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A cylinder of weight 500N is kept on two inclined planes as shown in the figure. Determine the reactions at the contact points A & B.
1 Answer
written 3.9 years ago by |
Let C be the centre and A, B be the contact points.
Reaction forces RA,RB and weight ′W′are under equilibriumat centre point C.
Hence,RAsin(180−30)=RBsin(180−50)=500sin(30+50)
Hence,RAsin(150)=RBsin(130)=500sin(80)
RA=500sin(150)sin(80)
=253.8567 N
RB=500sin(130)sin(80)
=388.9310 N
Hence, the reaction at contact point A =RA=253.8567 N with ∠50∘and the reaction at contact point B =RB=388.9310 N with ∠30∘