Classical-Mechanics · Unit 22 · Video 4 · Interactive Practice

An Unknown Force in an Unknown Direction: The Hinged Rod and the Person on a Hill

IKey Formulas

Formula	Name	What you need
$\sum_i \vec{F}_i = \vec{0}, \qquad \sum_i \vec{\tau}_{S,i} = \vec{0}$	The two equilibrium conditions	In a plane: two force equations, one torque equation
$\vec{F}_{\text{pivot}} = F\left(\cos\alpha\,\hat{i} + \sin\alpha\,\hat{j}\right)$	A force of unknown direction	Two unknowns: magnitude $F$ and angle $\alpha$
$T = \dfrac{mg}{2\sin\beta}$	Cable tension, from torques about the hinge	Rod weight $mg$ and cable angle $\beta$
$N_{1,2} = mg\left(\tfrac{1}{2}\cos\alpha \pm \dfrac{h}{d}\sin\alpha\right)$	Normal force on the lower / upper foot	Slope $\alpha$ , height $h$ of the cm, stance $d$

Key Insight: Take torques about the point where the force you know least about is applied — its moment arm is zero, so it leaves the problem before it is ever written down.

IIVisualization 1 — Closing the Forces at the Hinge

Exactly one magnitude and one direction of the hinge force leave the rod with zero net force.

Cable angle β 30°

IIIVisualization 2 — Choosing the Torque Point

Every force whose line of action passes through the torque point drops out of the equation.

Torque point

Torques about the hinge $P$

T\,l\sin\beta \;-\; mg\,\tfrac{l}{2} \;=\; 0

T = \frac{mg}{2\sin\beta} = \frac{(4.0\ \text{kg})(9.8\ \text{m}\,\text{s}^{-2})}{2\sin 30\degree} = 39.2\ \text{N}

The pivot force acts at

P

, so it never enters. The tension comes out at once — the full weight of the rod.

Torques about the wall point $A$ , a height $l\tan\beta$ above the hinge

F\cos\alpha\,\left(l\tan\beta\right) = mg\,\tfrac{l}{2}

F\cos\alpha = \frac{mg}{2}\cot\beta = 33.9\ \text{N}

This is the

x

force equation, reached without ever writing the tension.

Torques about the far end $B$

F\sin\alpha\,l = mg\,\tfrac{l}{2}

F\sin\alpha = \frac{mg}{2} = 19.6\ \text{N}

This is the

y

force equation: the hinge carries half the weight, the cable the other half.

💡 Dividing the $B$ equation by the $A$ equation gives $\tan\alpha = \tan\beta$ , hence $\alpha = \beta$ and $F = T$ — with the tension never computed.

IVVisualization 3 — Standing on a Hill

Where the vertical through the centre of mass meets the slope decides how the feet share the weight.

Slope α 20° Stance d 1.00 m

VQuiz Questions

Problem 1 · The Cable Tension

Given: a uniform rod of length $l = 2.0\ \text{m}$ and mass $m = 4.0\ \text{kg}$ , hinged to a wall at its left end and held horizontal by a massless cable running from the wall to the rod's right end, meeting the rod at $\beta = 30\degree$ . With $g = 9.8\ \text{m}\,\text{s}^{-2}$ , find the tension $T$ .

✅ Correct!

T = mg/(2\sin\beta)

, and

\sin 30\degree = \tfrac{1}{2}

makes the denominator

1

: the tension comes out equal to the rod's full weight. (The cable still supports only half of it — that is the vertical component

T\sin\beta = mg/2

❌ That is $mg/2$ . Half the weight is the cable's vertical component

T\sin\beta

, not the tension itself.

❌ Wrong trigonometric function. Only the component perpendicular to the rod,

T\sin\beta

, produces torque about the hinge.

❌ Check the moment arm of the weight. A uniform rod's weight acts at its centre, a distance

l/2

from the hinge — not

l

❌ Not quite. Take torques about the hinge: the pivot force then contributes nothing.

Show solution

Take torques about the hinge, counterclockwise positive. The pivot force is applied at the hinge, so its moment arm is zero and it never enters.

The weight acts at the centre, moment arm $l/2$ , turning the rod clockwise. Only the perpendicular component of the tension, $T\sin\beta$ , produces torque, at distance $l$ , turning the rod counterclockwise:

T\,l\sin\beta - mg\,\tfrac{l}{2} = 0 \quad\Longrightarrow\quad T = \frac{mg}{2\sin\beta}

The length cancels — a longer rod of the same mass needs the same tension. With the numbers:

T = \frac{(4.0\ \text{kg})(9.8\ \text{m}\,\text{s}^{-2})}{2\sin 30\degree} = \frac{39.2\ \text{N}}{1} = 39.2\ \text{N}

The tension grows with the weight of the rod and falls as the cable steepens.

Problem 2 · Which Torques Vanish

Given: the same rod, but with torques taken about the point $A$ where the cable meets the wall, a height $l\tan\beta$ above the hinge. The pivot force is written as its components $F\cos\alpha$ (along the rod) and $F\sin\alpha$ (up the wall). Which forces contribute zero torque about $A$ ?

✅ Correct! Both lines of action run through

A

: the cable along its own line, and

F\sin\alpha

along the wall. What survives is

F\cos\alpha\,(l\tan\beta) = mg\,l/2

❌ One more force also passes through $A$ .

F\sin\alpha

acts at the hinge and points straight up the wall — and

A

sits on that same vertical line.

❌ The weight does produce torque here. It acts at the rod's centre, a horizontal distance

l/2

from the wall, so its moment arm about

A

l/2

❌ Only one component drops out.

