Equilibrium of a Rigid Body

<<>>

Equilibrium of a Rigid Body

TL;DR

A rigid body is in equilibrium when all forces acting on it sum to zero AND all torques sum to zero. This means the body won't accelerate linearly or rotationally. You need to satisfy two independent conditions: ΣF = 0 and Στ = 0.

1. The Mental Model

Here's the key insight: a rigid body can move in two ways—it can slide around (translate) or it can spin (rotate). To stop both, you need two separate rules. The first rule stops translation: make sure all the pushes and pulls cancel out. The second rule stops rotation: make sure all the twists and turns cancel out. When both happen at once, nothing moves and nothing spins.

2. The Core Material

Condition 1: Zero Net Force

For a rigid body not to accelerate, the vector sum of all external forces must be zero:

$$\sum \vec{F} = 0$$

In component form (taking x and y as the horizontal and vertical directions):

$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$

This is exactly Newton's first law. If you have four people pushing on a box from different angles, and the box doesn't slide across the floor, their pushes must balance. That's this condition.

Condition 2: Zero Net Torque

This is the twist you might miss. Even if all forces balance, the body can still spin. Imagine a rod with two equal opposite forces pulling at opposite ends—they cancel, but the rod rotates. To prevent rotation, the sum of all torques (about any point) must also be zero:

$$\sum \vec{\tau} = 0$$

The torque of a force depends on where it acts. If force F acts at distance r from a pivot point, perpendicular to the force, the magnitude is:

$$\tau = rF$$

More generally: $\tau = r_\perp F$ where $r_\perp$ is the perpendicular distance from the pivot to the line of action of the force (this is called the moment arm).

Important: you can choose any point as your pivot to calculate torques, and if the body is in equilibrium, the total torque will be zero about every point. This is incredibly useful—pick a smart point to make the algebra easier.

How to Solve Equilibrium Problems

1. Draw a free-body diagram showing every force and its location on the body.
2. Choose a coordinate system (typically x horizontal, y vertical).
3. Write out the force balance equations: set ΣF_x = 0 and ΣF_y = 0.
4. Choose a pivot point (often where two unknowns act, to eliminate them).
5. Write the torque equation: set Στ = 0 about that pivot.
6. Solve the system of equations for unknowns (usually support forces or distances).

graph TD
  A["Rigid body at rest<br/>or moving at constant velocity"] --> B["Apply ΣF = 0<br/>No acceleration"]
  A --> C["Apply Στ = 0<br/>No angular acceleration"]
  B --> D["Sum all force components<br/>in x and y directions"]
  C --> E["Choose pivot point<br/>Sum torques<br/>⊙ = counterclockwise+<br/>⊗ = clockwise−"]
  D --> F["Solve simultaneous equations"]
  E --> F
  F --> G["Unknown forces/distances<br/>Body in equilibrium"]

3. Worked Example

A uniform plank of length L = 4 m and weight W = 100 N rests on two supports: one at the left end (point A) and one 1 m from the right end (point B). A person of weight 600 N stands 1 m from the left end. Find the normal forces at A and B.

Step 1: Draw the diagram.
- Support A at x = 0 (upward force N_A)
- Support B at x = 3 m (upward force N_B)
- Weight of plank: 100 N at centre, x = 2 m (downward)
- Person: 600 N at x = 1 m (downward)

Step 2: Force balance (vertical only).
$$N_A + N_B - 100 - 600 = 0$$
$$N_A + N_B = 700 \text{ N} \quad \text{...(1)}$$

Step 3: Torque balance about point A (choosing A eliminates N_A from the torque equation).

Taking counterclockwise as positive:
- Torque from N_A: 0 (acts at the pivot)
- Torque from plank weight: −100 × 2 = −200 N⋅m (clockwise, 2 m from A)
- Torque from person: −600 × 1 = −600 N⋅m (clockwise, 1 m from A)
- Torque from N_B: +N_B × 3 (counterclockwise, 3 m from A)

$$N_B \times 3 - 200 - 600 = 0$$
$$N_B = \frac{800}{3} \approx 267 \text{ N}$$

Step 4: Substitute back into (1).
$$N_A = 700 - 267 = 433 \text{ N}$$

Check: Torque about B should also be zero.
- Torque from N_A: 433 × 3 = 1299 N⋅m (counterclockwise)
- Torque from weight: 100 × 1 = 100 N⋅m (clockwise)
- Torque from person: 600 × 2 = 1200 N⋅m (clockwise)
- Sum: 1299 − 100 − 1200 = −1 N⋅m ≈ 0 ✓ (rounding error)

4. Key Takeaways

4.1 Most Important Concepts

• Two independent conditions: You cannot satisfy just ΣF = 0. You must satisfy Στ = 0 separately, or the body will spin.
• Torque depends on position: The same force creates different torques depending on where it acts. Move the force line closer to the pivot, torque shrinks.
• Choice of pivot is free: For a body in equilibrium, Στ = 0 about every point. Pick the smartest pivot (usually where unknowns cluster) to simplify algebra.
• Moment arm is key: The perpendicular distance from the pivot to the line of action is what matters, not the distance to the force's starting point.
• Equilibrium means constant velocity: The body can be stationary OR moving at constant velocity—both satisfy ΣF = 0 and Στ = 0.
• Free-body diagrams are essential: You cannot solve equilibrium problems without accurately drawing every force, its location, and direction.
• Units matter: Torque is force × distance (N⋅m), and must balance on both sides of the equation; forces (N) only balance forces.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Problem: A horizontal beam of negligible weight rests on two supports 6 m apart. Support A is at the left; support B is 6 m to the right. A load of 400 N hangs 2 m from the left end. A second load of 200 N hangs 5 m from the left end. Find N_A and N_B.

Work through this step by step:
1. Draw a free-body diagram labelling both support forces pointing up.
2. Write ΣF_y = 0 (the vertical force balance).
3. Choose pivot A. Write Στ = 0, taking counterclockwise as positive. Calculate the perpendicular distance from A to each load's line of action.
4. Solve the torque equation for N_B.
5. Substitute into the force equation to find N_A.
6. Check your answer by computing Στ about point B—it should equal zero.

Success looks like: you find N_A = 300 N and N_B = 300 N, and you can justify each number by showing your force and torque equations with correct signs and moment arms.

<<>>