Answer
$P=270N$, $V=170N$ and $H=210N$
Work Step by Step
Force $P$ has a horizontal component $P_x=P\cos39$, which points to the right, and a vertical component $P_y=P\sin39$, which points upward.
On the vertical side, the beam's weight $W$ points downward while $P_y$ points upward. We assume the force from the hinge $V$ points upward, too. So, $$\sum F_y=P_y+V-W=0$$ $$P_y+V=340N (1)$$
On the horizontal side, $P_x$ points rightward while $H$ points leftward. So, $$\sum F_x=P_x-H=0 (2)$$
We take the axis of rotation to be at the hinge, so $H$ and $V$ produce no torque. Taking the beam's length to be $l$, we have
- $W$ produces a clockwise torque $-\frac{Wl}{2}=-170l$
- $P$ produces a counterclockwise torque. The arm length is $l\sin39$, so its torque is $+Pl\sin39$
$$\sum\tau=Pl\sin39-170l=0$$ $$P\sin39=170$$ $$P=270N$$
Therefore, $P_x=210N$ and $P_y=170N$
(1): $V=340N-170N=170N$
(2): $H=210N$
