Classical-Mechanics · Unit 22 · Video 2 · Interactive Practice

Two Conditions, Not One: Static Equilibrium and Why the Lever Balances

IKey Formulas

Formula	Name	What it says
$\vec{F} = \vec{F}_1 + \vec{F}_2 + \cdots = \vec{0}$	Condition 1	The centre of mass stays at rest
$\vec{\tau}_S = \vec{\tau}_{S,1} + \vec{\tau}_{S,2} + \cdots = \vec{0}$	Condition 2, about a point $S$	The body does not rotate; $S$ is ours to choose
$F_{\text{pivot}} - m_b g - N_1 - N_2 = 0$	Beam, condition 1	The pivot carries the beam's weight and both normal forces
$d_2 N_2 - d_1 N_1 = 0 \ \Longleftrightarrow \ d_1 N_1 = d_2 N_2$	Lever law	Normal force times moment arm, equal on the two sides

Key Insight: Both conditions are needed and they are independent: a ruler pushed up at one end and down at the other has $\sum \vec{F} = \vec{0}$ and spins anyway. Torques taken about the pivot give $\vec{F}_{\text{pivot}}$ and $m_b \vec{g}$ a moment arm of zero, so the lever law falls out without the pivot force ever being computed.

IIForces That Cancel, a Body That Turns

Two equal and opposite forces sum to zero, yet where each one acts decides whether the rod rotates.

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IIIThe Balanced Beam

What must be true of the two loads for the beam not to rotate about its pivot?

Object 1

💡 Once the beam balances the torques sum to zero about every point, not only the pivot — but only the pivot keeps the unknown $\vec{F}_{\text{pivot}}$ out of the equation in the first place.

IVThree Bodies, Three Diagrams

The blocks press on the beam with contact forces; the weights $m_1\vec{g}$ and $m_2\vec{g}$ act on the blocks.

Free-body diagram

VQuiz Questions

Problem 1 · Balancing the Beam

Given: A uniform beam rests on a pivot placed at its centre of mass. Object 2 presses down on the beam with a normal force $N_2 = 30\ \text{N}$ at $d_2 = 0.40\ \text{m}$ to the left of the pivot; object 1 presses down at $d_1 = 0.60\ \text{m}$ to the right — find the normal force $N_1$ that keeps the beam static.

✅ Correct! The longer moment arm carries the smaller force:

0.60 \times 20 = 0.40 \times 30 = 12\ \text{N}\cdot\text{m}

on each side.

❌ The ratio is upside down. You computed

N_2 d_1 / d_2

. The lever law sets the products equal,

d_1 N_1 = d_2 N_2

, so the force at the longer arm must be smaller, not larger.

❌ Not quite. Equal forces would balance only at equal distances. Here

d_1 \ne d_2

, so equal forces give unequal torques.

❌ That is the torque, not the force.

d_2 N_2 = 12\ \text{N}\cdot\text{m}

is the torque about the pivot; divide it by

d_1

to get a force.

❌ Not quite. Set the torques about the pivot equal:

d_1 N_1 = d_2 N_2

Show solution

Take torques about the pivot, counterclockwise positive. The beam's weight and the pivot force both act at the pivot, so each has moment arm zero and contributes nothing. Only the two normal forces are left:

\tau_{\text{pivot}} = d_2 N_2 - d_1 N_1 = 0

so the lever law gives

N_1 = \frac{d_2 N_2}{d_1} = \frac{(0.40\ \text{m})(30\ \text{N})}{0.60\ \text{m}} = \frac{12\ \text{N}\cdot\text{m}}{0.60\ \text{m}} = 20\ \text{N}.

Condition 1 then fixes the pivot force, $F_{\text{pivot}} = m_b g + N_1 + N_2 = m_b g + 50\ \text{N}$ — but it was never needed to find $N_1$ .

Problem 2 · Where Did the Beam's Weight Go?

Given: The beam above is uniform with mass $m_b = 1.0\ \text{kg}$ , so its weight $m_b \vec{g}$ is certainly not negligible — yet neither $m_b g$ nor $F_{\text{pivot}}$ appears in $d_1 N_1 = d_2 N_2$ . Why?

✅ Correct! The pivot sits at the beam's centre of mass, so

m_b \vec{g}

and

\vec{F}_{\text{pivot}}

both act at the torque point:

\tau = r_\perp F

with

r_\perp = 0

for each.

❌ No assumption was made about $m_b$ . The beam's weight is in condition 1,

F_{\text{pivot}} - m_b g - N_1 - N_2 = 0

. It drops out of condition 2 only because of where it acts.

❌ Work is not the issue. Static equilibrium is a statement about forces and torques; a force with a nonzero moment arm would appear in condition 2 whether or not it does work.

❌ There is nothing to cancel. Each torque is individually zero — the two forces have different magnitudes (

9.8\ \text{N}

and

68.6\ \text{N}

in the visualization), so equal and opposite torques could not have been the reason.

❌ Not quite. Ask what the moment arm of each force about the pivot is.

