A non-uniform wooden pole of length 6.0\text{ m} and weight 150\text{ N} is balanced horiz… — JAMB UTME

A non-uniform wooden pole of length $6.0\text{ m}$ and weight $150\text{ N}$ is balanced horizontally on a pivot placed $2.4\text{ m}$ from its heavy end $P$ . The system achieves rotational equilibrium when a load of $50\text{ N}$ is hung directly from end $P$ . What is the distance of the center of gravity of the pole from end $P$ ?

Answer: 3.2 m

Answer

The distance of the center of gravity of the pole from end

P

3.2\text{ m}

Taking moments about the pivot at

2.4\text{ m}

from end

P

, the counter-clockwise moment created by the

50\text{ N}

load (

50\text{ N} \times 2.4\text{ m} = 120\text{ N}\cdot\text{m}

) must balance the clockwise moment created by the

150\text{ N}

weight of the pole acting at its center of gravity (

150\text{ N} \times (d - 2.4\text{ m})

). Equating these gives

120 = 150(d - 2.4)

, leading to

d - 2.4 = 0.8\text{ m}

, so

d = 3.2\text{ m}

Step-by-Step Solution

Identify force positions relative to the pivot

The

50\text{ N}

load is

2.4\text{ m}

to the left of the pivot. The

150\text{ N}

weight acts at the center of gravity, which is

(d - 2.4\text{ m})

to the right of the pivot.

Moments are evaluated relative to the fulcrum to eliminate the unknown normal reaction force at the pivot.

Apply the Principle of Moments

Anti-clockwise moment =

50 \times 2.4 = 120\text{ N}\cdot\text{m}

. Clockwise moment =

150 \times (d - 2.4)

. Setting them equal:

120 = 150(d - 2.4)

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anti-clockwise moment.

Solve the equation for distance

d

d - 2.4 = 0.8 \implies d = 3.2\text{ m}

Adding the displacement from the pivot (

0.8\text{ m}

) to the pivot position from end

P

(

2.4\text{ m}

) yields the position of the center of gravity from end

P

Key Concept

Rotational equilibrium and Principle of Moments for non-uniform rigid bodies

Estimated Time:1m 30s

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