Classical-Mechanics · Unit 21 · Video 2 · Interactive Practice

Torque by Components: A Force in Space, a Tilted Lever, and the Ankle on Tiptoe

IKey Formulas

Formula	Name	What it needs
$\vec{\tau}_S = \vec{r}_{S,F}\times\vec{F}$ , with $\hat{i}\times\hat{j} = \hat{k}$ , $\hat{j}\times\hat{k} = \hat{i}$ , $\hat{k}\times\hat{i} = \hat{j}$ , $\hat{i}\times\hat{i} = \vec{0}$	Torque by components	A right-handed set of unit vectors; $\vec{r}_{S,F}$ runs from $S$ to the point where $\vec{F}$ acts
$x\hat{i}\times(F_x\hat{i} + F_z\hat{k}) = -xF_z\,\hat{j}$	A force in space	$\hat{i}\times\hat{k} = -\hat{j}$ ; the component $F_x$ along $\vec{r}$ drops out
$\vec{\tau}_S = (L\cos\theta\,\hat{i} + L\sin\theta\,\hat{j})\times(-F\hat{j}) = -FL\cos\theta\,\hat{k}$	Tilted lever	$\theta$ above the horizontal; moment arm $L\cos\theta$ ; $-\hat{k}$ is clockwise
$\vec{\tau}_{S,T} = -bT\sin\alpha\,\hat{k}$ , $\vec{\tau}_{S,N} = sN\,\hat{k} = \tfrac{1}{2}s\,mg\,\hat{k}$ , $\vec{\tau}_{S,F} = \vec{0}$	Ankle on tiptoe	$\hat{i}$ right, $\hat{j}$ up, $\hat{k} = \hat{i}\times\hat{j}$ out of the screen; $\vec{F}$ acts at $S$ itself

Key Insight: Only the part of $\vec{F}$ perpendicular to $\vec{r}$ makes torque. Put the reference point on the line of action of an unknown force and that force leaves the calculation: with the foot at rest the two ankle torques cancel, $bT\sin\alpha = \tfrac{1}{2}mg\,s$ , so the tendon's vertical pull is $\tfrac{s}{b}$ times half the body weight.

IIVisualization 1 — A Force in Space

With $\vec{r}_{P,F} = x\hat{i}$ the torque is $-xF_z\,\hat{j}$ : it depends on $F_z$ and never on $F_x$ .

x (m) 3.0 F_x (N) 2.0 F_z (N) 1.5

IIIVisualization 2 — The Tilted Lever

A downward push $F$ at the tip of a lever tilted $\theta$ above the horizontal gives $|\vec{\tau}_S| = FL\cos\theta$ , clockwise.

θ (°) 35° Reading

IVVisualization 3 — Three Forces on the Ankle

About the ankle joint $S$ the tibia force drops out, $h$ never enters, and the tendon's torque cancels the floor's.

s (cm) 10 h (cm) 8 α (°) 55° β (°) 18°

VQuiz Questions

Problem 1 · A Force Out of the Line

Given: A force $\vec{F} = (3.0\,\hat{i} + 4.0\,\hat{k})\ \text{N}$ acts at the point $\vec{r}_{P,F} = 2.0\,\hat{i}\ \text{m}$ from $P$ , with $(\hat{i}, \hat{j}, \hat{k})$ right-handed — find the torque $\vec{r}_{P,F}\times\vec{F}$ about $P$ .

✅ Correct! Only

F_z = 4.0\ \text{N}

is perpendicular to

\vec{r}_{P,F}

xF_z = 8.0\ \text{N}\cdot\text{m}

, directed along

\hat{i}\times\hat{k} = -\hat{j}

❌ Check the sign. The cyclic product is

\hat{k}\times\hat{i} = +\hat{j}

, so reversing the order gives

\hat{i}\times\hat{k} = -\hat{j}

: the torque points along

-\hat{j}

❌ Not the whole force.

x|\vec{F}| = (2.0\ \text{m})(5.0\ \text{N})

would require

\vec{F}

perpendicular to

\vec{r}_{P,F}

. Here

F_x

lies along

\vec{r}_{P,F}

and

\hat{i}\times\hat{i} = \vec{0}

, so only

F_z

counts.

❌ That is the component along $\vec{r}$ .

x\hat{i}\times F_x\hat{i} = xF_x\,(\hat{i}\times\hat{i}) = \vec{0}

: the

3.0\ \text{N}

along the position vector makes no torque. The torque comes from

F_z

Show solution

Expand term by term, using $\hat{i}\times\hat{i} = \vec{0}$ and $\hat{i}\times\hat{k} = -\hat{j}$ :

\vec{r}_{P,F}\times\vec{F} = 2.0\,\hat{i}\times(3.0\,\hat{i} + 4.0\,\hat{k}) = (2.0)(3.0)(\hat{i}\times\hat{i}) + (2.0)(4.0)(\hat{i}\times\hat{k}) = \vec{0} + 8.0\,(-\hat{j})

\vec{r}_{P,F}\times\vec{F} = -8.0\,\hat{j}\ \text{N}\cdot\text{m}

The magnitude is the perpendicular component times the distance, $F_z\,x = (4.0\ \text{N})(2.0\ \text{m}) = 8.0\ \text{N}\cdot\text{m}$ ; the $3.0\ \text{N}$ along $\vec{r}_{P,F}$ contributes nothing.

