A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.
ID: 1445485 • Letter: A
Question
A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1) . The metal of this cable has a test strength of 1.20 kN , which means that it will break if the tension in it exceeds that amount.
Part A
What is the heaviest beam that the cable can support with the given configuration?
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Part B
Find the horizontal component of the force the hinge exerts on the beam.
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Part C
Find the vertical component of the force the hinge exerts on the beam.
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Figure 1 of 1
A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1) . The metal of this cable has a test strength of 1.20 kN , which means that it will break if the tension in it exceeds that amount.
Part A
What is the heaviest beam that the cable can support with the given configuration?
m = __ kgSubmitMy AnswersGive Up
Part B
Find the horizontal component of the force the hinge exerts on the beam.
Fh = __ NSubmitMy AnswersGive Up
Part C
Find the vertical component of the force the hinge exerts on the beam.
Fv = __ NSubmitMy AnswersGive Up
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Figure 1 of 1
Explanation / Answer
A)The torque of the cable = the torque of the weight of the beam (equilibrium).
The pivot point is where the beam is attached to the wall.
(1200N)(12m)(sin24) = (mg)(6)(sin90). Solving for m we have: m = 99.60 kg
B)Horizontal forces cancel out (equilibrium).
F(wall on beam- horizontal ) = F(horizontal of tension in the cable)
F(wall on beam-horizontal ) = (1200N)(cos24)
F( wall on beam-horizontal ) = 1096 N
C)Vertical forces cancel out.
F(wall on beam-vertical) + (1200N)(sin24) = (99.60kg)(9.8m/s^2).
F(wall on beam-vertical) = 488 N
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