A uniform rod with weight 40N rest on two supports p and q and three weights are suspended . calculate the reaction on x and y
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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)2R=1502+2002R=250 N
Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N
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