Answer
The magnitude of the vertical component of the force on the rod from the wall is $~~163~N$
Work Step by Step
In part (a), we found that the tension in the cable is $~~T = 408.5~N$ and the angle of the cable is $\theta = 53.1^{\circ}$
We can let "up" be the positive direction.
To find the vertical component of the force on the rod from the wall $F_{wy}$, we can consider the net vertical force on the rod:
$\sum F_y = 0$
$F_{wy} + T~sin~\theta - (50.0~kg)(9.8~m/s^2) = 0$
$F_{wy} = (50.0~kg)(9.8~m/s^2)- T~sin~\theta$
$F_{wy} = (50.0~kg)(9.8~m/s^2)- (408.5~N)~sin~53.1^{\circ}$
$F_{wy} = 163~N$
The magnitude of the vertical component of the force on the rod from the wall is $~~163~N$.
