A beam pivoted at one end has a force of 5.0N acting vertically upwards on it as shown. the beam is in equilibrium.
what is the weight of the beam.

The weight of the beam at the given equilibrium position is 2.0 N.

The given parameters;

The weight of the beam is calculated by taking moment about the pivot of the beam as follows;

clockwise moment = anticlockwise moment

Thus, the weight of the beam at the given equilibrium position is 2.0 N.

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Option A

Explanation:

Taking moments about the pivot, we know that the sum of clockwise moments equal the sum of anticlockwise moments at equilibrium. Therefore, at the pivot point

5N*2cm=Weight of beam*5 cm

Weight of beam = 5N*2cm/5cm=2 N

Clearly, the weight of the beam is 2N for it to remain in equilibrium

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