Question

# Two identical ladders are arranged as shown in the figure. Mass of each ladder is M and length L. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

A
f=(M+m2)gtanθ horizontally outwards
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B
f=(M+m2)gcotθ horizontally inwards
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C
f=(M+m2)gcosθ horizontally outwards
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D
None of these.
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Solution

## The correct option is B f=(M+m2)gcotθ horizontally inwardsDrawing the FBD of both rods As the rods are in equilibriumΣFX=0ΣFy=0τnet=0For right rodN2+mg2+Mg=N (1)N1=f (2)For left rodN2+N=Mg+mg2 (3)From (1)and(3)N1=f=N2=0=N=Mg+mg2Balancing torque about O,MgL2cosθ+fLsinθ=NLcosθMgcosθ2+fsinθ=(Mg+mg2)cosθf=(Mg2+mg2)cotθ=f=(M+m2)gcotθ

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