$\Sigma M_B = 0$
$6P = 2T \sin 10^\circ$
$3P = T \sin 10^\circ$ → Equation (1)
$\Sigma F_H = 0$
$B_H = T \cos 10^\circ$
From Equation (1),
$T = \dfrac{3P}{\sin 10^\circ}$
Thus,
$B_H = \left( \dfrac{3P}{\sin 10^\circ} \right) \cos 10^\circ$
$B_H = 3 \cot 10^\circ \, P$
$\Sigma F_V = 0$
$B_V = T \sin 10^\circ + P$
Again from Equation (1),
$T \sin 10^\circ = 3P$
Thus,
$B_V = 3P + P$
$B_V = 4P$
${R_B}^2 = {B_H}^2 + {B_V}^2$
${R_B}^2 = (3 \cot 10^\circ \, P)^2 + (4P)^2$
$R_B^2 = 305.47P^2$
$R_B = 17.48P$
$P = \dfrac{R_B}{17.48}$ → Equation (2)
Based on tension of rod (equation 1):
$P = \frac{1}{3} T \sin 10^\circ$
$P = \frac{1}{3} [ \, 5000 \times \frac{1}{4} \pi (0.5)^2 \, ] \sin 10^\circ$
$P = 56.83 \, \text{lb}$
Based on shear of rivet (equation 2):
$P = \dfrac{4000 [ \, \frac{1}{4} \pi (0.25)^2 \, ]}{17.48}$
$P = 11.23 \, \text{lb}$
Safe load P,
$P = 11.23 \, \text{ lb}$ answer
