Question

a uniform bar ab of length 5 cm is 60 Newton the bar is suspended in a horizontal position by two vertical strings X and Y.if string x is 0.6m fromA and string Y is 1.8 from B.find the tension in the string​

Answer 1

Explanation:

Since the system is in equilibrium;

The moment of forces about point A should be zero.

Thus;

10g×0.5+15g×1=Tsin53o×150+150=0.8TT=250 N

At point A, the hinge force is working which can be resolved as Ry & Rx

Rx=Tcos 53oRx=250×0.6=150 NRy=Tsin53oRy=250×0.8=200 N

R=(Rx)2+(Ry)2

R=1502+2002

R=250 N

Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N

Answer 2

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