A crate with a mass of 151.5 kg is suspended from the end of a uniform boom with a mass of 74.9 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.

Explanation:

Given data:

m₁ = mass of crate = 151.5 kg

m₂ = mass of the boom = 74.9 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:

From image you can see that

tan (θ) = 6/12 = 0.5

so

θ = arctan(0.5)= 26.57°

From the image you can see that

tan(φ) = 3/12 = 0.25

so

φ = arctan(0.25) = 14.04°

As the system is in equilibrium so the magnitude of clockwise torque must be equal to anticlockwise torque. so

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)

T = (m₁/2 + m₂)×g×cos(θ) / sin(θ + φ)

T = [(151.5 kg)/2 + (74.9 kg)]×(9.81 m/s²)×cos(26.57°)/ sin(26.7°) + (14.04°))

T = 2023.28 N

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer