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Question

A ladder of mass M and length L is supported in equilibrium against a smooth vertical wall and a rough horizontal surface, as shown in figure. If θ be the angle of inclination of the rod with the horizontal, then calculate
(a) normal reaction of the wall on the ladder
(b) normal reaction of the ground on the ladder
(c) net force applied by the ground on the ladder
981815_158a4515bc88422290abc9d5acd9fad0.png

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Solution

The free-body diagram of the ladder is shown in figure. Note that all the three forces acting on the ladder have to pass through a common point O, otherwise it cannot be in equilibrium.
The total reaction force F from the horizontal surface is inclined at an angle α(>θ) to the horizontal. The horizontal component of this force is friction force and its vertical component is the normal reaction applied by the ground on the ladder.
(b) Applying the conditions of equilibrium, we get
Fx=0fN=0f=N...(i)
Fy=0N1Mg=0N1=Mg...(ii)
Taking torque about centre of the rod C,τC=0
(N1)L2cosθ(f)L2sinθNL2sinθ=0...(iii)
Substituting the value of N1 and f from Eqs. (i) and (ii) in Eq. (iii), we get
MgL2cosθNLsinθ=0
N=Mg2tanθ=12Mgcotθ...(iv)
Thus, the net force applied by ground on the ladder is
F=N12+f2
F=12Mg4+cot2θ

1028870_981815_ans_fee9c8a9c88f41c6bb58d953c6b21e7d.png

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