A uniform rigid beam MN of length 2.0\text{ m} and weight 80\text{ N} is smoothly pivoted … — JAMB UTME

A uniform rigid beam $MN$ of length $2.0\text{ m}$ and weight $80\text{ N}$ is smoothly pivoted at end $M$ . A vertical load of $120\text{ N}$ is hung at a distance of $1.5\text{ m}$ from $M$ . The beam is held horizontally in equilibrium by a cable attached at end $N$ pulling upward at an angle of $30^\circ$ to the horizontal beam. What is the magnitude of the tension in the cable?

A
$100\text{ N}$
B
$130\text{ N}$
$260\text{ N}$ Answer
D
$340\text{ N}$

Answer

The magnitude of the tension in the cable is

260\text{ N}

For rotational equilibrium about pivot

M

, the sum of clockwise moments must equal the counterclockwise moment. Clockwise moment from the beam's weight and load is

(80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 260\text{ N}\cdot\text{m}

. Counterclockwise moment from the cable tension is

T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}

. Equating the two gives

T = 260\text{ N}

Step-by-Step Solution

Identify the positions and lines of action of all forces relative to pivot

M

The weight of the uniform beam (

80\text{ N}

) acts at its center of gravity (

1.0\text{ m}

from

M

). The suspended load (

120\text{ N}

) acts at

1.5\text{ m}

from

M

. Cable tension

T

acts at

2.0\text{ m}

from

M

at an angle of

30^\circ

to the beam.

Rotational equilibrium requires evaluating moments created by all forces about the pivot point.

Calculate the sum of clockwise moments about pivot

M

\sum \tau_{\text{clockwise}} = (80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 80\text{ N}\cdot\text{m} + 180\text{ N}\cdot\text{m} = 260\text{ N}\cdot\text{m}

Both downward forces exert clockwise turning effects about pivot

M

Determine the counterclockwise moment exerted by the inclined cable tension

T

\tau_{\text{counterclockwise}} = T \times d \sin\theta = T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}.

Only the component of tension perpendicular to the beam (

T \sin 30^\circ

) produces a moment about the pivot.

Apply the Principle of Moments to calculate tension

T

1.0 T = 260 \implies T = 260\text{ N}.

For the beam to remain horizontally balanced, total clockwise moment must equal total counterclockwise moment.

Key Concept

Principle of Moments and Rotational Equilibrium with Inclined Forces

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