Answer
$F = (-45~\hat{i}+200~\hat{j})~N$
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 $F_w$, 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:
$\sum \tau = 0$
$h~F_w-(5~cos~\theta)~mg - (2.0~sin~\theta)~F = 0$
$h~F_w = (5~cos~\theta)~mg + (2.0~sin~\theta)~F$
$F_w = \frac{(5~cos~\theta)~mg + (2.0~sin~\theta)~F}{h}$
$F_w = \frac{(5~cos~53.13^{\circ})~(200~N) + (2.0~sin~53.13^{\circ})~(150~N)}{8.0~m}$
$F_w = 105~N$
To find the horizontal component of the force on the ladder due to the ground, we can consider the horizontal forces on the ladder:
$F_x+150~N-105~N = 0$
$F_x = 105~N - 150~N$
$F_x = -45~N$
The vertical component of the force on the ladder due to the ground is equal in magnitude to the weight of the ladder:
$F_y = 200~N$
We can express the force on the ladder due to the ground in unit-vector notation:
$F = (-45~\hat{i}+200~\hat{j})~N$
