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 (Moments)

Learning Objective

Apply the principle of moments to situations that involve more than one force on each side of the pivot (fulcrum) and to determine unknown forces or distances required for rotational equilibrium. The core requirement – a single force on each side – is also reviewed.

Key Concepts

• Moment (also called torque in some texts) – the turning effect of a force about a pivot.

  \[

  \text{moment }(\tau)=F\,d

  \]

  where F is the magnitude of the force (N) and d is the perpendicular distance (m) from the line of action of the force to the pivot. The moment is a vector; in the IGCSE syllabus it is described by its sense: clockwise (CW) or counter‑clockwise (CCW).

• Direction of a moment – label each moment as CW or CCW. In algebraic work you may assign a sign (+ for CW, – for CCW) or keep separate columns for the two senses.
• Principle of moments (rotational equilibrium) – a lever is in equilibrium when the resultant moment about the pivot is zero:

  \[

  \sum \tau =0\qquad\text{or}\qquad\sum \tau{\text{CW}}=\sum \tau{\text{CCW}}

  \]

• Multiple forces on one side – the total moment on a side is the algebraic sum of the individual moments:

  \[

  \tau{\text{total}}=\sum{i}Fi di

  \]

General Procedure for Solving Moment Problems

1. Identify the pivot (fulcrum) and draw a clear, to‑scale diagram.
2. Mark the line of action of every force and note whether its moment is CW or CCW.
3. Measure the perpendicular distance d from each line of action to the pivot.
4. Calculate each moment using \(\tau =F d\). Record the sign (+ for CW, – for CCW) or place the value in the appropriate CW/CCW column.
5. Sum the moments on the CW side and on the CCW side.
6. Apply the equilibrium condition \(\sum \tau =0\) (or \(\sum \tau{\text{CW}}=\sum \tau{\text{CCW}}\)).
7. Solve the resulting equation for the unknown quantity (force, distance or mass).
8. Check units and confirm that the answer is physically reasonable.

Core Example – One Force on Each Side (Syllabus Requirement)

A simple lever is balanced about a pivot O.

• \(F_{L}=20\ \text{N}\) acting \(0.40\ \text{m}\) to the left of O (CW).
• \(F_{R}=15\ \text{N}\) acting \(0.53\ \text{m}\) to the right of O (CCW).

Equilibrium: \(\; +8.0 - 7.95 = 0\) (to the nearest 0.01 N·m). The moments are essentially equal, so the lever is balanced. This illustrates the core specification of “one force on each side”.

Worked Example 1 – Two Forces on Each Side

Horizontal beam pivoted at O.

• \(F_{1}=30\ \text{N}\) at \(0.40\ \text{m}\) left of O (CW)
• \(F_{2}=20\ \text{N}\) at \(0.60\ \text{m}\) left of O (CW)
• \(F_{3}=25\ \text{N}\) at \(0.50\ \text{m}\) right of O (CCW)
• \(F_{4}=?\) at \(0.30\ \text{m}\) right of O (CCW)

Apply \(\sum \tau =0\):

\[

+12.0+12.0-12.5-0.30F{4}=0\;\Longrightarrow\;0.30F{4}=11.5\;\Longrightarrow\;F_{4}\approx38.3\ \text{N}

\]

Worked Example 2 – Three Forces on One Side, Two on the Other

Uniform horizontal bar (negligible mass) pivots at P.

• \(F_{1}=12\ \text{N}\) upward, \(0.20\ \text{m}\) left of P (CW)
• \(F_{2}=8\ \text{N}\) upward, \(0.45\ \text{m}\) left of P (CW)
• \(F_{3}=5\ \text{N}\) downward, \(0.30\ \text{m}\) left of P (CW)
• \(F_{4}=15\ \text{N}\) upward, \(0.35\ \text{m}\) right of P (CCW)
• \(F_{5}=?\) upward, \(0.55\ \text{m}\) right of P (CCW)

Equilibrium:

\[

(+2.4+3.6+1.5)-5.25-0.55F{5}=0\;\Longrightarrow\;2.25=0.55F{5}\;\Longrightarrow\;F_{5}\approx4.1\ \text{N}

\]

Common Mistakes to Avoid

• Using the slant (hypotenuse) distance instead of the perpendicular distance to the pivot.
• Assigning the wrong sense (CW/CCW) to a moment.
• Adding moments without keeping track of their signs.
• Ignoring forces that act through the pivot – they produce zero moment but must still be shown.
• Forgetting to convert masses to forces (multiply by \(g\)) when the problem involves weights.

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 (\(d{L}\) and \(d{R}\)) change, altering the required balancing forces. The same procedure applies:

1. Measure the actual distances from the pivot to each line of action.
2. Calculate each moment \(F d\) with the correct sign.
3. Set \(\sum \tau =0\) and solve for the unknown quantity.