Question

In Fig. 12-33, one end of a uniform beam of mass 40.0 kg is hinged to a wall: the other end is supported by a wire that makes angles theta = 30.0degree with both wall and beam. Find the tension in the wire and the magnitude and angle from the horizontal of the force of the hinge on the beam.

Answer 1

We begin by setting the torque equal to zero.

W = 222N
T = ?
τ = 0 = L/2 ⋅ W ⋅sin2θ − L⋅ T sin(180 −θ)
= L/2 ⋅W sin2θ − L⋅ T sin(θ)
= 1/2 W sin2θ −T sin(θ)

T = W sin2θ/2sin(θ)
= 392sin60/2sin30
= 339.482 N

Now we balance the forces

Vertical
0 = Fv + T sin(90 − 2θ + θ) −W
Fv = W − Tsin(90 −θ)
= 392 −339.482⋅sin(90 −60)
= 222.26N

Horizontal
0 = Fh − Tcos(90 −2θ +θ)
Fh = Tcos(90 −θ)
=339.482⋅ cos(90 −60)
=294N