Classical-Mechanics · Unit 22 · Video 3 · Interactive Practice

The Generalized Lever Law: Forces Applied at an Angle

IKey Formulas

Formula	Name	What it needs
$F_{\text{pivot}} = (m_b + m_1 + m_2)\,g$	Force condition, Example 18.1	All forces vertical; the pivot holds beam and bodies: $(2.9\ \text{kg})(9.8\ \text{m}\cdot\text{s}^{-2}) \approx 28\ \text{N}$ (two significant figures)
$d_1 N_1 = d_2 N_2 \;\Longrightarrow\; d_2 = \dfrac{d_1 m_1}{m_2}$	Lever law	Torques about the pivot, where $F_{\text{pivot}}$ and $m_b\vec{g}$ have zero moment arm; $g$ cancels
$F_{i,\parallel} = F_i\cos\theta_i$ , $\quad F_{i,\perp} = F_i\sin\theta_i$	Resolving a tilted force	Each $\theta_i$ is measured from the horizontal pointing away from the pivot on that force's own side, with $0 \le \theta_i \le \pi$ , so $F_{i,\perp} \ge 0$ and the two parallel parts point opposite ways
$d_1\left\|F_{1,\perp}\right\| = d_2\left\|F_{2,\perp}\right\| \;\Longleftrightarrow\; d_1 F_1 \sin\theta_1 = d_2 F_2 \sin\theta_2$	Generalized lever law	Exactly the statement $(\tau_{S,\text{total}})_z = 0$ about the pivot, counterclockwise positive

Key Insight: The parallel part of a tilted force acts along the beam, and that line passes through the pivot, so its moment arm is zero: it enters only the sideways force condition $F_1\cos\theta_1 = F_2\cos\theta_2$ , never the rotation. At $\theta_1 = \theta_2 = \tfrac{\pi}{2}$ both sines equal $1$ and the generalized law collapses to the lever law $d_1F_1 = d_2F_2$ .

IIVisualization 1 — The Lever Law in Numbers

Where must body 2 sit for the beam to stay level, and does the pivot force care?

d₂ (m) 0.350 m₂ (kg)

💡 The beam's own length never enters the balance — it only has to be long enough to carry both bodies, $d_1, d_2 \le l/2 = 0.5\ \text{m}$ .

IIIVisualization 2 — Resolving a Tilted Force

How much of a tilted force actually turns the beam about the pivot?

θ₁ (°) 30° F₁ (N) 8.0

IVVisualization 3 — Two Tilted Forces in Equilibrium

Two tilted forces balance the beam when $d_1F_1\sin\theta_1 = d_2F_2\sin\theta_2$ — perpendicular parts only.

θ₁ (°) 90° θ₂ (°) 90° F₂ (N) 8.0

VQuiz Questions

Problem 1 · Torque of a Tilted Force

Given: a force of magnitude $F = 12\ \text{N}$ acts on a beam at $d = 0.50\ \text{m}$ from the pivot, at $\theta = 30^\circ$ above the beam — find the magnitude of its torque about the pivot.

✅ Correct! Only

F_\perp = F\sin\theta = (12\ \text{N})(\tfrac{1}{2}) = 6.0\ \text{N}

turns the beam, and

(0.50\ \text{m})(6.0\ \text{N}) = 3.0\ \text{N}\cdot\text{m}

❌ That is $dF$ — the whole magnitude times the distance. The part of

\vec{F}

along the beam has its line of action through the pivot, so it contributes nothing.

❌ Wrong component.

F\cos\theta = 10.4\ \text{N}

is the part along the beam; the part that turns it is

F\sin\theta

❌ Half the right answer. The factor

\sin 30^\circ = \tfrac{1}{2}

belongs in the calculation once. Resolve the force once to get

F_\perp = 6.0\ \text{N}

, then multiply by the full distance

d = 0.50\ \text{m}

❌ Not quite. With

\sin 30^\circ = \tfrac{1}{2}

the perpendicular part is

6.0\ \text{N}

, and the torque is

d

times that.

Show solution

Split $\vec{F}$ into a part along the beam and a part perpendicular to it:

F_\parallel = F\cos\theta = (12\ \text{N})(0.866) = 10.4\ \text{N}, \qquad F_\perp = F\sin\theta = (12\ \text{N})\left(\tfrac{1}{2}\right) = 6.0\ \text{N}

$F_\parallel$ acts along the beam, whose line passes through the pivot: zero moment arm, zero torque. Only $F_\perp$ is left:

\left|\tau_S\right| = d\,F\sin\theta = (0.50\ \text{m})(12\ \text{N})\left(\tfrac{1}{2}\right) = 3.0\ \text{N}\cdot\text{m}

A force tilted at $30^\circ$ delivers exactly half the torque the same force would deliver perpendicular to the beam.

