A uniform horizontal wooden beam AB of length 4.0\text{ m} and mass 8.0\text{ kg} is suppo… — JAMB UTME

A uniform horizontal wooden beam $AB$ of length $4.0\text{ m}$ and mass $8.0\text{ kg}$ is supported on a pivot placed $1.0\text{ m}$ from end $A$ . A mass of $5.0\text{ kg}$ is suspended from end $B$ . What is the magnitude of the downward vertical force $F$ in newtons that must be applied at end $A$ to keep the beam in horizontal equilibrium? (Take $g = 10\text{ m/s}^2$ ).

Answer: 230 N

Answer

The magnitude of the downward force required at end A to maintain horizontal equilibrium is 230 N.

Taking moments about the pivot, the downward force F at end A produces an anticlockwise moment of F × 1.0 m. This balances the clockwise moments produced by the weight of the beam (80 N × 1.0 m) and the load at end B (50 N × 3.0 m). Equating anticlockwise and clockwise moments gives F × 1.0 = 80 + 150 = 230 N.

Step-by-Step Solution

Determine the weights and distance of each force from the pivot point

Weight of beam = 80 N acting at 1.0 m right of pivot; weight at B = 50 N acting at 3.0 m right of pivot; force F acts at 1.0 m left of pivot.

The center of gravity of a uniform 4.0 m beam is at its midpoint (2.0 m from end A).

Set up the moment equilibrium equation about the pivot

F × 1.0 = (80 × 1.0) + (50 × 3.0)

For static rotational equilibrium, total anticlockwise moments equal total clockwise moments.

Solve for the force magnitude F

F = 230 N

Summing clockwise moments yields 80 + 150 = 230 N m, which divided by 1.0 m gives F = 230 N.

Key Concept

Principle of Moments and Static Equilibrium

Estimated Time:1m 30s

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