A 13 m long ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the coefficient of friction between the ladder and the floor so that the ladder remains in equilibrium?

Explanation:

The free body diagram of the ladder is shown in the figure below

\(\tan \theta = \frac{{12}}{5}\)

Balancing of force in the vertical direction

\(\sum {F_V} = 0\) , N₁ = W

The moment about point A is zero for equilibrium

\(\sum {M_A} = 0\),

\( - W\frac{l}{2}\cos \theta - μ {N_1}l\sin \theta + {N_1}l\cos \theta = 0\)

\( \Rightarrow μ \sin \theta = \;\frac{{\cos \theta }}{2}\;\)

\( \Rightarrow μ = \frac{1}{{2\tan \theta }} = \frac{1}{{2 \times \frac{{12}}{5}}} = {\rm{\;}}0.208{\rm{\;}}\)

μ  = 0.208 ∝ 0.21