Question

hangingbeam A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.7 kg and the sign has a mass of ms = 17.5 kg. The length of the beam is L = 2.86 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = 34.9°.

1)What is the tension in the wire?

2) What is the net force the hinge exerts on the beam?

Answer 1

A)
As the beam is in equilibrium, net force and net torque ating on the beam must be zero.

Apply net torque about pivot = 0

mb*g*(L/2)*sin(90-theta) + ms*g*L*sin(90-theta) - T*(2*L/3)*sin(theta) = 0

T = (0.5*mb + ms)*g/(2*tan(theta)/3)

= (0.5*6.7 + 17.5)*9.8/(2*tan(34.9)/3)

= 439 N <<<<<<<<-------------Answer
B) let Fx and Fy are the componet of forces exerted by the hinge on the beam

Apply, Fnetx = 0

Fx - T = 0

Fx = T = 439 N

Apply, Fnety = 0

Fy - mb*g - ms*g = 0

Fy = (mb + ms)*g

= (6.7 + 17.5)9.8

= 237.2 N

Fh = sqrt(Fx^2 + Fy^2)

= sqrt(439^2 + 237.2^2)

= 499 N <<<<<<<<<<----------------------Answer