A light rigid lever of length 2.5\text{ m} is pivoted horizontally at one end O. A downwar… — JAMB UTME

A light rigid lever of length $2.5\text{ m}$ is pivoted horizontally at one end $O$ . A downward vertical weight of $60\text{ N}$ is hung from the lever at a distance of $1.5\text{ m}$ from $O$ . An upward force $F$ inclined at an angle of $30^\circ$ to the lever is applied at the free end. What is the magnitude of the force $F$ required to maintain horizontal equilibrium?

A
$36\text{ N}$
$72\text{ N}$ Answer
C
$104\text{ N}$
D
$180\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

72\text{ N}

According to the Principle of Moments, for a body in rotational equilibrium, the total clockwise moment about a pivot equals the total counterclockwise moment. The load creates a clockwise moment of

60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

. The force

F

applied at

30^\circ

has a perpendicular component of

F \sin 30^\circ = 0.5F

. The counterclockwise moment is

0.5F \times 2.5\text{ m} = 1.25F

. Setting

1.25F = 90\text{ N}\cdot\text{m}

gives

F = 72\text{ N}

Step-by-Step Solution

Calculate the clockwise moment about the pivot

O

caused by the hanging load.

\text{Clockwise Moment} = 60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

The force of

60\text{ N}

acts perpendicularly at a distance of

1.5\text{ m}

from the pivot.

Express the counterclockwise moment about the pivot

O

exerted by the force

F

\text{Counterclockwise Moment} = F \sin(30^\circ) \times 2.5\text{ m} = 1.25 F\text{ N}\cdot\text{m}

Only the perpendicular component of the force,

F \sin(30^\circ)

, contributes to the moment about pivot

O

Equate the clockwise and counterclockwise moments according to the Principle of Moments.

1.25 F = 90 \implies F = \frac{90}{1.25} = 72\text{ N}

For rotational equilibrium, the sum of clockwise moments must equal the sum of counterclockwise moments.

Key Concept

Principle of Moments and Perpendicular Force Components

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