A light rigid rod OP of length 1.2\text{ m} is pivoted smoothly at end O. A vertical downw… — JAMB UTME

A light rigid rod $OP$ of length $1.2\text{ m}$ is pivoted smoothly at end $O$ . A vertical downward load of $30\text{ N}$ is suspended from end $P$ . An upward force $F$ is applied at the midpoint of the rod at an angle of $30^\circ$ to the horizontal rod to keep it in horizontal equilibrium. What is the magnitude of the force $F$ ?

A
$60\text{ N}$
$120\text{ N}$ Answer
C
$240\text{ N}$
D
$40\text{ N}$

Answer

The magnitude of the force

F

120\text{ N}

The correct answer is

120\text{ N}

. For rotational equilibrium about the pivot, the clockwise moment created by the suspended load (

30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

) must equal the counterclockwise moment generated by force

F

. Because force

F

acts at an angle of

30^\circ

at the midpoint (

0.6\text{ m}

), its effective perpendicular component is

F \sin(30^\circ) = 0.5F

. Equating the moments gives

0.5F \times 0.6\text{ m} = 36\text{ N}\cdot\text{m}

, which yields

0.3F = 36

, resulting in

F = 120\text{ N}

Step-by-Step Solution

Calculate the clockwise moment about the pivot

O

due to the load at end

P

\tau_{\text{clockwise}} = 30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

The force of

30\text{ N}

acts vertically downwards at a perpendicular distance of

1.2\text{ m}

from the pivot.

Determine the perpendicular distance (or perpendicular component of force) for

F

applied at the midpoint.

Midpoint distance =

\frac{1.2\text{ m}}{2} = 0.6\text{ m}

; Perpendicular force component =

F \sin(30^\circ) = 0.5F

Only the component perpendicular to the line of action contributes to the moment about pivot

O

Set up the counterclockwise moment expression about pivot

O

\tau_{\text{counterclockwise}} = F \sin(30^\circ) \times 0.6\text{ m} = 0.3F\text{ N}\cdot\text{m}

The moment is the product of the perpendicular force component and the distance from the pivot.

Equate clockwise and counterclockwise moments to solve for

F

0.3F = 36 \implies F = \frac{36}{0.3} = 120\text{ N}

According to the principle of moments, total clockwise moments must equal total counterclockwise moments for rotational equilibrium.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 15s

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