Problem 270 | Resultant of Non-Concurrent Force System

For horizontal resultant, R_y = 0 and R_x = R
 

$M_R = \Sigma M_B$

$R(1) = F_x(2) + F_y(1)$

$R(1) = 316(\frac{1}{\sqrt{10}})(2) + 316(\frac{3}{\sqrt{10}})(1)$

$R = 499.64 \, \text{ lb to the right at A}$
 

 

$\Sigma M_C = M_R$

$F_x(4) - F_y(2) + P_y(4) = R(3)$

$316(\frac{1}{\sqrt{10}})(4) - 316(\frac{3}{\sqrt{10}})(2) + P(\frac{2}{\sqrt{5}})(4) = 499.64(3)$

$P = 474.82 \, \text{ lb}$           answer
 

$\Sigma M_D = M_R$

$F_x(4) + F_y(2) - T_y(4) = R(3)$

$316(\frac{1}{\sqrt{10}})(4) + 316(\frac{3}{\sqrt{10}})(2) - T(\frac{2}{\sqrt{13}})(4) = 499.64(3)$

$T = -225.18 \, \text{ lb}$           answer