A uniform horizontal rod AB of length 2.0\text{ m} and mass 6.0\text{ kg} is suspended hor… — JAMB UTME

A uniform horizontal rod $AB$ of length $2.0\text{ m}$ and mass $6.0\text{ kg}$ is suspended horizontally by two light vertical strings attached at end $A$ and at a point $C$ located $0.5\text{ m}$ from end $B$ . Taking $g = 10\text{ m/s}^2$ , what is the tension in the string at point $C$ in Newtons?

Answer: 40 N

Answer

The tension in the string at point C is 40 N.

The weight of the rod (

60\text{ N}

) acts at its midpoint (

1.0\text{ m}

from end

A

). Point

C

is located

1.5\text{ m}

from end

A

. Taking moments about end

A

gives

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

, which evaluates to

T_C = 40\text{ N}

Step-by-Step Solution

Calculate total weight and identify the center of gravity position

Weight

W = 60\text{ N}

acting at

1.0\text{ m}

from end

A

For a uniform rod, the weight acts vertically downwards at its geometric center.

Set up the moment equilibrium equation taking end A as pivot

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

Taking moments about point

A

eliminates the force at

A

and equates clockwise moment from weight to counter-clockwise moment from tension at

C

Solve for the tension force at point C

T_C = 40\text{ N}

Dividing the total moment of

60\text{ N}\cdot\text{m}

by the moment arm of

1.5\text{ m}

yields

40\text{ N}

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

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