understand that, when there is no resultant force and no resultant torque, a system is in equilibrium

⚖️ Equilibrium of Forces

Key Concepts

In physics, a system is in equilibrium when it experiences no net change in motion.

Think of a perfectly balanced seesaw: the children on either side are at rest, and the seesaw stays level.

For a system to be in equilibrium, two conditions must be met:

• Resultant force \$\,\displaystyle\sum \mathbf{F} = \mathbf{0}\,\$ (no net push or pull).
• Resultant torque \$\,\displaystyle\sum \tau = 0\,\$ (no net twisting).

Force Equilibrium

Forces are vectors, so they have both magnitude and direction.

The vector sum of all forces acting on a body must be zero.

Example: A book resting on a table experiences two forces: the weight \$mg\$ downward and the normal force \$N\$ upward.

Since \$N = mg\$, the forces cancel: \$N - mg = 0\$.

1. Identify all forces.
2. Resolve each force into components (usually horizontal and vertical).
3. Sum the components: \$\sum Fx = 0\$ and \$\sum Fy = 0\$.
4. If both sums are zero, the system is in force equilibrium.

Torque Equilibrium

Torque (or moment) is the tendency of a force to rotate an object about a point.

It is calculated as \$\tau = r \times F\$, where \$r\$ is the lever arm (distance from the pivot) and \$F\$ is the force perpendicular to \$r\$.

For equilibrium, the algebraic sum of all torques about any point must be zero.

Example: A door held open by a hinge and a hand.

The hand applies a force at a distance \$rh\$ from the hinge, creating a torque \$\tauh = rh Fh\$.

The door’s weight acts at its centre of gravity, distance \$rg\$, producing \$\taug = r_g mg\$.

If \$\tauh = \taug\$, the door stays open without moving.

Illustrative Example

⚙️ Balance Beam

Two masses \$m1\$ and \$m2\$ hang from a beam of length \$L\$ at distances \$d1\$ and \$d2\$ from the pivot.

The beam is in equilibrium when:

\$ m1 g \, d1 = m2 g \, d2 \$

Since \$g\$ cancels, the condition simplifies to \$m1 d1 = m2 d2\$.

This is the classic “lever” rule: heavier mass must be closer to the pivot to balance a lighter mass farther away.

Exam Tips

• Always start by drawing a clear diagram with all forces and points of application.

• Use a consistent sign convention (e.g., counter‑clockwise torques positive).

• Check both force and torque equilibrium; missing one can lead to a wrong answer.

• Remember that the point about which you sum torques can be any point, but choosing the pivot simplifies calculations.

• In multiple‑choice questions, look for the option that satisfies both \$\sum \mathbf{F}=0\$ and \$\sum \tau=0\$.

Summary Table