Question

A uniform horizontal beam is attached to a vertical wall by a frictionless hinge and supported from below at an angle è = 38o by a brace that is attached to a pin. The beam has a weight of 347 N. Three additional forces keep the beam in equilibrium. The brace applies a force P to the right end of the beam that is directed upward at the angle with respect to the horizontal. The hinge applies a force to the left end of the beam that has a horizontal component H and a vertical component V. Find the magnitudes of these three forces.

I do not know how to start this.

Answer 1

Net torque about the hinge is
mg•(L/2)- P(y) •L = 0,
( Forces H and V have zero moment about this point)
P(y) = mg/2 =347/2 =173.5 N.
P(y)/P(x) =tanα.
P(x) = P(y)/tanα=173.5/tan38 =222 N.
P(y)/P =sin α,
P =P(y)/sin α =173.5/sin38 =282 N.
As the system is in equilibrium, horizontal forces are balanced:
Force H acts into the wall (to the left)= P(x) (to the right).
H = P(x) = 222 N.
Vertical forces are also in balance
mg – V - P(y) =0
V = mg – P(y) =347 – 173.5 =173.5 N.