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Question

For the truss shown in figure, the force in the member PQ is

A
4600 N
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B
4807.4 N
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C
5605.55 N
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D
3000 N
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Solution

The correct option is B 4807.4 N

Fx=0

Hp=0...............(1)

Fy=0

VP50002000+RA=0........(2)

MP=0

5000×2+2000×4RA×6=0

RA=180006=3000 N........(3)

Substituting (3) in (2)

VP=4000 N

Take the joint, P

Fy=0

VP=FPQsinθ

(Assume FPQ in downward direction)

4000=FPQ×313

FPQ=4807.4 N

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