Moments, Equilibrium and Centre of Gravity: Question 2

Worked solution

Part (a): The principle of moments

For an object in equilibrium (here, the balanced seesaw), the principle of moments states:

$\text{sum of clockwise moments about the pivot} = \text{sum of anticlockwise moments about the pivot}$

Part (b): Moment of the first child

$\text{moment} = \text{force} \times \text{perpendicular distance}$

$\text{moment} = 300\text{ N} \times 1.5\text{ m}$

$\text{moment} = \boxed{450\text{ N m}}$

Part (c): Finding the second child’s weight

Since the seesaw is balanced, the moment of the second child must equal the moment of the first child found in part (b):

$W_2 \times 1.8\text{ m} = 450\text{ N m}$

Rearrange to make $W_2$ the subject:

$W_2 = \frac{450\text{ N m}}{1.8\text{ m}}$

$W_2 = \boxed{250\text{ N}}$

Part (d): Why the seesaw’s own weight can be ignored

The seesaw is described as balanced and pivoted at its centre, and it is uniform (its mass is spread evenly along its length). This means its centre of gravity, the single point where its whole weight can be considered to act, is located exactly at the pivot.

Since the perpendicular distance from the pivot to the line of action of the seesaw’s own weight is zero, the moment this weight produces about the pivot is also zero ($\text{moment} = \text{weight} \times 0 = 0$). A force with zero moment has no effect on the balance, so the seesaw’s own weight can be left out of the calculation entirely.

Final answers

• (a) Clockwise moments about the pivot $=$ anticlockwise moments about the pivot
• (b) Moment of the first child $=$ $450\text{ N m}$
• (c) Weight of the second child $=$ $250\text{ N}$
• (d) The seesaw’s weight acts at the pivot, so its perpendicular distance from the pivot is zero, giving zero moment