Show solution

The magnitude of a torque about $S$ is $\tau_S = r_\perp F$ , where $r_\perp$ is the perpendicular distance from $S$ to the line of action of the force. The pivot is placed directly beneath the beam's centre of mass, and the beam is thin, so both $\vec{F}_{\text{pivot}}$ and $m_b \vec{g}$ act at the pivot point itself:

r_\perp = 0 \quad \Longrightarrow \quad \tau_{\text{pivot}} = 0 \ \text{ for each of them.}

Only $N_1$ and $N_2$ survive in condition 2, giving $d_2 N_2 - d_1 N_1 = 0$ . This is the reason $S$ is worth choosing carefully: the torque point was placed exactly where the force we know least about acts, and that force left the equation before it was ever computed.

Problem 3 · Slide Object 1 Outward

Given: A beam pivoted at its centre of mass balances with $m_2 = 4.0\ \text{kg}$ at $d_2 = 0.25\ \text{m}$ on the left and $m_1 = 2.0\ \text{kg}$ at $d_1 = 0.50\ \text{m}$ on the right ( $g = 9.8\ \text{m}\,\text{s}^{-2}$ ). Object 1 is now slid outward to $d_1 = 0.70\ \text{m}$ — find the net torque about the pivot, counterclockwise positive.

✅ Correct!

d_1 N_1

grew to

13.72\ \text{N}\cdot\text{m}

while

d_2 N_2

stayed at

9.8\ \text{N}\cdot\text{m}

, so the right side wins and the right end goes down.

❌ Right size, wrong sense.

N_1

pushes down to the right of the pivot, which turns the beam clockwise — a negative torque with counterclockwise positive. The larger term is

d_1 N_1

, so the sum is negative.

❌ That is condition 1 talking. The four forces are indeed unchanged, so

F_{\text{pivot}} - m_b g - N_1 - N_2 = 0

still holds — but the position of

N_1

changed, and condition 2 sees positions. This is exactly the case where force balance alone fails.

❌ One term is missing.

d_1 N_1 = 13.72\ \text{N}\cdot\text{m}

is only object 1's contribution; object 2 still supplies

+d_2 N_2 = 9.8\ \text{N}\cdot\text{m}

the other way.

❌ Not quite. Compute

\tau = +d_2 N_2 - d_1 N_1

with

N_i = m_i g

Show solution

Step 1: the normal forces. Each object is itself in equilibrium and every force on it is vertical, so the beam pushes up on it with $N_i = m_i g$ , and by Newton's third law it presses down on the beam with the same magnitude:

N_2 = (4.0)(9.8) = 39.2\ \text{N}, \qquad N_1 = (2.0)(9.8) = 19.6\ \text{N}.

Step 2: check the original balance. $d_2 N_2 = (0.25)(39.2) = 9.8\ \text{N}\cdot\text{m}$ and $d_1 N_1 = (0.50)(19.6) = 9.8\ \text{N}\cdot\text{m}$ — equal, so the beam was balanced.

Step 3: torques after the slide. Nothing about the forces changed, only $d_1$ :

\tau_{\text{pivot}} = +d_2 N_2 - d_1 N_1 = 9.8 - (0.70)(19.6) = 9.8 - 13.72 = -3.92\ \text{N}\cdot\text{m}.

The sign is negative, so the rotation is clockwise: the right end swings down. Condition 1 never noticed.

Problem 4 · Crowbar

Given: A crowbar rests on a fulcrum. You press down on the handle with $15\ \text{N}$ at $0.60\ \text{m}$ from the fulcrum, and the bar presses up on a load sitting $0.050\ \text{m}$ from the fulcrum on the other side. The bar is uniform, its centre of mass is at the fulcrum, and it is in static equilibrium — find the magnitude of the force the bar exerts on the load.

✅ Correct! A moment arm twelve times longer multiplies the force twelve-fold:

0.60/0.050 = 12

and

12 \times 15\ \text{N} = 180\ \text{N}

❌ The ratio is inverted. You multiplied by

0.050/0.60

. The short arm carries the large force, since

d_1 N_1 = d_2 N_2

with

d_1

small forces

N_1

up.

❌ Equal forces are not what equilibrium requires. Condition 2 equates torques, not forces; equal forces at unequal arms give unequal torques and the bar turns.

❌ Those are newton-metres.

(15\ \text{N})(0.60\ \text{m}) = 9.0\ \text{N}\cdot\text{m}

is the torque your hand applies; divide it by the load's moment arm to get a force.

❌ Not quite. Take torques about the fulcrum: the force at the fulcrum has zero moment arm, so only the hand and the load appear.

Show solution

Take torques about the fulcrum. The force of the fulcrum on the bar and the bar's own weight both act there, so both have moment arm zero and drop out — the same choice that removed $\vec{F}_{\text{pivot}}$ from the beam problem. Only the hand force $N_2 = 15\ \text{N}$ at $d_2 = 0.60\ \text{m}$ and the load force $N_1$ at $d_1 = 0.050\ \text{m}$ remain, and they turn the bar opposite ways:

d_2 N_2 - d_1 N_1 = 0 \quad \Longrightarrow \quad N_1 = \frac{d_2 N_2}{d_1} = \frac{(0.60\ \text{m})(15\ \text{N})}{0.050\ \text{m}} = 180\ \text{N}.

By Newton's third law the bar pushes up on the load with $180\ \text{N}$ and the load presses down on the bar with the same $180\ \text{N}$ — equal magnitudes on two different bodies. That factor of $d_2/d_1 = 12$ is the mechanical advantage of the lever.

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