Problem 271 | Resultant of Non-Concurrent Force System

For vertical resultant, R_x = 0 and R_y = R
 

$M_R = \Sigma M_E$

$R(2) = T_x(4) - T_y(2)$

$R(2) = 361(\frac{3}{\sqrt{13}})(4) - 361(\frac{2}{\sqrt{13}})(2)$

$R = 400.49 \, \text{ lb downward at point A}$
 

 

$\Sigma M_B = M_R$

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

$F(\frac{1}{\sqrt{10}})(2) + F(\frac{3}{\sqrt{10}})(1) = 400.49(1)$

$F = 253.29 \, \text{ lb}$           answer
 

$\Sigma F_V = R$

$P_y + T_y - F_y = R$

$P(\frac{2}{\sqrt{5}}) + T(\frac{2}{\sqrt{13}}) - F(\frac{3}{\sqrt{10}}) = R$

$P(\frac{2}{\sqrt{5}}) + 361(\frac{2}{\sqrt{13}}) - 253.29(\frac{3}{\sqrt{10}}) = 400.49$

$P = 492.53 \, \text{ lb}$           answer