Answer
The tension in the wire is 642N. The horizontal reaction at the wall is 526N. The vertical reaction at the wall is 1.76N.
Work Step by Step
1) Sum the torques about the hinge.
$(+ \circlearrowleft) \sum \tau=0$
$Tsin35^{\circ}*1.35m-155N*.85m-215N*1.70m=0$
$T=(155N*.85m+215N*1.70m)/(sin35^{\circ}1.35m)$
$T\approx642N$
The tension in the wire is 642N
2) Sum the forces vertically
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$R_Y+sin35^{\circ}*T-155N-215N=0$
$R_Y=-sin35^{\circ}*T+155N+215N$
$R_Y\approx1.76N$
The vertical reaction at the wall is 1.76N.
3) Sum the forces horizontally.
$(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $
$R_X-T*cos35^{\circ}=0$
$R_X=T*cos35^{\circ}$
$R_X=642N*cos35^{\circ}$
$R_X\approx526N$
The horizontal reaction at the wall is 526N.
