Classical-Mechanics · Unit 22 · Video 5 · Interactive Practice

The Knee Joint, and Why Any Torque Point Will Do

IKey Formulas

Formula	Name	What you need
$\vec{\tau}_{O,i} = \vec{r}_{O,i} \times \vec{F}_i$	Torque about a point $O$	The position vector from $O$ to where $\vec{F}_i$ acts
$d\,T\cos\theta + \tfrac{1}{10}x\,mg - \tfrac{1}{2}s\,mg = 0$	Torque balance at the knee	$d$ , $x$ , $s$ , $\theta$
$T = \dfrac{mg\left(\tfrac{s}{2} - \tfrac{x}{10}\right)}{d\cos\theta}$	Ligament tension	Net load arm $\div$ ligament arm
$\vec{\tau}_A = \vec{\tau}_B \ \text{ whenever } \ \sum_i \vec{F}_i = \vec{0}$	Torques about any two points	Zero net force — nothing else

Key Insight: Take torques about the point where the force you know least about acts — its position vector vanishes, so that unknown leaves the equation before you write it.

IIVisualization 1 — Where to Take Torques

Every point gives a torque equation that sums to zero; only a few of them drop an unknown out.

Torque point O: x (cm) 0.0 Torque point O: y (cm) −3.8

IIIVisualization 2 — Why the Ligament Pulls Five Body Weights

The ligament resists at an arm of only $d\cos\theta$ against a net load arm of $\tfrac{s}{2} - \tfrac{x}{10}$ .

Ligament angle θ (deg) 40 Attachment distance d (cm) 3.8

IVVisualization 3 — Any Two Points

Two points, two torque sums — they agree exactly when the forces add to zero.

Forces A: x (m) −1.5 A: y (m) −1.0 B: x (m) 1.5 B: y (m) 1.5

💡 Gravity counts once in that sum: the weights of all the particles are replaced by a single force at the centre of gravity, so a body of $10^{25}$ particles still enters the proof as a handful of forces.

VQuiz Questions

Problem 1 · The Ligament Tension

Given: the crouching sprinter of the video, $m = 70\ \text{kg}$ so $mg = 686\ \text{N}$ , with $s = 36\ \text{cm}$ , $x = 18\ \text{cm}$ , $d = 3.8\ \text{cm}$ , $\theta = 40\degree$ and $N = \tfrac{1}{2}mg$ — find the patellar ligament tension $T$ .

✅ Correct! A net load arm of

16.2\ \text{cm}

against a ligament arm of

2.91\ \text{cm}

multiplies the weight by

5.57

T = 3.8 \times 10^{3}\ \text{N}

, about five and a half body weights.

❌ That is $mg(s/2 - x/10)/d$ — the $\cos\theta$ is missing. The tension pulls at

40\degree

to the horizontal, so only the component perpendicular to

\vec{r}_{P,S}

turns the shin: the effective arm is

d\cos\theta = 2.91\ \text{cm}

, not

d = 3.8\ \text{cm}

❌ The leg's own weight was left out. It acts

x = 18\ \text{cm}

behind the line of the femur contact and turns the shin the opposite way from

\vec{N}

, cutting the load arm from

s/2 = 18\ \text{cm}

s/2 - x/10 = 16.2\ \text{cm}

❌ That used $N = mg$ . The sprinter's weight is shared equally by two legs, so the ground pushes on this foot with

N = \tfrac{1}{2}mg = 343\ \text{N}

❌ Not quite. Take torques about

S

, where the tension acts, then use

F\sin\alpha = T\cos\theta

from the horizontal force equation.

Show solution

Step 1: torques about the ligament attachment $S$ . The tension acts at $S$ , so $\vec{\tau}_{S,T} = \vec{0}$ . The femur force acts at $d\,\hat{j}$ from $S$ , and the two vertical forces contribute only through their horizontal distances $x$ and $s$ :

d\,F\sin\alpha + \tfrac{1}{10}x\,mg - \tfrac{1}{2}s\,mg = 0

Step 2: eliminate $F\sin\alpha$ . The horizontal force equation is $-F\sin\alpha + T\cos\theta = 0$ , so $F\sin\alpha = T\cos\theta$ :

d\,T\cos\theta + \tfrac{1}{10}x\,mg - \tfrac{1}{2}s\,mg = 0 \quad \Longrightarrow \quad T = \frac{mg\left(\tfrac{s}{2} - \tfrac{x}{10}\right)}{d\cos\theta}

Step 3: the numbers. $\tfrac{s}{2} = 18.0\ \text{cm}$ , $\tfrac{x}{10} = 1.8\ \text{cm}$ , and $d\cos 40\degree = (3.8\ \text{cm})(0.766) = 2.91\ \text{cm}$ :

T = \frac{(686\ \text{N})(16.2\ \text{cm})}{2.91\ \text{cm}} = 3.8 \times 10^{3}\ \text{N}

The centimetres cancel, so there is no need to convert to metres.

