Verizon LTE 5:08 PM a flipitphysics.com A purple beam is hinged to a wall to hol
ID: 2035307 • Letter: V
Question
Verizon LTE 5:08 PM a flipitphysics.com A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb -6.5 kg and the sign has a mass of m 16.3 kg. The length of the beam is L-2.56 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 8 30.9 What is the tension in the wire? What is the net force the hinge exerts on the beam O"The maximum tension the wire can have without breaking is T 894 N. What is the maximum mass sign that can be hung from the beam? What else could be done in order to be able to hold a heavier sign? Owhile still keeping it horizontal, attach the wire to the end of the beam Okeeping the wire attached at the same location on the beam, make the wire perpendicular to the beam Oattach the sign on the beam closer to the wall Oshorten the length of the wire attaching the box to the beamExplanation / Answer
In equlibrium Net torque = 0
mb*g*L/2*costheta + ms*g*L*costheta - T*(2/3)L*sintheta = 0
6.5*9.8*2.56/2*cos30.9 + 16.3*9.8*2.56*cos30.9 - T*(2/3)*2.56*sin30.9 = 0
Tension T = 480 N
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2)
along horizontal Fnet = 0
T - Fx = 0
Fx = T = 480 N
along vertical
______________
Fnet = 0
Fy - mb*g - ms*g = 0
Fy = mb*g + ms*g
Fy = (6.5*9.8) + (16.3*9.8) = 223.44 N
net force exerted by hinge on the beam F = sqrt(Fx^2+Fy^2)
F = sqrt(480^2 + 223.44^2) = 529.5 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 = 894 N
6.5*9.8*2.56/2*cos30.9 + ms*9.8*2.56*cos30.9 - 894*(2/3)*2.56*sin30.9 = 0
ms = 33.15 kg
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4)
1 , 2, 3
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