Problem 346 | Equilibrium of Non-Concurrent Force System

$\Sigma M_A = 0$

$4(\frac{2}{\sqrt{5}}T) = 2(200) + 6(100)$

$T = 279.51 \, \text{ lb}$           answer
 

 

$\Sigma F_V = 0$

$A_V + \frac{2}{\sqrt{5}}T = 200 + 100$

$A_V + \frac{2}{\sqrt{5}}(279.51) = 300$

$A_V = 50 \, \text{ lb}$           answer
 

$\Sigma F_H = 0$

$A_H = \frac{1}{\sqrt{5}}T$

$A_H = \frac{1}{\sqrt{5}}(279.51)$

$A_H = 125 \, \text{ lb}$           answer