Problem 2 · The Offsets That Never Appear

Given: taking torques about $S$ , the leg's weight acts at $-x\,\hat{i} - y_L\,\hat{j}$ and the normal force at $-s\,\hat{i} - y_N\,\hat{j}$ , yet neither $y_L$ nor $y_N$ survives into $d\,F\sin\alpha + \tfrac{1}{10}x\,mg - \tfrac{1}{2}s\,mg = 0$ — why?

✅ Correct! A vertical force's torque is

r_\perp F

with

r_\perp

the horizontal distance to its line of action — the vertical part of the position vector is parallel to the force and crosses to zero.

❌ Nothing was approximated.

y_L

and

y_N

cancel exactly, however large they are — slide the weight up or down its own line of action and its torque about

S

does not change at all.

❌ That choice removed a different term. Putting

S

on that vertical line is what makes the horizontal distances of

\vec{N}

and the weight equal to

s

and

x

; it is the verticality of the forces that kills

y_L

and

y_N

❌ The terms are not cancelling against each other. Each of the two torques loses its own offset on its own, before anything is added:

\vec{\tau}_{S,2} = \tfrac{1}{10}x\,mg\,\hat{k}

and

\vec{\tau}_{S,3} = -\tfrac{1}{2}s\,mg\,\hat{k}

contain no offsets at all.

❌ Not quite. Write out

(-x\,\hat{i} - y_L\,\hat{j}) \times (-\tfrac{1}{10}mg\,\hat{j})

term by term and see which piece dies.

Show solution

Expand the cross product for the leg's weight, using $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = \vec{0}$ :

\vec{\tau}_{S,2} = (-x\,\hat{i} - y_L\,\hat{j}) \times \left(-\tfrac{1}{10}mg\,\hat{j}\right) = \tfrac{1}{10}x\,mg\,\hat{k} + \vec{0}

The $-y_L\,\hat{j}$ piece of the position vector is parallel to the force, so it contributes nothing. The same happens for $\vec{N}$ :

\vec{\tau}_{S,3} = (-s\,\hat{i} - y_N\,\hat{j}) \times (N\hat{j}) = -sN\,\hat{k} = -\tfrac{1}{2}s\,mg\,\hat{k}

This is the general statement that a force's torque depends only on the perpendicular distance from the point to its line of action: sliding a force along that line changes no torque. The femur force is not vertical, which is why its own vertical distance $d$ from $S$ is exactly what does appear, as $d\,F\sin\alpha$ .

Problem 3 · Direction and Size of the Femur Force

Given: the same knee with $T = 3.8 \times 10^{3}\ \text{N}$ , $\theta = 40\degree$ , $mg = 686\ \text{N}$ and $N = \tfrac{1}{2}mg$ , so that $F\sin\alpha = T\cos\theta = 2911\ \text{N}$ and $F\cos\alpha = \tfrac{2}{5}mg + T\sin\theta = 274.4\ \text{N} + 2443\ \text{N}$ — find the angle $\alpha$ from the vertical and the magnitude $F$ .

What is the angle $\alpha$ ?

What is the magnitude $F$ ?

✅ Correct!

\cot\alpha = 2717.4/2911 = 0.933

gives

\alpha = 47\degree

, and

F = 2911\ \text{N}/\sin 47\degree = 4.0 \times 10^{3}\ \text{N}

— nearly six body weights pressed through the joint.

❌ That is $\arctan(0.933)$ . Dividing the vertical equation by the horizontal one gives

\dfrac{F\cos\alpha}{F\sin\alpha} = \cot\alpha

, not

\tan\alpha

. Since

\cot\alpha = 0.933

\tan\alpha = 1/0.933 = 1.072

and

\alpha = 47\degree

❌ The $\tfrac{2}{5}mg$ term was dropped. Without it

\cot\alpha = T\sin\theta/T\cos\theta = \tan\theta

, which would give

\alpha = 90\degree - \theta = 50\degree

. The ground and the leg's weight leave

\tfrac{2}{5}mg = 274.4\ \text{N}

for the femur to carry as well.

❌ The $T\sin\theta$ term was dropped. That leaves

\cot\alpha = 274.4/2911 = 0.094

and

\alpha = 85\degree

. The ligament pulls upward at

40\degree

, and the femur must press down against that

2443\ \text{N}

as well as against the ground.

❌ That is $T\cos\theta$ , the horizontal component alone. It equals

F\sin\alpha

, so divide by

\sin 47\degree

to recover the whole vector:

F = 2911/0.731

❌ $\sin$ and $\cos$ are swapped.