Problem 2 · The Same Lever on the Moon

Given: the worked example carried to the Moon, where $g = 1.6\ \text{m}\cdot\text{s}^{-2}$ : a uniform beam of mass $m_b = 2.0\ \text{kg}$ on a pivot beneath its centre, $m_1 = 0.3\ \text{kg}$ at $d_1 = 0.40\ \text{m}$ to the right, and $m_2 = 0.6\ \text{kg}$ at $d_2$ to the left.

What is $d_2$ ?

What force does the pivot exert on the beam?

✅ Correct! The balance distance is untouched because

g

cancels, but the pivot force is a weight and falls with

g

(2.9\ \text{kg})(1.6\ \text{m}\cdot\text{s}^{-2}) = 4.6\ \text{N}

❌ $g$ cancels. The lever law reads

d_1 m_1 g = d_2 m_2 g

; divide through by

g

and the arrangement that balances on Earth balances unchanged on the Moon.

❌ The ratio is inverted.

d_2 = d_1 m_1 / m_2

: the heavier body sits closer to the pivot, not further away.

❌ That is the Earth value.

28\ \text{N}

used

g = 9.8\ \text{m}\cdot\text{s}^{-2}

; the pivot force is a weight, so it changes with

g

❌ Something is missing from the sum. The pivot holds up everything above it — the beam and both bodies,

m_b + m_1 + m_2

Show solution

Distance. Torques about the pivot: $F_{\text{pivot}}$ and $m_b\vec{g}$ act at the pivot with zero moment arm, so

d_1 N_1 = d_2 N_2 \;\Longrightarrow\; d_2 = \frac{d_1 m_1 g}{m_2 g} = \frac{d_1 m_1}{m_2} = \frac{(0.40\ \text{m})(0.3\ \text{kg})}{0.6\ \text{kg}} = 0.20\ \text{m}

$g$ cancels, so $d_2 = 0.20\ \text{m}$ on the Moon exactly as on Earth. Check: $d_1m_1 = d_2m_2 = 0.12\ \text{kg}\cdot\text{m}$ .

Pivot force. Every force is vertical, so $F_{\text{pivot}} - m_b g - N_1 - N_2 = 0$ with $N_1 = m_1 g$ and $N_2 = m_2 g$ :

F_{\text{pivot}} = (m_b + m_1 + m_2)\,g = (2.9\ \text{kg})(1.6\ \text{m}\cdot\text{s}^{-2}) = 4.64\ \text{N} \approx 4.6\ \text{N}

The torque condition is blind to $g$ ; the force condition is proportional to it.

Problem 3 · Find the Balancing Angle

Given: $\vec{F}_1$ of magnitude $6.0\ \text{N}$ acts $d_1 = 0.40\ \text{m}$ to the right of the pivot at $\theta_1 = 90^\circ$ , and $\vec{F}_2$ of magnitude $16\ \text{N}$ acts $d_2 = 0.30\ \text{m}$ to the left — find the angle $\theta_2$ that leaves the beam in rotational equilibrium.

What is $d_1 F_1 \sin\theta_1$ ?

What is $\theta_2$ ?

✅ Correct!

\sin\theta_2 = 2.4 / 4.8 = \tfrac{1}{2}

, so the

16\ \text{N}

force must be tilted until only half of it acts across the beam.

❌ $\sin 90^\circ = 1$ , not $0$ . It is

\cos 90^\circ

that vanishes: a vertical force on a horizontal beam is entirely perpendicular to it.

❌ Check the decimal point.

(0.40\ \text{m})(6.0\ \text{N})(1) = 2.4\ \text{N}\cdot\text{m}

;

24\ \text{N}\cdot\text{m}

is ten times too large.

❌ Multiply, do not divide. The product is

d_1 F_1 \sin\theta_1

, so

(0.40\ \text{m})(6.0\ \text{N})(1)

❌ Too much torque. At

90^\circ

the second force delivers

(0.30)(16) = 4.8\ \text{N}\cdot\text{m}

, twice what is needed.

❌ That solves $\cos\theta_2 = \tfrac{1}{2}$ . The law uses the sine of the angle from the beam, so

\sin\theta_2 = \tfrac{1}{2}

❌ Along the beam it does nothing. At

\theta_2 = 0^\circ

the perpendicular part vanishes and

\vec{F}_1

is left unopposed.

