A rectangular wall 4 meters high and 5 meters wide is to be painted. If the cost of painting the entire wall is Rs. 15 per square meter, what will be the total cost?

The ladder has weight $W = 200\ \mathrm{N}$ acting downward at its center.
Normal reaction $N_g$ from ground upward at lower end A.
Friction $f \leq \mu N_g$ at A, toward the wall ($\mu = 0.4$).
Normal reaction $N_w$ from wall, horizontal away from wall, at upper end B.
Ladder length $L = 5\ \mathrm{m}$.

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$$N_g = W$$ $$N_g = 200\ \mathrm{N}$$

$$f = N_w$$

For no slipping at minimum $\theta$, $f = \mu N_g$.

$$\mu N_g = N_w$$ $$N_w = 0.4 \times 200\ \mathrm{N} = 80\ \mathrm{N}$$

Clockwise moment by weight: $W \times \frac{L}{2} \cos \theta$.
Anticlockwise moment by $N_w$: $N_w \times L \sin \theta$.
For equilibrium:

$$N_w L \sin \theta = W \frac{L}{2} \cos \theta$$

Cancel $L$:

$$N_w \sin \theta = W \frac{1}{2} \cos \theta$$ $$\sin \theta = \frac{W}{2 N_w} \cos \theta$$

Substitute $W = 200\ \mathrm{N}$, $N_w = 80\ \mathrm{N}$:

$$\sin \theta = \frac{200}{2 \times 80} \cos \theta$$ $$\sin \theta = \frac{200}{160} \cos \theta$$ $$\sin \theta = 1.25 \cos \theta$$ $$\tan \theta = 1.25$$ $$\theta = \tan^{-1}(1.25)$$

$$1.25 = \frac{5}{4}$$ $$\theta = \tan^{-1}\left(\frac{5}{4}\right) \approx 51.34^\circ$$

$\theta_\mathrm{min} = \tan^{-1}(1.25) \approx 51.3^\circ$
(or exactly $\theta_\mathrm{min} = \cot^{-1}(2 \times 0.4) = \cot^{-1}(0.8)$)
\theta_\min \approx 51.3^\circ