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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.

a beam pivoted at one end has a force of 50N acting vertically upwards on it as shown the beam is in equilibrium what is the weight of the beam class=

Respuesta :

opudodennis

Answer:

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

onyebuchinnaji

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

The given parameters;

  • force at the end of the pivot, F = 5 N
  • position of the force, = 2 cm
  • center of gravity of the beam, C.G = 5 cm

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

clockwise moment = anticlockwise moment

[tex]W(5 \ cm) = F(2\ cm)\\\\W = \frac{F \times 2}{5} \\\\W = \frac{5 \times 2}{5} \\\\W = 2.0 \ N[/tex]

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

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