Answer
$F = 190~N$
Work Step by Step
We can find the angle $\theta$ the ladder makes with the ground:
$sin~\theta = \frac{8.0~m}{10~m}$
$\theta = sin^{-1}~(\frac{8.0~m}{10~m})$
$\theta = 53.13^{\circ}$
To find an expression for $F_w$, the magnitude of the force of the wall on the ladder, we can consider the torque about a rotation axis at the bottom of the ladder:
$\sum \tau = 0$
$h~F_w-(5~cos~\theta)~mg - (2.0~sin~\theta)~F = 0$
$h~F_w = (5~cos~\theta)~mg + (2.0~sin~\theta)~F$
$F_w = \frac{(5~cos~\theta)~mg + (2.0~sin~\theta)~F}{h}$
$F_w = \frac{(5~cos~53.13^{\circ})~(200~N) + (2.0~sin~53.13^{\circ})~F}{8.0~m}$
$F_w = \frac{(5~cos~53.13^{\circ})~(200~N)}{8.0~m} + \frac{(2.0~sin~53.13^{\circ})~F}{8.0~m}$
$F_w = 75+0.20~F$
We can find the minimum value of $F$:
$mg~\mu_s+F_w-F = 0$
$F = mg~\mu_s+75+0.20~F$
$0.80~F = mg~\mu_s+75$
$F = \frac{mg~\mu_s+75}{0.80}$
$F = \frac{(200~N)(0.38)+(75~N)}{0.80}$
$F = 190~N$
