Can someone help with a physics problem about tension and horizontal & vertical forces of a sign hanging from a beam? A shop sign weighing 250 N...

This is a problem in equilibrium and statics. You must
look at both the forces, and the torques, acting to solve this problem. All the forces
must total zero, and all the torques must also total
zero.

Looking first at the forces, you
have:

The force of the hinge in the y-direction 
F(hy)

The force of the hinge in the x-direction 
F(hx)

The force from the wire tension in the y-direction 
F(ty)

The force from the wire tension in the x-direction 
F(tx)

The weight of the beam = 150
N

The weight of the sign = 250
N

Summing these forces in the y-direction you
get:

F(hy) + F(ty) - 150 - 250 =
0

Then F(hy) +F(ty) = 400
N

Summing in the
x-direction:

F(hx)-F(tx) = 0 or F(hx) =
F(tx)

Now consider the torques, with the point of rotation
where the wire is attached to the beam:

clockwise torque of
.35 m * 250 N

clockwise torque of .35 m * (150*.35/1.35)
(this is the portion of the beam itself to the right of the point of
attachment)

counterclockwise torque of 1.35 *
F(hy)

summing:  1.35 F(hy) - .35*250 - .35 (.35*150/1.35) =
0

Solving for F(hy) = 74.9
N

Go back to the first equation and solve for
F(ty)

F(ty) = 400 - F(hy) =
325.1N

Since the sin(35) = F(ty)/F(t), then F(t) = 566.8
N

since tan35 = F(hy)/F(hx) then F(hx) =
106.97N

Summarizing:

F(ty) =
325.1 N

F(hy) = 74.9 N

F(t) =
566.8 N

F(hx = 106.97 N