Problem 2 · Which Angle Goes in the Cosine?

Given: A lever of length $L = 0.40\ \text{m}$ pivots at $S$ and leans to the right, making $30\degree$ with the vertical. A force of magnitude $F = 50\ \text{N}$ pushes straight down on its far end. With $\hat{i}$ right, $\hat{j}$ up and $\hat{k} = \hat{i}\times\hat{j}$ out of the screen — find $\vec{\tau}_S$ .

✅ Correct! Measured from the horizontal the lever sits at

\theta = 60\degree

, so the moment arm is

L\cos 60\degree = 0.20\ \text{m}

, and the downward push turns the lever clockwise, along

-\hat{k}

❌ Wrong angle in the cosine. In

FL\cos\theta

the angle

\theta

is measured from the horizontal:

\theta = 90\degree - 30\degree = 60\degree

. Using

\cos 30\degree

gives

FL\cos 30\degree \approx 17.3\ \text{N}\cdot\text{m}

, the torque of a lever

30\degree

above the horizontal.

❌ Check the direction.

L\cos\theta\,\hat{i}\times(-F\hat{j}) = -FL\cos\theta\,\hat{k}

: a downward push on a tip to the right of

S

turns the lever clockwise, along

-\hat{k}

, into the screen.

❌ That is the horizontal-lever value.

FL

uses the whole length as the moment arm. The perpendicular distance from

S

to the vertical line of action is only

L\cos 60\degree = 0.20\ \text{m}

Show solution

First convert the angle: $30\degree$ from the vertical is $\theta = 60\degree$ above the horizontal, with $\cos 60\degree = \tfrac{1}{2}$ and $\sin 60\degree = \tfrac{\sqrt{3}}{2}$ . Then

\vec{r}_{S,F} = L\cos 60\degree\,\hat{i} + L\sin 60\degree\,\hat{j} = \left(0.20\,\hat{i} + 0.20\sqrt{3}\,\hat{j}\right)\ \text{m}, \qquad \vec{F} = -50\,\hat{j}\ \text{N}

\vec{\tau}_S = (0.20)(-50)(\hat{i}\times\hat{j}) + (0.20\sqrt{3})(-50)(\hat{j}\times\hat{j}) = -10\,\hat{k} + \vec{0} = -10\,\hat{k}\ \text{N}\cdot\text{m}

Check with the moment arm: the line of action is vertical, a horizontal distance $L\cos 60\degree = 0.20\ \text{m}$ from $S$ , so $|\vec{\tau}_S| = (50\ \text{N})(0.20\ \text{m}) = 10\ \text{N}\cdot\text{m}$ , clockwise.

Problem 3 · The Tendon on Tiptoe

Given: A person of mass $m = 60\ \text{kg}$ crouches on tiptoe with the weight shared evenly between both feet ( $g = 9.8\ \text{m/s}^2$ ). For one foot, the floor contact is $s = 10\ \text{cm}$ in front of the ankle joint $S$ and $h = 7.5\ \text{cm}$ below it; the Achilles tendon attaches $b = 4.0\ \text{cm}$ behind $S$ , level with it, and pulls at $\alpha = 60\degree$ above the horizontal. With $\hat{i}$ toward the toes, $\hat{j}$ up and $\hat{k}$ out of the screen — find the torque of the normal force about $S$ , then the tendon tension $T$ that keeps the foot at rest.

Torque of the normal force about S

Tendon tension T

✅ Correct! The floor's torque

+29.4\,\hat{k}\ \text{N}\cdot\text{m}

must be cancelled by the tendon's

-bT\sin\alpha\,\hat{k}

, which demands

T \approx 849\ \text{N}

, more than the whole body weight

mg = 588\ \text{N}

❌ Check the sign.

s\,\hat{i}\times N\hat{j} = sN\,\hat{k}

: the floor pushes the toes up, in front of

S

, turning the foot counterclockwise, along

+\hat{k}

❌ That uses $|\vec{r}_{S,N}| = 0.125\ \text{m}$ as the moment arm. The part

-h\hat{j}

\vec{r}_{S,N}

is parallel to

\vec{N}

and

\hat{j}\times\hat{j} = \vec{0}

, so

h

drops out; the moment arm is the horizontal distance

s = 0.10\ \text{m}

❌ That is the full weight. The weight is shared by two feet, so the floor pushes on one foot with

N = \tfrac{1}{2}mg = 294\ \text{N}

, not

mg = 588\ \text{N}

❌ That is only the vertical part of the pull. The balance fixes

T\sin\alpha = 735\ \text{N}

; divide by

\sin 60\degree

to get

T

itself.

