Question

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.
1. 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.)

Answer 1

The horizontal rope is under 7,019.4 N of tension.

The pulling force conveyed axially by a string, cable, chain, or another similar device, or by the ends of a rod, is referred to as tension.

The action-reaction pair of forces acting at the ends of the element can also be referred to as tension.

The tension unit is the Newton.

the horizontal rope is tense.

The following formula is used to determine the horizontal rope's tension:

Apply the torque principle;

Mg (L/2) sin + mg L sin = T(L/2) cos

M + 2mg sin/cos = T

(M + 2m)g tan = T

let θ = 66⁰ (The question should include this value)

(M + 2m)g tan = T

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

T = 7,019.4 N

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

Learn more about tension here: brainly.com/question/918617

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