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Q)

Two identical ladders are arranged as shown in the figure, resting on a horizontal surface. Mass of each ladder is M and length L. A block of mass m hangs from the apex point P. If the system is in equilibrium, find direction and magnitude of friction force acting at A or B.

a)Mmg b)(M-m)g c) (M+m)g d) (M/m)g

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A)
since the system is in equilibrium,

mgsintheta +Mgsintheta - N2costheta = 0, ( N2 is the component of the frictional force acting parallel to the ground through P)

therefore N2 = gsintheta(M+m)/Lcostheta

The frictional forces N1at A and B is perpendicular to the ground in the  upward direction = (M+m)g
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