A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of
ID: 1467473 • Letter: A
Question
A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.8 kg and the sign has a mass of ms = 15.2 kg. The length of the beam is L = 2.52 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 = 32°.
1)
What is the tension in the wire?
N
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536
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536
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Wednesday, November 18 at 6:40 PM
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2)
What is the net force the hinge exerts on the beam?
N
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480
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480
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Wednesday, November 18 at 6:01 PM
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Correct!
3)
The maximum tension the wire can have without breaking is T = 885 N.
What is the maximum mass sign that can be hung from the beam?
kg
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4)
What else could be done in order to be able to hold a heavier sign?
while still keeping it horizontal, 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 beam
Explanation / Answer
1)
As the beam is in equibrium, net force and net torque acting on the must be zero.
Let T is the tension in the wire.
Apply, Net torque about pivot = 0
T*(2*L/3)sin(32) - mb*g*(L/2)*sin(90 - 32) - ms*g*L*sin(90 - 32) = 0
T*(2/3)sin(32) = mb*g*(1/2)*sin(58) + ms*g*sin(58)
T = (mb*g*(1/2)*sin(58) + ms*g*sin(58))/((2/3)*sin(32))
= (6.8*9.8*0.5*sin(58) + 15.2*9.8*sin(58))/((2/3)*sin(32))
= 437.6 N <<<<<<<<<-------------------Answer
2) Let Fx and Fy are forces exerted by wall on th beam.
Apply, Fnetx = 0
T*cos(32) - Fx = 0
Fx = T*cos(32)
= 437.6*cos(32)
= 371.1 N
Apply, Fnety - 0
Fy - mb*g - ms*g = 0
Fy = (mb + ms)*g
= (6.8 + 15.2)*9.8
= 215.6 N
Fnet = sqrt(Fx^2 + Fy^2)
= sqrt(371.1 + 215.6^2)
= 216.4 N <<<<<<<<<<<<<<---------------------------Answer
3)
Tmax*(2*L/3)sin(32) - mb*g*(L/2)*sin(90 - 32) - ms_max*g*L*sin(90 - 32) = 0
Tmax*(2/3)sin(32) - mb*g*(1/2)*sin(58) - ms_max*g*sin(58) = 0
ms_max*g*sin(58) = Tmax*(2/3)sin(32) - mb*g*(1/2)*sin(58)
ms_max = (Tmax*(2/3)sin(32) - mb*g*(1/2)*sin(58))/(g*sin(58))
= (885*(2/3)*sin(32) - 6.8*9.8*0.5*sin(58))/(9.8*sin(58))
= 34.2 kg <<<<<<<<<<<<<<---------------------------Answer
4)
while still keeping it horizontal, 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