Question

In the figure, a uniform beam of weight 470 N and length 3.6 m is suspended horizontally. On the left it is hinged to a wall; on the right it is supported by a cable bolted to the wall at distance D above the beam. The least tension that will snap the cable is 1400 N. What value of D corresponds to that tension?

Answer #1

To solve this problem, first of all draw a right-angled triangle.

Let the hypotenuse be the cable with tension 1400 N forming an angle with the beam.

So, the upward force = 1400N sinΘ

and the downward force = 470 N

Under balanced condition -

1400N sinΘ * 3.6m = 470N *(3.6 / 2) m

=> 1400N sinΘ * 3.6m = 470N * 1.8 m

=> sinΘ = (470*1.8)/(1400*3.6) = 0.168

=> Θ = 9.66 degree

The cable is bolted to the wall at a distance D above the beam.

So -

Tan Θ = D/3.6

D = 3.6 * tan(9.66)

D = 0.61 m

Therefore, the value of D corresponding to a tension of 1400 N is 0.61 m (Answer)

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