Answer
$P = 200~N$
Work Step by Step
We can find the angle the beam makes above the horizontal:
$sin~\theta = \frac{d}{L}$
$\theta = sin^{-1}~\frac{d}{L}$
$\theta = sin^{-1}~\frac{1.50~m}{2.50~m}$
$\theta = 36.87^{\circ}$
To find the magnitude of $P$, we can consider the torque about a rotation axis at the bottom of the beam:
$\sum~\tau = 0$
$L~P-(\frac{L}{2}~cos~\theta)~mg = 0$
$L~P = (\frac{L}{2}~cos~\theta)~mg$
$P = \frac{mg}{2}~cos~\theta$
$P = \frac{500~N}{2}~cos~36.87^{\circ}$
$P = 200~N$
