Torque and Equilibrium

Overview

Torque is the rotational equivalent of force. It measures the effectiveness of a force in causing rotation about an axis.

Torque

Definition

$\tau = \vec{r} \times \vec{F} = rF\sin\theta$

Where:

• $r$ = distance from axis to point where force is applied
• $F$ = magnitude of force
• $\theta$ = angle between $\vec{r}$ and $\vec{F}$

Equivalent Forms

$\tau = r \times F = r_\perp \times F = r \times F_\perp$

Where:

• $r_\perp = r\sin\theta$ = moment arm (perpendicular distance from axis to line of force)
• $F_\perp = F\sin\theta$ = component of force perpendicular to $r$

Units

• N·m (newton-meter)
• Note: Same units as work/energy, but torque is NOT energy

Sign Convention

• Counterclockwise (CCW): positive
• Clockwise (CW): negative

Newton's Second Law for Rotation

$\sum\tau = I\alpha$

Where $I$ is moment of inertia

Moment of Inertia

Resistance to rotational acceleration (rotational mass):

$I = \sum m_i r_i^2$

For continuous objects:

$I = \int r^2 \, dm$

Common Moments of Inertia

Object	Axis	$I$
Point mass	Distance $r$	$mr^2$
Solid cylinder/disk	Through center	$\frac{1}{2}MR^2$
Hollow cylinder	Through center	$MR^2$
Solid sphere	Through center	$\frac{2}{5}MR^2$
Hollow sphere	Through center	$\frac{2}{3}MR^2$
Thin rod	Through center	$\frac{1}{12}ML^2$
Thin rod	Through end	$\frac{1}{3}ML^2$

Parallel Axis Theorem

$I = I_{cm} + Md^2$

Where $d$ is distance from center of mass to new axis

Static Equilibrium

An object is in static equilibrium when:

Translational Equilibrium

$\sum\vec{F} = 0$

(No net force → no acceleration)

Rotational Equilibrium

$\sum\tau = 0$

(No net torque → no angular acceleration)

Conditions for Static Equilibrium

1. $\sum F_x = 0$
2. $\sum F_y = 0$
3. $\sum\tau = 0$ (about any point)

Problem-Solving Strategy for Equilibrium

1. Draw a free body diagram
2. Choose a convenient origin for torques (often where unknown forces act)
3. Apply equilibrium conditions
4. Solve the resulting equations

Center of Gravity

The point where weight effectively acts:

$x_{cg} = \frac{\sum(m_i g x_i)}{\sum m_i g} = x_{cm}$

For uniform gravitational field, center of gravity = center of mass

Examples

Example 1: Seesaw

Two children on a seesaw: 30 kg at 2 m from pivot, unknown mass at 1.5 m. Find mass for balance.

$\sum\tau = 0$

$m_1 g r_1 - m_2 g r_2 = 0$

$30(2) = m_2(1.5)$

$m_2 = 40 \text{ kg}$

Example 2: Ladder Against Wall

A 10 kg, 4 m ladder leans at 60° against a frictionless wall. Find forces.

Taking torques about the base:

$N_{\text{wall}} \times 4\sin(60°) = mg \times 2\cos(60°)$

$N_{\text{wall}} = \frac{mg \times \cos(60°)}{2\sin(60°)} = \frac{10(9.8) \times 0.5}{2 \times 0.866} = 28.3 \text{ N}$

Example 3: Beam with Load

A 5 m beam (20 kg) is supported at both ends. A 50 kg load is placed 2 m from the left end. Find support forces.

Taking torques about left end:

$R_{\text{right}} \times 5 = 20g \times 2.5 + 50g \times 2$

$R_{\text{right}} = \frac{(20 \times 2.5 + 50 \times 2)g}{5} = 294 \text{ N}$

From $\sum F_y = 0$:

$R_{\text{left}} + R_{\text{right}} = 20g + 50g = 686 \text{ N}$

$R_{\text{left}} = 392 \text{ N}$

Example 4: Angular Acceleration

A disk ($I = 0.5$ kg·m²) has a tangential force of 10 N applied at radius 0.2 m. Find angular acceleration.

$\tau = rF = 0.2 \times 10 = 2 \text{ N·m}$

$\alpha = \frac{\tau}{I} = \frac{2}{0.5} = 4 \text{ rad/s}^2$

Stability and Balance

• Stable equilibrium: Object returns to equilibrium if displaced
• Unstable equilibrium: Object moves away from equilibrium if displaced
• Neutral equilibrium: Object remains in new position

Conditions for Stability

• Low center of gravity
• Wide base of support
• Center of gravity over the base