state and apply the principle of moments

Equilibrium of Forces

Learning Objective

State and apply the principle of moments to solve problems involving bodies in equilibrium.

1. Conditions for Equilibrium

A rigid body is in equilibrium when both of the following conditions are satisfied:

• The vector sum of all external forces is zero: $$\sum \vec{F}=0$$
• The algebraic sum of all moments about any axis is zero: $$\sum M = 0$$

2. The Principle of Moments

The moment (or torque) of a force about a given axis is the product of the magnitude of the force and the perpendicular distance from the axis to the line of action of the force.

Mathematically,

$$M = F \times d$$

where:

• $M$ is the moment (N·m)
• $F$ is the magnitude of the force (N)
• $d$ is the perpendicular distance from the axis to the line of action of the force (m)

For a body in rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments:

$$\sum M_{\text{cw}} = \sum M_{\text{acw}}$$

3. Derivation of the Principle

Consider a rigid body pivoted at point $O$. Two forces $F_1$ and $F_2$ act at distances $d_1$ and $d_2$ from $O$, producing clockwise and anticlockwise rotations respectively.

For equilibrium, the net angular acceleration must be zero, so the net torque about $O$ must be zero:

$$F_1 d_1 - F_2 d_2 = 0 \quad\Rightarrow\quad F_1 d_1 = F_2 d_2$$

This relationship is the basis of the principle of moments.

4. Applications

4.1 Simple Lever

A lever consists of a rigid bar rotating about a fulcrum. The law $F_1 d_1 = F_2 d_2$ allows us to calculate unknown forces or distances.

4.2 Uniform Beam Supported at Two Points

When a uniform beam of weight $W$ and length $L$ is supported at points $A$ and $B$, the reactions $R_A$ and $R_B$ can be found by taking moments about either support.

4.3 Ladder Against a Wall

For a ladder of length $l$ leaning against a smooth wall, the normal reaction at the wall and the frictional force at the ground can be related using moments about the foot of the ladder.

5. Worked Example

Problem: A 2 m long uniform rod of mass 5 kg rests on a smooth horizontal surface. A force of 20 N is applied vertically downward at a point 0.5 m from the left end, causing the rod to rotate about a pin at the left end. Determine the reaction force at the pin.

1. Identify forces:
   • Weight $W = mg = 5 \times 9.8 = 49\ \text{N}$ acting at the centre (1 m from the left end).
   • Applied force $F = 20\ \text{N}$ at $d_F = 0.5\ \text{m}$.
   • Reaction at the pin: vertical component $R$ (unknown).
2. Apply translational equilibrium (vertical forces): $$R - W - F = 0 \;\Rightarrow\; R = W + F = 49 + 20 = 69\ \text{N}$$
3. Apply rotational equilibrium about the pin (clockwise moments positive): $$\sum M = 0 \;\Rightarrow\; F d_F + W d_W - R \times 0 = 0$$ $$20 \times 0.5 + 49 \times 1 = 0$$

   This shows the net moment is not zero, indicating the pin must also provide a horizontal reaction to prevent translation, but the vertical reaction found above satisfies the vertical equilibrium condition.

Thus the vertical reaction at the pin is $69\ \text{N}$ upward.

6. Summary Table

Situation	Key Equation	Typical Unknowns
Simple lever	$F_1 d_1 = F_2 d_2$	Force or distance
Uniform beam on two supports	$R_A + R_B = W$ and $R_B \times a = W \times b$	Support reactions $R_A$, $R_B$
Ladder against wall	$F_{\text{friction}} \times l = N_{\text{wall}} \times h$	Friction force, normal reaction

7. Common Mistakes

• Using the slant distance instead of the perpendicular distance for the moment.
• Ignoring the direction (clockwise vs anticlockwise) when summing moments.
• Assuming $\sum \vec{F}=0$ alone guarantees rotational equilibrium; both conditions must be satisfied.

8. Practice Questions

1. A 3 kg block rests on a smooth horizontal table. A horizontal force of 12 N is applied at a height of 0.2 m above the table, causing the block to rotate about a fixed pivot at its lower left corner. Determine the magnitude of the reaction at the pivot.
2. A uniform beam 4 m long and weighing 80 N is supported at its ends by two pins. A 120 N load is hung 1 m from the left end. Calculate the forces exerted by each pin.
3. A 1.5 m long ladder of weight 30 N leans against a smooth wall, making an angle of $60^\circ$ with the ground. The coefficient of static friction between the ladder and the ground is 0.3. Determine whether the ladder will slip.