Apurple beam is hinged to a wall to hold up abie sig. The beam has a mass of -6.
ID: 2032738 • Letter: A
Question
Apurple beam is hinged to a wall to hold up abie sig. The beam has a mass of -6.9 kg and the sign has mass of m- 153 kg. The length of the beam s L-2.54 m. The sign is attached at the very end of the beam, but the hortzontal 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 8-32.3 1) what ts the tension in the wire? 145.33 2) What is the net force the hinge exerts on the beam N Submit 3)The monum tension the wire can have without breaking is T.?N. What is the maximum mass sign that can be hung from the beam Submit 4) What else could be done in order to be able to hold a heavier sign while still keepling it hortzontal, attach the wire to the end of the beam keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam attach the sign on the beam closer to the wall shorten the length of the wire attaching the box to the beanm 5) Below is some space to write notes on this problem Q Search or enter website nameExplanation / Answer
In equlibrium Net torque = 0
mb*g*L/2*costheta + ms*g*L*costheta - T*(2/3)L*sintheta = 0
6.9*9.8*2.54/2*cos32.3 + 15.3*9.8*2.54*cos32.3 - T*(2/3)*2.54*sin32.3 = 0
Tension T = 436 N
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2)
along horizontal Fnet = 0
T - Fx = 0
Fx = T = 436 N
along vertical Fnet = 0
Fy - mb*g - ms*g = 0
Fy = mb*g + ms*g
Fy = (6.9*9.8) + (15.3*9.8) = 217.56 N
net force exerted by hinge on the beam F = sqrt(Fx^2+Fy^2) = 487.3 N
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3)
In equlibrium Net torque = 0
mb*g*L/2*costheta + ms*g*L*costheta - T*(2/3)L*sintheta = 0
if T = 840 N
6.9*9.8*2.54/2*cos32.3 + ms*9.8*2.54*cos32.3 - 895*(2/3)*2.54*sin32.3 = 0
6.5*9.8*2.43/2*cos34.1 + ms*9.8*2.43*cos34.1 - 840*(2/3)*2.43*sin34.1 = 0
ms = 35.0 kg
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4)
1 , 2, 3
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