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

the distance from hinge to the cable tied on the beam is sqrt(5^2-4^2) = 3 m

Torque due to weight of the beam = torque due to tension in the cable

m*g*(12/2) = 1.2*1000*3*sin(theta)

sin(theta) = 4/5

m*9.8*6 = 1.2*1000*3*(4/5)

m = 49 kg

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horizontal component of force is Fh = 1.2*1000*cos(theta) = 1.2*1000*(3/5) = 720 N

Vertical component of force is Fv = 1.2*1000*sin(theta) - mg = (1.2*1000*(4.5)) - (49*9.8) = 4919.8 N

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