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