❌ Wrong component. With

\alpha

measured from the horizontal, the part of

\vec{T}

perpendicular to

\vec{r}_{S,T} = -b\,\hat{i}

is the vertical

T\sin\alpha

, not

T\cos\alpha

❌ Check the normal force. Balancing against

mg = 588\ \text{N}

doubles the answer; one foot carries only

N = \tfrac{1}{2}mg = 294\ \text{N}

Show solution

Normal force. One foot carries half the weight, $N = \tfrac{1}{2}mg = \tfrac{1}{2}(60\ \text{kg})(9.8\ \text{m/s}^2) = 294\ \text{N}$ , acting at $\vec{r}_{S,N} = (0.10\,\hat{i} - 0.075\,\hat{j})\ \text{m}$ :

\vec{\tau}_{S,N} = (0.10\,\hat{i} - 0.075\,\hat{j})\times 294\,\hat{j} = (0.10)(294)\,\hat{k} + \vec{0} = +29.4\,\hat{k}\ \text{N}\cdot\text{m}

Tendon. With $\vec{r}_{S,T} = -b\,\hat{i}$ and $\vec{T} = T\cos\alpha\,\hat{i} + T\sin\alpha\,\hat{j}$ , $\ \vec{\tau}_{S,T} = -bT\sin\alpha\,\hat{k}$ . The tibia force acts at $S$ , so it contributes $\vec{0}$ . At rest the two torques cancel:

bT\sin\alpha = sN \quad\Longrightarrow\quad T\sin\alpha = \frac{s}{b}\,N = \frac{0.10\ \text{m}}{0.040\ \text{m}}\,(294\ \text{N}) = 735\ \text{N}

T = \frac{735\ \text{N}}{\sin 60\degree} = 490\sqrt{3}\ \text{N} \approx 849\ \text{N}

Problem 4 · Moving the Reference Point

Given: The same foot, but torques are now taken about the tendon's attachment point $H$ on the heel. The ankle joint $S$ lies a horizontal distance $b$ in front of $H$ , at the same height, and the tibia pushes on the foot at $S$ with $\vec{F} = -F\sin\beta\,\hat{i} - F\cos\beta\,\hat{j}$ (down and backward, $\beta$ from the vertical) — find $\vec{\tau}_{H,F}$ , and which of the three forces now makes no torque.

Torque of the tibia force about H

Which force makes no torque about H?

✅ Correct! About

H

the tendon drops out but the unknown

F

and

\beta

enter through

-bF\cos\beta\,\hat{k}

. About

S

it is the other way round, and there the only unknown left is

T

, which is why

S

makes the ankle problem solvable.

❌ Zero only about $S$ .

\vec{F}

acts at

S

, and from

H

that point sits at

\vec{r}_{H,F} = b\,\hat{i} \neq \vec{0}

, so about

H

the tibia force has the nonzero moment arm

b\cos\beta

❌ Wrong component.

-F\sin\beta\,\hat{i}

lies along

\vec{r}_{H,F} = b\,\hat{i}

, and

\hat{i}\times\hat{i} = \vec{0}

. The vertical part

-F\cos\beta\,\hat{j}

does the turning.

❌ Check the sign.

b\,\hat{i}\times(-F\cos\beta\,\hat{j}) = -bF\cos\beta\,\hat{k}

: a downward push in front of

H

turns the foot clockwise.

❌ Not any more.

\vec{F}

acts at

S

, a distance

b

in front of

H

, so its torque about

H

-bF\cos\beta\,\hat{k}

❌ $\vec{N}$ still has a moment arm. Its vertical line of action is the horizontal distance

d = b + s

from

H

, so

\vec{\tau}_{H,N} = dN\,\hat{k}

❌ One force acts at $H$ itself. The tendon pulls at

H

, so

\vec{r}_{H,T} = \vec{0}

and

\vec{\tau}_{H,T} = \vec{0}

Show solution

Tibia. From $H$ the point of application is $\vec{r}_{H,F} = b\,\hat{i}$ :

\vec{\tau}_{H,F} = b\,\hat{i}\times(-F\sin\beta\,\hat{i} - F\cos\beta\,\hat{j}) = -bF\sin\beta\,(\hat{i}\times\hat{i}) - bF\cos\beta\,(\hat{i}\times\hat{j}) = -bF\cos\beta\,\hat{k}

Tendon. It pulls at $H$ , so $\vec{r}_{H,T} = \vec{0}$ and $\vec{\tau}_{H,T} = \vec{0}$ .

Floor. $\vec{r}_{H,N} = (b + s)\,\hat{i} - h\,\hat{j}$ , so $\vec{\tau}_{H,N} = (b + s)N\,\hat{k} = dN\,\hat{k}$ .

The torque equation about $H$ therefore contains the two unknowns $F$ and $\beta$ and loses $T$ , the quantity we wanted. About $S$ the tibia force is the one that drops out, leaving $T$ as the single unknown.

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