is perpendicular to $\vec{r}_{S,P}$ ( $\theta = 90\degree$ ) and the moment arm is the radius $R = 0.50\ \text{m}$ :

\tau_S = RF = (0.50\ \text{m})(12\ \text{N}) = 6.0\ \text{N}\cdot\text{m}

Step 2 — direction. Curl the fingers of your right hand counterclockwise as seen from above; the thumb points up. So $\vec{\tau}_S$ is $6.0\ \text{N}\cdot\text{m}$ vertically up, along the axle — never in the plane of $\vec{r}_{S,P}$ and $\vec{F}_P$ .

Step 3 — the viewer below. She sees the mirror image: for her the table turns clockwise. Her declared positive sense is clockwise as she sees it; curling the fingers that way points the thumb away from her, which is up. The torque points along her positive normal, so she records

+6.0\ \text{N}\cdot\text{m}

Contrast. A viewer above who declared clockwise positive would use the downward normal and record $-6.0\ \text{N}\cdot\text{m}$ . The torque vector is identical in both cases; only the declared normal differs, which is why a convention must say which side the plane is viewed from.

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