Moments, Equilibrium and Centre of Gravity: Question 6

Worked solution

Step 1: Find the perpendicular distance of each force from the pivot

The pivot is at the $50\text{ cm}$ mark.

• The $6.0\text{ N}$ force acts at the $20\text{ cm}$ mark, a perpendicular distance of $50 - 20 = 30\text{ cm} = 0.30\text{ m}$ from the pivot.
• The $9.0\text{ N}$ force acts at the $70\text{ cm}$ mark, a perpendicular distance of $70 - 50 = 20\text{ cm} = 0.20\text{ m}$ from the pivot.

Step 2: Calculate each moment

$\text{moment}_1 = 6.0\text{ N} \times 0.30\text{ m} = 1.8\text{ N m}$

$\text{moment}_2 = 9.0\text{ N} \times 0.20\text{ m} = 1.8\text{ N m}$

Step 3: Compare the two moments

The force at the $20\text{ cm}$ mark acts to the left of the pivot and pulls down, so it produces an anticlockwise moment. The force at the $70\text{ cm}$ mark acts to the right of the pivot and pulls down, so it produces a clockwise moment.

Both moments have the same size, $1.8\text{ N m}$, but act in opposite rotational directions. By the principle of moments, the clockwise moment equals the anticlockwise moment, so the rule stays in equilibrium.

Note that the rule’s own weight does not need to be considered: since the rule is uniform and pivoted exactly at its centre ($50\text{ cm}$ mark), its weight acts through the pivot itself and produces zero moment.

Why the other options are wrong

Final answers

• The rule remains balanced and horizontal, option C