Apply the principle of moments to other situations, including those with more than one force each side of the pivot

1.5.2 Turning Effect of Forces

Learning Objective

Apply the principle of moments to situations that involve more than one force on each side of the pivot (fulcrum).

Key Concepts

• Moment (torque): The turning effect of a force about a pivot.

  \$M = F \times d\$

  where \$F\$ is the magnitude of the force and \$d\$ is the perpendicular distance from the line of action of the force to the pivot.

• Clockwise (CW) and Counter‑clockwise (CCW) moments: Moments are classified by the direction they would rotate the lever.
• Principle of Moments (Rotational Equilibrium):

  \$\sum M{CW} = \sum M{CCW}\$

  A lever is in rotational equilibrium when the total clockwise moments equal the total counter‑clockwise moments about the pivot.

• Multiple forces on each side: The total moment on a side is the algebraic sum of the individual moments produced by each force on that side.

Step‑by‑Step Procedure for Solving Problems

1. Identify the pivot (fulcrum) and draw a clear diagram.
2. Mark the direction (CW or CCW) of each force’s moment about the pivot.
3. Measure or note the perpendicular distance \$d\$ from each force to the pivot.
4. Calculate each individual moment using \$M = Fd\$.
5. Sum the moments on the clockwise side and on the counter‑clockwise side.
6. Apply the principle of moments: set the total CW moments equal to the total CCW moments.
7. Solve the resulting equation for the unknown quantity (force, distance, etc.).
8. Check the units and verify that the answer is physically reasonable.

Typical Example

Two forces act on a horizontal beam that pivots about a point \$O\$.

Given:

• \$F_1 = 30\ \text{N}\$ acting \$0.40\ \text{m}\$ left of \$O\$ (CW)
• \$F_2 = 20\ \text{N}\$ acting \$0.60\ \text{m}\$ left of \$O\$ (CW)
• \$F_3 = 25\ \text{N}\$ acting \$0.50\ \text{m}\$ right of \$O\$ (CCW)
• \$F_4 = ?\$ acting \$0.30\ \text{m}\$ right of \$O\$ (CCW)

Solution:

1. Calculate known moments:

   • \$M{CW} = F1 d1 + F2 d_2 = 30(0.40) + 20(0.60) = 12 + 12 = 24\ \text{N·m}\$
   • \$M{CCW\ (known)} = F3 d_3 = 25(0.50) = 12.5\ \text{N·m}\$

2. Let \$M{CCW\ (unknown)} = F4 d_4\$.
3. Apply equilibrium:

   \$\sum M{CW} = \sum M{CCW}\$

   \$24 = 12.5 + F_4(0.30)\$

4. Solve for \$F_4\$:

   \$F_4(0.30) = 24 - 12.5 = 11.5\$

   \$F_4 = \frac{11.5}{0.30} \approx 38.3\ \text{N}\$

Common Mistakes to Avoid

• Using the slant distance instead of the perpendicular distance to the pivot.
• Forgetting to assign the correct direction (CW or CCW) to each moment.
• Adding moments algebraically without considering their signs.
• Neglecting forces that act directly through the pivot (they produce zero moment).

Summary Table

Practice Questions

1. A seesaw is balanced on a pivot at its centre. A child of mass \$30\ \text{kg}\$ sits \$1.2\ \text{m}\$ from the pivot on the left. What mass must sit \$0.8\ \text{m}\$ from the pivot on the right to keep the seesaw level? (Take \$g = 9.8\ \text{m s}^{-2}\$.)
2. Three forces act on a horizontal bar that pivots about point \$P\$:

   • \$F_1 = 15\ \text{N}\$ upward, \$0.25\ \text{m}\$ left of \$P\$ (CW)
   • \$F_2 = 10\ \text{N}\$ downward, \$0.40\ \text{m}\$ right of \$P\$ (CCW)
   • \$F_3 = ?\$ upward, \$0.35\ \text{m}\$ right of \$P\$ (CCW)

   Find \$F_3\$ for rotational equilibrium.

3. A uniform beam \$2.0\ \text{m}\$ long and mass \$8\ \text{kg}\$ rests on a support \$0.5\ \text{m}\$ from its left end. A load of \$20\ \text{N}\$ hangs \$1.5\ \text{m}\$ from the left end. Determine the reaction forces at the support and at the left end.

Extension: Varying the Pivot Position

When the pivot is not at the centre, the distances on each side change, affecting the required balancing forces. The same procedure applies: calculate each moment using the actual distances, then equate total clockwise and counter‑clockwise moments.