Question

A crate with a mass of 193.5 kg is suspended from the end of a uniform boom with a mass of 91.7 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.

Answer 1

Answer:

2538.72 N

Explanation:

Given data:

There are some missing values, so, I will assume them, and then you affix them. It's basically just substitution....

m₁ = mass of crate = 193.5 kg

m₂ = mass of the boom = 91.7 kg

g = acceleration by gravity = 9.81 m/s²

L = length of the boom

θ = angle of the boom to the horizontal

φ = angle of the cable to horizontal

Solution:

I'm assuming that

tan (θ) = 6/12 = 0.5

This then makes

θ = arctan(0.5)= 26.57°

The second and last assumption I'm making is that

tan(φ) = 3/12 = 0.25

This also makes

φ = arctan(0.25) = 14.04°

Now, since the system is in equilibrium, the magnitude of clockwise torque has to be equal to anticlockwise torque. This means that

m₁ * g * cos(θ) *L/2 + m₂ * g *cos(θ) * L = T * sin(θ+φ) * L

(m₁/2 + m₂) * g * cos(θ) = T * sin(θ+φ), making T subject of formula, we have

T = (m₁/2 + m₂) * g * cos(θ) / sin(θ + φ)

T = [(193.5)/2 + (91.7)] * (9.81) * cos(26.57°) / sin(26.7°) + (14.04°))

T = [(96.75 + 91.7) * 9.81 * 0.894] / 0.651

T = 188.45 * 8.77 / 0.651

T = 1652.7065 / 0.651

T = 2538.72 N