Moments, Equilibrium and Centre of Gravity: Question 10

Worked solution

Step 1: Combine the moments on the left of the pivot

Both the $40\text{ N}$ and $20\text{ N}$ forces act on the same side of the pivot (the left), so they produce moments in the same rotational direction (anticlockwise), and their moments add together:

$\text{moment of } 40\text{ N} = 40\text{ N} \times 0.50\text{ m} = 20\text{ N m}$

$\text{moment of } 20\text{ N} = 20\text{ N} \times 0.30\text{ m} = 6\text{ N m}$

$\text{total anticlockwise moment} = 20\text{ N m} + 6\text{ N m} = 26\text{ N m}$

Step 2: Apply the principle of moments

The bar is in equilibrium, so the total anticlockwise moment must equal the clockwise moment produced by $F$:

$F \times 0.80\text{ m} = 26\text{ N m}$

Step 3: Solve for $F$

$F = \frac{26\text{ N m}}{0.80\text{ m}}$

$F = \boxed{32.5\text{ N}}$

Why the other options are wrong

Final answers

• $F =$ $32.5\text{ N}$, option C