## A uniform, 6.9-m-long beam weighing 5180 N is hinged to a wall and supported by a thin cable attached 1.5 m from the free end of the beam. T

Question

A uniform, 6.9-m-long beam weighing 5180 N is hinged to a wall and supported by a thin cable attached 1.5 m from the free end of the beam. The cable runs between the beam and the wall and makes a 45° angle with the beam. What is the tension in the cable when the beam is at an angle of 27° above the horizontal?

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1 week 2021-07-22T11:45:32+00:00 1 Answers 10 views 0

The tension is the cable when the beam is at an angle of 27° above the horizontal is 3100.5 N

Explanation:

Weight of beam = 5180 N

Angle of beam = 27 °

Location of of thin cable attachment = 1.5 m from free end of beam

Angle of cable = 45° with the beam

Angle of beam with horizontal = 45 – 27 = 18°

Taking moment about the wall and noting that the weight of the beam acts at the center, we get;

6.9/2 × cos 27 × 5180 = (6.9-1.5)× cos 18 × T

Therefore, T  = 15923.1776/5.136 = 3100.5 N

Therefore, the tension is the cable when the beam is at an angle of 27° above the horizontal = 3100.5 N.