❌ Not quite. Set

d_1F_1\sin\theta_1 = d_2F_2\sin\theta_2

and solve for

\sin\theta_2

Show solution

Step 1 — the torque that must be matched. With $\sin 90^\circ = 1$ ,

d_1 F_1 \sin\theta_1 = (0.40\ \text{m})(6.0\ \text{N})(1) = 2.4\ \text{N}\cdot\text{m}

Step 2 — impose the generalized lever law.

d_2 F_2 \sin\theta_2 = 2.4\ \text{N}\cdot\text{m} \;\Longrightarrow\; \sin\theta_2 = \frac{2.4\ \text{N}\cdot\text{m}}{(0.30\ \text{m})(16\ \text{N})} = \frac{2.4}{4.8} = \frac{1}{2}

Step 3 — solve on $0 \le \theta_2 \le \pi$ . $\theta_2 = 30^\circ$ satisfies this, and so does $\theta_2 = 150^\circ$ : the sine is symmetric about $90^\circ$ , so the force may lean towards the pivot or away from it and still carry the same perpendicular part $F_2\sin\theta_2 = 8.0\ \text{N}$ .

The two leanings differ in their parallel parts, $\pm F_2\cos\theta_2$ , which is settled by the sideways force condition, not by the torque.

Problem 4 · Along the Beam, and the Sense of the Turn

Given: $\vec{F}_1$ of magnitude $5.0\ \text{N}$ acts $d_1 = 0.60\ \text{m}$ to the right of the pivot, directed along the beam away from the pivot ( $\theta_1 = 0^\circ$ ); $\vec{F}_2$ acts $d_2 = 0.20\ \text{m}$ to the left at $\theta_2 = 90^\circ$ .

What magnitude $F_2$ leaves the beam in rotational equilibrium?

With $\hat{k}$ out of the page, a force of magnitude $F$ at distance $d$ to the left of the pivot, at angle $\theta$ above the beam, contributes which $(\tau_S)_z$ ?

✅ Correct!

\sin 0^\circ = 0

kills the first torque entirely, so the second must vanish too; and a force lifting the left end turns the beam clockwise, into the page, which is negative

z

❌ That is $d_1F_1 = d_2F_2$ without the sines.

\vec{F}_1

lies along the beam, so

F_1\sin\theta_1 = 0

and it makes no torque however large it is.

❌ Magnitudes are not what balance. Equilibrium matches the products

dF\sin\theta

, and the first of them is zero here.

❌ Half right.

\vec{F}_1

indeed contributes nothing, but

\vec{F}_2

is perpendicular at

d_2 = 0.20\ \text{m}

, so it makes torque

d_2F_2

unless

F_2

itself is zero.

❌ Check the sense. Lifting the left end swings the beam clockwise, and counterclockwise is the positive direction.

❌ Wrong component.

F\cos\theta

points along the beam, through the pivot: zero moment arm.

❌ Two slips at once.

F\cos\theta

is the part along the beam, whose line of action runs through the pivot and makes no torque at all; and lifting the left end turns the beam clockwise, which is negative

z

❌ Not quite. Take torques about the pivot: each force contributes

\pm\,dF\sin\theta

, with the sign set by whether it turns the beam counterclockwise or clockwise.

Show solution

The magnitude. $\vec{F}_1$ is directed along the beam, so its line of action passes through the pivot:

d_1 F_1 \sin\theta_1 = (0.60\ \text{m})(5.0\ \text{N})\sin 0^\circ = 0

Rotational equilibrium then demands $d_2 F_2 \sin\theta_2 = 0$ , and with $d_2 = 0.20\ \text{m}$ and $\sin 90^\circ = 1$ the only solution is $F_2 = 0$ . A force along the beam cannot be balanced by a perpendicular one — it needs no balancing at all.

The sign. Put $\hat{i}$ to the right, $\hat{j}$ up and $\hat{k} = \hat{i}\times\hat{j}$ out of the page. For a force at $d$ to the left, $\vec{r}_{S,F}$ points along $-\hat{i}$ , the angle between $\vec{r}_{S,F}$ and $\vec{F}$ is $\theta$ , so the magnitude of the torque is $dF\sin\theta$ . That force lifts the left end, which is a clockwise turn, into the page:

(\tau_{S,2})_z = -\,d F \sin\theta

The two conventions together give the generalized lever law as a single statement: $(\tau_{S,\text{total}})_z = d_1F_1\sin\theta_1 - d_2F_2\sin\theta_2 = 0$ .

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