$\Sigma F_H = 0$
$B_H = P \cos 40^\circ$
$\Sigma M_A = 0$
$3B_V + 0.1P \cos 40\circ = 2P \sin 40^\circ$
$B_V = \frac{2}{3}P \sin 40^\circ - \frac{1}{30}P \cos 40^\circ$
$B_V = \frac{1}{30}(20 \sin 40^\circ - \cos 40^\circ)P ~ \text{N}$
$M_C = \Sigma M_\text{to the right of C}$
$M_C = 1 \times B_V$
$M_C = \frac{1}{30}(20 \sin 40^\circ - \cos 40^\circ)P ~ \text{N}\cdot\text{m}$
At the bottom of the section through C
$\sigma = \sigma_a + \sigma_f$
$\sigma = \dfrac{P}{A} + \dfrac{6M}{bd^2}$
$10 = \dfrac{P\cos 40^\circ}{100(200)} + \dfrac{6\left[ \frac{1}{30}(20 \sin 40^\circ - \cos 40^\circ)P \right] (1000)}{100(200^2)}$
$10 = \dfrac{\cos 40^\circ}{20\,000}P + \dfrac{20 \sin 40^\circ - \cos 40^\circ}{20\,000}P$
$10 = \dfrac{\cos 40^\circ}{20\,000}P + \dfrac{\sin 40^\circ}{1\,000}P - \dfrac{\cos 40^\circ}{20\,000}P$
$10 = \dfrac{\sin 40^\circ}{1\,000}P$
$P = 15\,557.24 ~ \text{N} = 15.56 ~ \text{kN}$ answer
