An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a length of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.
Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)

Tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod.

Tension can also be described as the action-reaction pair of forces acting at each end of said elements.

The unit of tension is Newton (same unit as force, since tension is also a force).

Tension in the horizontal rope

The tension in the horizontal rope is calculated as follows;

Apply the principle of torque;

T(L/2) cosθ = Mg (L/2) sinθ + mg L sinθ

T = (M + 2m)g sinθ/cosθ

T = (M + 2m)g tanθ

let θ = 66⁰ (this value should be given in the question)

T = (M + 2m)g tanθ

T = (74.9 + 2 x 122)(9.8)(tan 66)

T = 7,019.4 N

Thus, the the tension in the horizontal rope is 7,019.4 N.

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