Apply the principle of moments to situations with one force each side of the pivot, including balancing of a beam

1.5.2 Turning Effect of Forces

Learning Objective

Apply the principle of moments to situations with one force on each side of the pivot, including the balancing of a beam.

Key Concepts

• Force – a push or pull acting at a point.
• Pivot (or fulcrum) – the point about which a body rotates.
• Moment (or torque) – the turning effect of a force, calculated as the product of the force magnitude and its perpendicular distance from the pivot.
• Principle of Moments – a body is in rotational equilibrium when the clockwise moments equal the anticlockwise moments about the pivot.

Definition of a Moment

The moment \$M\$ of a force \$F\$ about a pivot is given by

\$M = F \times d\$

where \$d\$ is the perpendicular distance from the line of action of the force to the pivot (the lever arm).

Direction of Moments

Moments are classified as:

• Clockwise (CW)
• Anticlockwise (ACW)

When solving problems, assign a sign convention (e.g., CW = positive, ACW = negative) and keep it consistent.

Principle of Moments

A rigid body will remain at rest (or rotate at constant speed) when the sum of all moments about any pivot is zero:

\$\sum M{\text{CW}} = \sum M{\text{ACW}} \quad\text{or}\quad \sum M = 0\$

Balancing a Beam with One Force on Each Side

Consider a uniform beam supported at a pivot. A force \$F1\$ acts at a distance \$d1\$ on the left side, and a force \$F2\$ acts at a distance \$d2\$ on the right side.

For the beam to be in equilibrium:

\$F1 \times d1 = F2 \times d2\$

This equation can be rearranged to find any unknown quantity.

Worked Example

Problem: A 2.0 m long uniform beam is pivoted 0.8 m from its left end. A 30 N weight is hung at the left end. At what distance from the right end must a 20 N weight be placed to keep the beam balanced?

1. Identify the pivot and distances:

   • Distance from pivot to 30 N weight: \$d_1 = 0.8\ \text{m}\$ (left side).
   • Let \$x\$ be the distance from the right end to the 20 N weight. The distance from the pivot to this weight is \$d_2 = 2.0 - 0.8 - x = 1.2 - x\$ (right side).

2. Apply the principle of moments (clockwise moments = anticlockwise moments):

   \$30\ \text{N} \times 0.8\ \text{m} = 20\ \text{N} \times (1.2 - x)\$

3. Solve for \$x\$:

   \$24 = 20(1.2 - x)\$

   \$24 = 24 - 20x\$

   \$20x = 0\$

   \$x = 0\ \text{m}\$

4. Interpretation: The 20 N weight must be placed at the very right end of the beam (i.e., \$x = 0\$) for equilibrium.

Common Mistakes

• Using the total length of the beam instead of the perpendicular distance from the pivot.
• Forgetting to keep the direction (sign) of each moment consistent.
• Assuming the beam’s own weight contributes a moment when the pivot is at the centre of a uniform beam (in that case the weight acts through the pivot and produces no moment).

Summary Table

Practice Questions

1. A uniform 1.5 m beam is supported at its centre. A 40 N force is applied 0.3 m to the left of the pivot. Where should a 25 N force be applied on the right side to keep the beam in equilibrium?
2. A seesaw is 4.0 m long and pivots at its centre. A child weighing 300 N sits 0.9 m from the pivot. How far from the pivot must a second child weighing 200 N sit on the opposite side to balance the seesaw?
3. In a laboratory setup, a metre rule (mass 0.5 kg, uniform) is balanced on a knife‑edge at the 30 cm mark. A 2.0 kg mass is hung at the 0 cm end. Calculate the distance from the knife‑edge at which a 1.0 kg mass must be placed on the right side to achieve equilibrium. (Take \$g = 9.8\ \text{m s}^{-2}\$.)

Answers to Practice Questions

1. Using \$40 \times 0.3 = 25 \times d\$, we get \$d = 0.48\ \text{m}\$ to the right of the pivot.
2. Balance condition: \$300 \times 0.9 = 200 \times d \;\Rightarrow\; d = 1.35\ \text{m}\$ from the pivot on the opposite side.
3. Weight of the metre rule acts at its centre (50 cm). Moment of rule about knife‑edge: \$0.5g \times 20\ \text{cm}=0.5\times9.8\times0.20=0.98\ \text{N·m}\$ (anticlockwise). Moment of 2 kg mass: \$2g \times 0.30 = 2\times9.8\times0.30 = 5.88\ \text{N·m}\$ (clockwise). Total clockwise moment = \$5.88 + 0.98 = 6.86\ \text{N·m}\$. For the 1 kg mass: \$1g \times d = 6.86 \;\Rightarrow\; d = \frac{6.86}{9.8}=0.70\ \text{m}\$ to the right of the knife‑edge.