Question

A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of m_b = 6.5 kg and the sign has a mass of m_s = 16 kg. The length of the beam is L = 2.49 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 = 30°.

1) What is the tension in the wire?

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

3)The maximum tension the wire can have without breaking is T = 1025 N.

What is the maximum mass sign that can be hung from the beam?

Answer 1

On horizontal

Fx + T = 0


On Vertical

Fy - Mb*g - Ms*g = 0

Fy = (Mb+Ms)*g

    = (6.5+16)*9.8

   = 220.5 N

1) Torque = 0

=> T*(2/3)*L*sin theta = Mb*g*(L/2)*cos theta + Ms*g*L*cos theta
=> T*(2/3)*2.49*sin30 = [6.5*9.8*(2.49/2)*cos30] + (16*9.8*2.49*cos30)
=> T*0.83 = [68.68]+[338.114]

=> T*0.83 = 406.8

=> T = 490.11 N --> Answer to (1)

2) Fx = - T = -490.11 N

Net Force, F = sqrt (Fx^2 + Fy^2)

            F = 537.42 N --> Answer to (2)

3) Given T = 1025 N.

Max Mass, Ms = [ 2*T*Tan theta / 3*g ] - [mb / 2]

                     = [ 2x1025*Tan 30 / 3*9.8 ] - [ 6.5 / 2]

                   = 40.23 - 3.25

                     = 36.99 kg --> Answer