Answer in Brief

In the rocket arm shown in the figure the moment of ‘F’ about ‘O’ balances that P=250 N.Find F.

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#### 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**

Concept: Resultant of concurrent forces

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