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 mb = 6.3 kg and the sign has a mass of ms = 15.0 kg. The length of the beam is L = 2.73 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 ? = 31.6

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 34.3, so the angle of the forces with the beam are 90 - 34.3 = 55.7° moments normal to the beam: mb*g*sin55.7*2.73/2 + ms*g*sin55.4*2.73 = F*2/3*2.73 where F = T/cos55.7 with T = tension in the line mb*g*sin55.7*2.73/2 + ms*g*sin55.4*2.73 = T*2/3*2.73/cos55.7 6.6*9.81*sin(55.7°)*2.73/2 + 16.5*9.87*sin(55.7°)*2.73 = T*2*2.73*cos(55.7°)/3 ---> T = 429.23 N = tension in the line ------- b) forces in direction along the beam in direction of the hinge: F = 6.6*9.81*cos(55.7°) + 16.5*9.87*cos(55.7°) + 429.23*sin(55°) F = 479.9 N this is also the force of the hinge on the beam ---------- c) put T = 846 N into the moment equation, and look for the mass of the sign: 6.6*9.81*sin(55.7°)*2.73/2 + ms*9.87*sin(55.7°)*2.73 = 846*2*2.73*cos(55.7°)/3 ms (max) = 35.700 kg ------ d) attach the sign on the beam closer to the wall, --> good keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam

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