§12.3 — Rigid Body Equilibrium

The equilibrium conditions are straightforward: forces must balance and torques must balance. But where these forces act matters!

Example: Torque and Moment Arm

Two equal and opposite forces: F = 100 N.

Case 1: Both forces pass through the same point: \[ \sum F = 100 - 100 = 0 \quad ✓ \] \[ \sum \tau = 0 \quad ✓ \]

The body stays still.

Case 2: Forces act at different points (e.g., the two ends of a rod):

If the separation = d, about the center:

\[ \sum F = 100 - 100 = 0 \quad ✓ \]

But: \[ \sum \tau = F \cdot (d/2) + F \cdot (d/2) = F \cdot d = 100 \times d \neq 0 \quad ✗ \]

The body rotates! This is called a couple (or torque pair).

Definition: A Couple

A couple is two equal, opposite, and non-collinear forces:

\[ \sum F = 0 \quad \text{(no net force)} \] \[ \sum \tau = F \cdot d \neq 0 \quad \text{(net torque)} \]

The body doesn't translate but does rotate. The torque of a couple:

\[ \tau_{\text{couple}} = F \cdot d \]

where d is the perpendicular distance between the lines of action.

Practical Example: A Lever

A lever is a rigid rod balanced on a pivot (fulcrum).

Simple Setup

• Pivot at position x = 0
• Load W = 100 N at distance d_W = 0.5 m (load arm)
• Effort F at distance d_F = 2 m (effort arm)

Equilibrium Condition (Torques)

About the pivot:

\[ F \cdot d_F = W \cdot d_W \]

\[ F \times 2 = 100 \times 0.5 \]

\[ F = 25 \text{ N} \]

Mechanical Advantage: \[ \text{MA} = \frac{d_F}{d_W} = \frac{2}{0.5} = 4 \]

With just 25 N of effort, you lift a 100 N load!

More Complex Example: A Horizontal Beam

Uniform beam, mass m = 50 kg, length L = 4 m, on two supports:

• Left support at x = 0
• Right support at x = L = 4 m
• Additional load W = 200 N at x = 1 m

Forces acting:

1. Weight of beam: W_beam = mg = 50 × 10 = 500 N at center (x = 2 m)
2. Additional load: W = 200 N at x = 1 m
3. Normal force from left support: N_L at x = 0
4. Normal force from right support: N_R at x = 4 m

First Condition: Forces

\[ N_L + N_R - W_{\text{beam}} - W = 0 \]

\[ N_L + N_R = 500 + 200 = 700 \text{ N} \]

Second Condition: Torques (About Left Support)

\[ N_L \times 0 + N_R \times 4 - W_{\text{beam}} \times 2 - W \times 1 = 0 \]

\[ N_R \times 4 = 500 \times 2 + 200 \times 1 \]

\[ N_R \times 4 = 1000 + 200 = 1200 \]

\[ N_R = 300 \text{ N} \]

Solution

From the first condition:

\[ N_L = 700 - 300 = 400 \text{ N} \]

Verification (torques about right support):

\[ N_L \times 4 - W_{\text{beam}} \times 2 - W \times 3 = 400 \times 4 - 500 \times 2 - 200 \times 3 \]

\[ = 1600 - 1000 - 600 = 0 \quad ✓ \]

Strategy for Solving Rigid Body Equilibrium Problems

1. Draw forces (free body diagram — FBD)
2. Choose coordinates
3. Choose rotation axis (usually at the point of unknown force)
4. Write equations:
   • \( \sum F_x = 0 \) (if forces in x-direction)
   • \( \sum F_y = 0 \) (y-direction)
   • \( \sum \tau = 0 \) (torque about chosen axis)
5. Solve the simultaneous equations

Practical Tips

Smart Axis Choice

Choosing the rotation axis wisely simplifies calculations:

• If there's a pivot point, choose it as the axis (its normal force produces no torque)
• If the unknown force acts there, choose that point

Distributed Load

If load is distributed (e.g., beam weight):

• Treat it as a point load at the center of mass
• Then solve normally

Common Mistakes

• Ignoring friction: unless stated as frictionless, include friction
• Forgetting object weight: if the object isn't massless, include its weight
• Sign errors: be consistent with signs; up is positive, torque counterclockwise is positive

What You Should Know

• Rigid body equilibrium requires both force and torque balance
• A couple: two equal opposite forces (non-collinear) produce only torque
• Levers and mechanical advantage: lift heavier loads with less effort
• Strategy: FBD → choose axis → write equations → solve

Preview of §12.4

Now we know how to analyze objects in equilibrium. But what happens when equilibrium breaks? Objects deform. Let's define stress and strain.

📚 See also: Halliday Vol 1, Ch 12, §12.3 — Rigid body equilibrium. 🔗 Reference: §12.1 (Equilibrium conditions), §10.6 (Torque).

⇧ Back to chapter