\alpha

is measured from the vertical, so the horizontal component is

F\sin\alpha

and

F = T\cos\theta/\sin\alpha

, not

T\cos\theta/\cos\alpha

❌ That is the tension again. The femur force also carries

\tfrac{2}{5}mg

, so it must come out larger than

T

5.8

body weights against

5.5

❌ Check the angle. Divide

F\cos\alpha

F\sin\alpha

; the

F

cancels and what is left is

\cot\alpha

❌ Check the magnitude. Use either component with its own trigonometric function:

F = T\cos\theta/\sin\alpha

Show solution

Step 1: divide one force equation by the other. The $F$ cancels:

\cot\alpha = \frac{F\cos\alpha}{F\sin\alpha} = \frac{\tfrac{2}{5}mg + T\sin\theta}{T\cos\theta} = \frac{274.4\ \text{N} + 2443\ \text{N}}{2911\ \text{N}} = 0.933

\alpha = \operatorname{arccot}(0.933) = 47\degree

Step 2: put $\alpha$ back into either component.

F = \frac{T\cos\theta}{\sin\alpha} = \frac{2911\ \text{N}}{\sin 47\degree} = 4.0 \times 10^{3}\ \text{N}

Rounding check. Carrying the unrounded tension instead, $\cot\alpha = \tan\theta + \dfrac{2d/5}{\tfrac{s}{2} - \tfrac{x}{10}} = 0.839 + \dfrac{1.52\ \text{cm}}{16.2\ \text{cm}} = 0.933$ , the same value: $mg$ and $\cos\theta$ cancel out of the ratio, so the rounding of $T$ changed nothing.

Problem 4 · Moving the Torque Point

Given: a rigid body carries forces with vector sum $\sum_i \vec{F}_i = -12\ \text{N}\,\hat{j}$ ; the torque about a point $A$ is $\vec{\tau}_A = 8\ \text{N}\cdot\text{m}\,\hat{k}$ and the vector from $A$ to $B$ is $\vec{r}_{A,B} = 0.5\ \text{m}\,\hat{i}$ — find $\vec{\tau}_B$ .

✅ Correct!

\vec{r}_{A,B} \times \sum_i \vec{F}_i = (0.5\,\hat{i}) \times (-12\,\hat{j}) = -6\,\hat{k}

, so

\vec{\tau}_B = \vec{\tau}_A - (-6\,\hat{k}) = 14\ \text{N}\cdot\text{m}\,\hat{k}

❌ The correction was added instead of subtracted. The substitution runs

\vec{\tau}_A = \vec{r}_{A,B} \times \sum_i \vec{F}_i + \vec{\tau}_B

, so

\vec{\tau}_B = \vec{\tau}_A - \vec{r}_{A,B} \times \sum_i \vec{F}_i = 8 - (-6) = 14

❌ The theorem does not apply here.

\vec{\tau}_A = \vec{\tau}_B

needs

\sum_i \vec{F}_i = \vec{0}

; this body has a net force of

12\ \text{N}

downward, so the torque really does depend on the point.

❌ That is only the correction term $\vec{r}_{A,B} \times \sum_i \vec{F}_i$ . It still has to be removed from

\vec{\tau}_A = 8\ \text{N}\cdot\text{m}\,\hat{k}

❌ Not quite. Start from

\vec{r}_{A,i} = \vec{r}_{A,B} + \vec{r}_{B,i}

and sum the cross products with

\vec{F}_i

Show solution

Step 1: the general relation. Substituting $\vec{r}_{A,i} = \vec{r}_{A,B} + \vec{r}_{B,i}$ and pulling the fixed vector $\vec{r}_{A,B}$ out of the sum:

\vec{\tau}_A = \sum_i \vec{r}_{A,i} \times \vec{F}_i = \vec{r}_{A,B} \times \sum_i \vec{F}_i + \vec{\tau}_B

Step 2: the correction term. With $\hat{i} \times \hat{j} = \hat{k}$ ,

\vec{r}_{A,B} \times \sum_i \vec{F}_i = (0.5\ \text{m}\,\hat{i}) \times (-12\ \text{N}\,\hat{j}) = -6\ \text{N}\cdot\text{m}\,\hat{k}

Step 3: solve for $\vec{\tau}_B$ .

\vec{\tau}_B = \vec{\tau}_A - \vec{r}_{A,B} \times \sum_i \vec{F}_i = 8\ \text{N}\cdot\text{m}\,\hat{k} + 6\ \text{N}\cdot\text{m}\,\hat{k} = 14\ \text{N}\cdot\text{m}\,\hat{k}

If the body were in equilibrium the correction term would vanish and both torques would be $8\ \text{N}\cdot\text{m}\,\hat{k}$ — which is exactly the licence the knee problem used to move its torque point from $S$ to the femur contact $P$ .

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