A uniform horizontal beam XY of length 3.0\text{ m} and weight 80\text{ N} is hinged at en… — JAMB UTME

A uniform horizontal beam $XY$ of length $3.0\text{ m}$ and weight $80\text{ N}$ is hinged at end $X$ to a vertical wall. It is held in horizontal equilibrium by a light cable attached to end $Y$ that makes an angle of $30^\circ$ with the beam. What is the tension in the cable?

A
$40\text{ N}$
$80\text{ N}$ Answer
C
$160\text{ N}$
D
$240\text{ N}$

Answer

The tension in the cable is

80\text{ N}

Taking moments about the hinge at end

X

, the clockwise moment due to the weight of the beam acting at its midpoint (

1.5\text{ m}

) is

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

. The counterclockwise moment provided by the cable tension

T

at end

Y

(

3.0\text{ m}

) is

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

. Setting clockwise moments equal to counterclockwise moments gives

1.5 T = 120

, yielding

T = 80\text{ N}

Step-by-Step Solution

Identify the center of gravity and calculate the clockwise moment about the hinge at end

X

Since the beam is uniform, its weight of

80\text{ N}

acts at its center of gravity, which is at a distance of

1.5\text{ m}

from

X

. Clockwise moment =

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

For a uniform body, weight acts precisely at the midpoint.

Determine the counterclockwise moment exerted by the cable tension

T

about point

X

The perpendicular component of the tension is

T \sin(30^\circ)

. Counterclockwise moment =

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

Only the component of force perpendicular to the beam produces a moment about the pivot.

Apply the principle of moments for rotational equilibrium to solve for

T

1.5 T = 120 \implies T = \frac{120}{1.5} = 80\text{ N}

For equilibrium, total counterclockwise moments about any pivot must equal total clockwise moments.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

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