An object with a mass of m = 132 kg is suspended by a rope from the end of a uni

Question

An object with a mass of m = 132 kg is suspended by a rope from the end of a uniform boom with a mass of M = 85.5 kg and a length of l = 7.38 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.The boom makes an angle of ? = 52.1° with the vertical wall. Calculate the tension in the vertical rope.

2.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.)

Explanation / Answer

1)

tension in vertical rope ,

as the m mass is in equilibrium

Tv = m*g

Tv = 132 * 9.8

Tv = 1293.6 N

the tension in the vertical rope is 1293.6 N

2)

Here ,

let the tension in the cable is T

T * L * cos(52.1) - m*g * L * sin(52.1) - M*g*L*sin(52.1)/2 = 0

T * cos(52.1) - 132 * 9.8 * sin(52.1) - 85.5 * 9.8 * sin(52.1)/2 = 0

solving for T

T = 2120 N

the tension in the horizontal rope is 2120 N

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