Question

A crate with a mass of M = 60.5 kg is suspended by a rope from the midpoint of a uniform boom. The boom has a mass of m = 119 kg and a length of l = 9.64 m. The end of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 60.2° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?

Answer 1

The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as

$T_1=730.85 \mathrm{~N}$

$T_1'=\frac{1980.51 \mathrm{~N}}{}$

What is the tension in the vertical rope.?

Generally, the equation for is  mathematically given as

1) Tension in vertical rope,

$T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}$

2) Tension in horizontal rope,

$\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta$

$T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51$

$T_1=\frac{1980.51 \mathrm{~N}}{}$

In conclusion, the tension in the horizontal rope is

$T_1=\frac{1980.51 \mathrm{~N}}{}$

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