A 5 m long ladder is resting on a smooth vertical wall with its lower end 3 m from the wall. What should be the coefficient of friction between the ladder and the floor for equilibrium?

Concept:

The resting on between any frictional floor and a vertical wall will always satisfy all the static equilibrium condition i.e.

∑ F_x = ∑ F_y = ∑ M_{at any point} = 0

Calculation:

Given:

Length of ladder (AB) = 5 m, OB = 3 m

Let W will be the weight of the ladder, N_B and N_A will be support reaction, θ is the angle between ladder and floor and μ is the friction coefficient between ladder and floor.

Free body diagram of the ladder;

OA² = AB² - OB² , OA2 = 52 - 32 

OA2 = 16, OA = 4 m

From Δ OAB,

\(\cos θ = \frac{3}{5}\)

Now apply ∑ Fy = 0

N_B = W 
Now take moment about point A, which should be equal to zero

∑ MA = 0

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {W \times \frac{5}{2} \times \cos \theta } \right) = {N_B} \times 3\)

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {{N_B} \times \frac{5}{2} \times \frac{3}{5}} \right) = {N_B} \times 3\)

\(\left( {\mu \times 4} \right) + \left( {\frac{3}{2}} \right) =~3\)

\(\mu = \frac{3}{8}\)

Hence the value of the coefficient of friction between ladder and floor will be 3/8