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?

SubmitMy AnswersGive Up

Part B

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

SubmitMy AnswersGive Up

Part C

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

Explanation / Answer

The torque of the cable about pivot point = the torque of the weight of the beam (equilibrium) about piviot point
(1200N)(3m)(sin53.13) = (mg)(6)(sin90).

m =48.9295 kg

Horizontal forces cancel out (equilibrium).
F(wall on beam- horizontal ) = F(horizontal of tension in the cable)
F(wall on beam-horizontal ) = (1200N)(cos53.13)
F( wall on beam-horizontal ) = 720 N

Vertical forces cancel out.
F(wall on beam-vertical) + (1200N)(sin53.13) = (48.9295kg)(9.81m/s^2).
F(wall on beam-vertical) = - 480 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.