Answer
The magnitude of the net force of the floor on the beam is $~~376~N$
Work Step by Step
In part (a), we found that $~~P = 200~N$
Also, we found that the beam makes an angle of $\theta = 36.87^{\circ}$ above the horizontal.
We can find the vertical component of the force of the floor on the beam:
$F_y+P~cos~\theta-mg = 0$
$F_y = mg - P~cos~\theta$
$F_y = (500~N) - (200~N)~cos~36.87^{\circ}$
$F_y = 340~N$
We can find the horizontal component of the force of the floor on the beam:
$F_x-P~sin~\theta = 0$
$F_x = P~sin~\theta$
$F_x = (200~N)~sin~36.87^{\circ}$
$F_x = 160~N$
We can find the magnitude of the net force of the floor on the beam:
$F = \sqrt{(160~N)^2+(340~N)^2} = 376~N$
