IP Physics Notes (Upper Secondary, Year 3-4): 4) Turning Forces

Quick recap -- A turning effect appears when a force acts with an offset from a pivot. Balance comes from matching clockwise and anticlockwise moments while keeping the net force zero.

Moment of a Force

• The moment (torque) of a force measures its turning effect about a pivot: \( \tau = F d_\perp \), where \( d_\perp \) is the perpendicular distance from the pivot to the force's line of action.
• SI unit: \( \pu{N.m} \). It is a vector quantity whose sense is clockwise or anticlockwise around the pivot.
• Increase moment by raising the force magnitude, pushing further from the pivot, or pushing perpendicular to the lever arm. Oblique pushes need resolving into the perpendicular component first.
• Three classic lever classes:
  • Class 1 (pivot between effort and load, e.g., seesaw) can magnify force or distance.
  • Class 2 (load between pivot and effort, e.g., wheelbarrow) always multiplies effort.
  • Class 3 (effort between pivot and load, e.g., tweezers) trades force multiplication for larger movement.

Principle of Moments

• For a body in rotational equilibrium about a point: \[ \sum \tau_{\text{clockwise}} = \sum \tau_{\text{anticlockwise}} \]
• Equilibrium also requires zero resultant force: \( \sum \vec{F} = \vec{0} \). Both translational and rotational conditions must hold simultaneously.
• Problem-solving flow:
  1. Draw the lever, marking all forces and pivot locations.
  2. Replace angled forces with perpendicular components about the pivot.
  3. Take moments about the pivot (or any convenient point) and equate clockwise/anticlockwise terms.
  4. Use \( \sum \vec{F} = \vec{0} \) if additional unknown forces remain.

Common Moment Scenarios

• Balanced beam: equal masses on a metre rule satisfy \( m_1 g d_1 = m_2 g d_2 \).
• Door or wrench: wider grip (larger \( d_\perp \)) reduces required force for the same torque.
• Human arm: biceps attach close to the elbow, so muscles exert large forces to generate modest torques -- a Class 3 lever case study.

Centre of Gravity

• The centre of gravity (COG) is the point where the entire weight of a body may be considered to act.
• Uniform, symmetrical objects have their COG at the geometric centre. Irregular shapes need experimental determination.
• Suspend-and-plumb method for a flat irregular lamina:
  1. Drill three small holes near the edge, far apart.
  2. Suspend the lamina from the first hole using a pin and let it hang freely.
  3. Hang a plumb line from the same pin; mark the vertical traced by the string.
  4. Repeat from the other holes; the intersection of the lines locates the COG.
• The object settles with its COG vertically beneath the suspension point because any offset would create an unbalanced moment and cause rotation.

Stability and Equilibrium Types

• Stability depends on how the COG moves relative to the base when the object is tilted.
• Stable equilibrium: after a small tilt, the COG rises and its vertical line falls inside the base. Weight creates a restoring moment back to the original position.
• Unstable equilibrium: the COG shifts beyond the base, so weight produces an overturning moment that topples the object.
• Neutral equilibrium: the COG remains at the same height regardless of position (e.g., a ball on a flat surface); no net moment develops.
• Improve stability by lowering the COG or widening the base so the vertical line through the COG remains within support boundaries.

Worked Example: Uniform Beam with Hanging Masses

A light, uniform \( \pu{1.2 m} \) beam is pivoted \( \pu{0.30 m} \) from its left end. A ( \pu{25 N} ) weight hangs \( \pu{0.10 m} \) from the left end. Find the distance from the pivot to the point where a ( \pu{15 N} ) weight must hang on the right to balance the beam.

1. The beam's own weight acts at its centre; because it is light, neglect it (or treat it as already balanced by the pivot).
2. Distance of the ( \pu{25 N} ) weight from the pivot: \( d_1 = \pu{0.30 m} - \pu{0.10 m} = \pu{0.20 m} \).
3. Take moments about the pivot. Define clockwise as positive. The left weight produces an anticlockwise moment, the right weight produces clockwise. \[ 15 \times d_2 = 25 \times 0.20 \]
4. Solve: \( d_2 = \dfrac{25 \times 0.20}{15} \approx \pu{0.33 m} \).
5. Measure \( \pu{0.33 m} \) to the right of the pivot (i.e., about ( \pu{0.63 m} ) from the left end) to achieve balance.

Key Takeaways

• Treat every moment problem with a clear diagram and perpendicular distances.
• Equilibrium requires both zero net force and matched clockwise/anticlockwise moments.
• Locate the COG experimentally with the suspend-and-plumb method, then reason about stability via how the COG shifts relative to the object's base.