Question #ed224

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Question #ed224

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Answer 1 · 2017-01-06T14:20:57

Option -A

Explanation:

given

Considering the equilibrium of forces acting on the beam we can say that the moment of the applied upward force $F = 5 N$ $F=5N$ acting at P about the pivot O will be equal to the moment of the weight $W$ $W$ about O, but they will be oppositely directed.

So

$O Q \times W = O P \times F$ $OQxxW=OPxxF$

$\Rightarrow W = \frac{O P \times F}{O Q} = \frac{O P \times F}{O P + P Q} = \frac{2 \times 5}{(2 + 3)} = 2 N$ $=>W=(OPxxF)/(OQ)=(OPxxF)/(OP+PQ)=(2xx5)/((2+3))=2N$

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Question #ed224

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