A uniform horizontal beam AB of length 4.0\text{ m} and mass 10\text{ kg} is hinged smooth… — JAMB UTME

A uniform horizontal beam $AB$ of length $4.0\text{ m}$ and mass $10\text{ kg}$ is hinged smoothly to a vertical wall at end $A$ . It is held horizontally in static equilibrium by a light cable attached to end $B$ and anchored to the wall above $A$ , making an angle of $30^\circ$ with the beam. A mass of $5\text{ kg}$ is suspended from the beam at a distance of $3.0\text{ m}$ from hinge $A$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the cable?

$175\text{ N}$ Answer
B
$87.5\text{ N}$
C
$275\text{ N}$
D
$17.5\text{ N}$

Answer

The tension in the cable is

175\text{ N}

Applying the principle of moments about the hinge at end A, the clockwise moments due to the beam's center of mass (

100\text{ N}

2.0\text{ m}

) and the suspended load (

50\text{ N}

3.0\text{ m}

) are balanced by the counterclockwise moment of the cable tension (

T \sin 30^\circ

4.0\text{ m}

). Solving

(100 \times 2.0) + (50 \times 3.0) = 2.0 T

yields

T = 175\text{ N}

Step-by-Step Solution

Calculate the downward gravitational forces (weights) acting on the system.

Weight of beam

W_{\text{beam}} = m_{\text{beam}} g = 10\text{ kg} \times 10\text{ m/s}^2 = 100\text{ N}

acting at

2.0\text{ m}

from

A

. Weight of load

W_{\text{load}} = m_{\text{load}} g = 5\text{ kg} \times 10\text{ m/s}^2 = 50\text{ N}

acting at

3.0\text{ m}

from

A

Forces causing clockwise moments must be expressed in force units (newtons) and located at their respective lines of action.

Formulate the equilibrium condition using the Principle of Moments about hinge

A

\sum \tau_A = 0 \implies (W_{\text{beam}} \times 2.0\text{ m}) + (W_{\text{load}} \times 3.0\text{ m}) = T \sin(30^\circ) \times 4.0\text{ m}

The hinge

A

eliminates reaction forces at the hinge from the moment equation.

Substitute numerical values and solve for tension

T

(100 \times 2.0) + (50 \times 3.0) = T \times 0.5 \times 4.0 \implies 200 + 150 = 2.0 T \implies 350 = 2.0 T \implies T = 175\text{ N}$.

Perpendicular distance from

A

to line of action of tension is

4.0 \sin 30^\circ = 2.0\text{ m}

Key Concept

Equilibrium of rigid bodies and Principle of Moments under non-perpendicular forces

Estimated Time:2m 0s

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