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Question
In the rocket arm shown in the figure the moment of ‘F’ about ‘O’ balances that P=250 N.Find F. 
Solution
Solution :
Given : P = 250 N
To find : Magnitude of force F
Solution :
`tan alpha =1/2`
=0.5
`alpha =26.5651°`
`tan θ (DE)/(AD)=(DE)/(BC)=3/4=0.75`
θ = 36.87°
∠CBD = ∠PBD = θ = 36.87°
∠CBP=2 θ = 2 x 36.87 = 73.74°
It is given that at O the moment of F about O balances the moment of P
Fcos α x OA = Psin2 θ x OB
Fcos26.5651 x 6 = 250sin 73.74 x 5
F=223.6068 N
Magnitude of force F= 223.6068 N
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