A 12.0 m uniform beam is hinged to a vertical wall and held horizontally by a 5.

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?

Part B

Find the horizontal component of the force the hinge exerts on the beam

Part C

Find the vertical component of the force the hinge exerts on the beam

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)(12)(sin24) = (mg)(6)(sin90). => m = 99.6 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 ) = 488.08 N
C)
Vertical forces cancel out.
F(wall on beam-vertical) + (1200N)(sin24) = (99.6kg)(9.8m/s^2).
F(wall on beam-vertical) = 487.99 N =488 N

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