Answer
The magnitude of the horizontal component of the force on the rod from the wall is $~~245~N$
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 above the horizontal is $\theta = 53.1^{\circ}$
To find the horizontal component of the force on the rod from the wall $F_{wx}$, we can consider the net horizontal force on the rod:
$\sum F_x = 0$
$F_{wx} - T~cos~\theta = 0$
$F_{wx} = T~cos~\theta$
$F_{wx} = (408.5~N)~cos~53.1^{\circ}$
$F_{wx} = 245~N$
The magnitude of the horizontal component of the force on the rod from the wall is $~~245~N$.