F\cos\alpha

acts along the rod, a perpendicular distance

l\tan\beta

below

A

— that is its moment arm.

❌ Not quite. A force has zero torque about a point when its line of action passes through that point.

Show solution

A force contributes no torque about $A$ exactly when its line of action passes through $A$ .

Tension: the cable runs from the rod's end to $A$ , so its line of action passes through $A$ . Zero torque.
$F\sin\alpha$ : applied at the hinge, directed along the wall. $A$ lies on the wall, directly above the hinge. Zero torque.
$F\cos\alpha$ : applied at the hinge, directed along the rod. Its line is the rod itself, a perpendicular distance $l\tan\beta$ below $A$ .
Weight: vertical through the rod's centre, a horizontal distance $l/2$ from the wall.

Equating the two surviving magnitudes and cancelling $l$ :

F\cos\alpha\,(l\tan\beta) = mg\,\tfrac{l}{2} \quad\Longrightarrow\quad F\cos\alpha = \frac{mg}{2}\cot\beta

which is the horizontal force equation — obtained without the tension appearing at all.

Problem 3 · Sharing the Weight on a Slope

Given: a person of mass $m$ stands on a hillside sloping at $\alpha = 30\degree$ . The centre of mass sits a height $h = 1.0\ \text{m}$ above the hillside, measured perpendicular to the slope, midway between the feet, and the feet do not slip.

With the feet $d = 1.5\ \text{m}$ apart, what does the lower foot carry?

At what separation does the upper foot leave the ground?

✅ Correct! The two normals sum to

mg\cos\alpha = 0.87\,mg

, split

0.77\,mg

and

0.10\,mg

; they become

0.87\,mg

and

0

d = 2h\tan\alpha = 1.15\ \text{m}

❌ Check the lower foot's share.

N_1 = mg\left(\tfrac{1}{2}\cos\alpha + \tfrac{h}{d}\sin\alpha\right)

: the level-ground half, plus the tilt term

\tfrac{h}{d}\sin\alpha

the lower foot gains and the upper foot loses.

❌ Check the tipping condition.

N_2 = 0

requires

\tfrac{1}{2}\cos\alpha = \tfrac{h}{d}\sin\alpha

, so

d = 2h\tan\alpha

— tangent, and with the factor

2

Show solution

Step 1 — the force condition across the slope gives $N_1 + N_2 = mg\cos\alpha$ ; along the slope it gives $f_1 + f_2 = mg\sin\alpha$ .

Step 2 — torques about the centre of mass (the weight then drops out), with the friction pair acting a perpendicular distance $h$ below it and each normal force $d/2$ from it:

h\,(f_1 + f_2) + (N_2 - N_1)\,\tfrac{d}{2} = 0 \quad\Longrightarrow\quad N_1 - N_2 = \frac{2h\,mg\sin\alpha}{d}

Step 3 — add and subtract the sum and difference, then halve:

N_{1,2} = mg\left(\tfrac{1}{2}\cos\alpha \pm \frac{h}{d}\sin\alpha\right)

With $\alpha = 30\degree$ , $h = 1.0\ \text{m}$ , $d = 1.5\ \text{m}$ : $\tfrac{1}{2}\cos 30\degree = 0.433$ and $\tfrac{h}{d}\sin 30\degree = 0.333$ , so

N_1 = 0.77\,mg, \qquad N_2 = 0.10\,mg, \qquad N_1 + N_2 = 0.87\,mg = mg\cos 30\degree\ \checkmark

Step 4 — lift-off is $N_2 = 0$ :

\tfrac{1}{2}\cos\alpha = \frac{h}{d}\sin\alpha \quad\Longrightarrow\quad d = 2h\tan\alpha = 2(1.0\ \text{m})\tan 30\degree = 1.15\ \text{m}

Geometrically: the vertical through the centre of mass meets the slope a distance $h\tan\alpha = 0.58\ \text{m}$ downhill of the midpoint, and that is exactly $d/2$ when $d = 1.15\ \text{m}$ — the vertical then passes through the lower foot, the edge of the base of support.

Problem 4 · Why No Coefficient of Friction

Given: the hill problem was solved without ever using $\mu_s$ , and $f_1$ and $f_2$ were never found separately. Why was no friction model needed?

✅ Correct! The torque equation about the centre of mass contains friction only as

h\,(f_1 + f_2)

, and equilibrium along the slope already fixes that sum — the split between the feet is never needed.

❌ They do not cancel. Both frictions point uphill a perpendicular distance

h

below the centre of mass, so both turn the person the same way: their torque is

+h(f_1 + f_2)

❌ Friction is essential here. Without it the person slides:

f_1 + f_2 = mg\sin\alpha

is exactly what holds them in place.

❌ $\mu_s N$ is only the maximum. Static friction takes whatever value equilibrium demands, up to that limit; here the equations themselves fix it.

❌ Not quite. Ask which combination of

f_1

and

f_2

actually appears in the three equilibrium equations.

Show solution

Write the three equations and look at where friction appears:

\text{across the slope:}\quad N_1 + N_2 = mg\cos\alpha

\text{along the slope:}\quad f_1 + f_2 = mg\sin\alpha

\text{torques about the cm:}\quad h\,(f_1 + f_2) + (N_2 - N_1)\,\tfrac{d}{2} = 0

Friction enters only through the combination $f_1 + f_2$ , and the second equation already gives that sum. Substituting it into the third leaves

N_1 - N_2 = \frac{2h\,mg\sin\alpha}{d}

with no coefficient anywhere. The coefficient would matter only for a different question: whether the required $f_1 + f_2 = mg\sin\alpha$ is available at all, which needs $\mu_s \ge \tan\alpha$ .

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