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प्रश्न
A non uniform beam of weight 120 N pivoted at one end is shown in the diagram below. Calculate the value of F to keep the beam in equilibrium.
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उत्तर
Given:
Weight of the beam, W = 120 N
The center of gravity (C.G.) is at a distance from the pivot = 0.20 m
F = 0.80 m
Formula:
W × distance to C.G. = F × distance to F
Solution:
∴ Anticlockwise moment = Clockwise moment
W × distance to C.G. = F × distance to F
120 × 0.20 = F × 0.80
24 = 0.80 F
`F = 24/0.80`
= 30 N
The value of F required to keep the beam in equilibrium is 30 N.
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