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?

Step 1: Identify forces acting on the ladder.
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 $L = 5\ \mathrm{m}$.

Step 2: Resolve vertically at A.
$N_g = W$
$N_g = 200\ \mathrm{N}$

Step 3: Resolve horizontally.
$f = N_w$
For no slipping at minimum $\theta$, $f = \mu N_g$.
$\mu N_g = N_w$
$N_w = 0.4 \times 200\ N = 80\ \mathrm{N}$

Step 4: Take moments about A (eliminates $N_g$ and $f$).
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 $W = 200\ \mathrm{N}$, $N_w = 80\ \mathrm{N}$:
$\sin \theta = \frac{200}{2 \times 80} \cos \theta$
$\sin \theta = \frac{200}{160} \cos \theta$
$\sin \theta = 1.25 \cos \theta$
$\tan \theta = 1.25$
$\theta = \tan^{-1}(1.25)$

Step 5: Calculate numerical value.
$1.25 = \frac{5}{4}$
$\theta = \tan^{-1}\left(\frac{5}{4}\right) \approx 51.34^\circ$

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