A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of

Question

A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of m_b = 6.3 kg and the sign has a mass of m_s = 17.9 kg. The length of the beam is L = 2.69 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 theta = 31.1. What is the tension in the wire? What is the net force the hinge exerts on the beam? The maximum tension the wire can have without breaking is T = 1078 N. What is the maximum mass sign that can be hung from the beam?

Explanation / Answer

a)

the moments around the hinge, parallel to the beam, and normal to it, sum up to 0.

the angle of the horizontal line with the beam is 31.1, so the

angle of the forces with the beam are 90 - 31.1 = 58.9 degree

moments normal to the beam:

mb*g*sin58.9*2.69/2 + ms*g*sin58.9*2.69 = F*2/3*2.69

where F = T/cos58.9 with T = tension in the line

mb*g*sin58.9*2.69/2 + ms*g*sin58.9*2.69 = T*2/3*2.69/cos58.9

6.3*9.81*sin(58.9 degree)*2.69/2 + 17.9*9.81*sin(58.9 degree)*2.69 = T*2*2.69*cos(58.9 degree)/3

475.5 = T * 0.925

T = 513.85 N = tension in the line

b)

forces in direction along the beam in direction of the hinge:

F = 6.3*9.81*cos(58.9 degree) + 17.9*9.81*cos(58.9 degree) + 513.85*sin(58.9 degree)

F =31.89 + 90.61 + 439.86

F = 562.36 N

this is also the force of the hinge on the beam

c)

put T = 1078 N into the moment equation, and look for the mass of the sign:

6.3*9.81*sin(58.9 degree)*2.69/2 + ms*9.87*sin(58.9 degree)*2.69 = 1078*2*2.69*cos(58.9 degree)/3

6.3*9.81*0.856*2.69/2 + ms*9.87*0.856*2.69 = 1078*2*2.69*0.516/3

71.16 + ms*22.73 = 997.54

ms (max) = (997.54 - 71.16)/22.73

ms (max) = 40.755 kg

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