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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.

A crate with a mass of 1755 kg is suspended from the end of a uniform boom with a mass of 947 kg The upper end of the boom is supported by a cable attached to t class=

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mickymike92

The tension in the cable is equal to 323.5 N.

What is the tension in the cable?

The tension, T in the cable is determined by taking moments about the pivot  marked X.

The angles of the boom and the cable with the horizontal are first calculated.

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

Tension = 241.68/0.747

Tension = 323.5 N

In conclusion, the tension in the cable helps to suspend the crate.

Learn more about tension at: https://brainly.com/question/24994188

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