Q.Four strings pull on a single point P, which stays at rest (equilibrium). Taking the upward vertical as reference: one string pulls with a force of $2\ \text{N}$ directed up and to the left, at $45^\circ$ from the upward vertical; a second string pulls with $1\ \text{N}$ directed up and to the right, at $45^\circ$ from the upward vertical; a third string pulls with an unknown force $\mathbf{F}_1$ directed horizontally to the right (it makes $45^\circ$ with the $1\ \text{N}$ force); and a fourth string pulls with an unknown force $\mathbf{F}_2$ directed vertically downward (at $90^\circ$ from $\mathbf{F}_1$). Find the forces $\mathbf{F}_1$ and $\mathbf{F}_2$.

Static Equilibrium: The Art of Staying Put

Imagine a book lying flat on a table. It doesn't move. A lamp hanging from the ceiling — still. A bridge holding up cars — motionless. What do all these have in common? They are in static equilibrium.

The word "static" means unchanging or stationary. "Equilibrium" comes from Latin aequilibrium — "equal balance." Put them together: a state where an object is completely at rest, with no tendency to start moving or rotating.

But here's the key insight: being at rest doesn't mean nothing is happening. Forces are still acting on that book — gravity pulls it down, the table pushes it up. The lamp feels tension from the ceiling pulling up and gravity pulling down. These forces are cancelling each other out perfectly.

The Two Conditions for Static Equilibrium

For an object to be truly static (not moving or rotating), two separate things must be true simultaneously.

Condition 1: No Net Force (Translational Equilibrium)

The sum of all forces acting on the object must be zero. In vector form:

This means:

• All upward forces equal all downward forces
• All leftward forces equal all rightward forces
• All forward forces equal all backward forces

If you break it into components (the standard exam approach):

Why this alone isn't enough: Imagine pushing a door at its handle — it rotates open. The forces might balance (you push, the hinges push back), but the door still moves. That's why we need the second condition.

Condition 2: No Net Torque (Rotational Equilibrium)

The sum of all torques (twisting effects) about any point must be zero:

Torque depends on three things: the force applied, the distance from the pivot point, and the angle at which you push. For a force $F$ applied at distance $r$ from the pivot, at angle $\theta$:

Putting It All Together

For an object to be in static equilibrium:

Both conditions must hold simultaneously. If either fails, the object will either accelerate (move in a straight line) or start rotating (or both).

A Simple Example: The Book on the Table

Consider a 2 kg book on a horizontal table. Gravity pulls down with force $F_g = mg = 2 \times 9.8 = 19.6 \text{ N}$.

The table pushes up with a normal force $N = 19.6 \text{ N}$.

Check condition 1: $\sum F_y = N - F_g = 19.6 - 19.6 = 0$ ✓ …