Question

A crate with a mass of 161.5 kg is suspended from the end of a uniform boom with a mass of 72.3 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

The correct answer is "2205.72 N".

Step-by-step explanation:

Given:

m₁ = 161.5 kg

m₂ = 72.3 kg

By taking moment about point A, we get

⇒
`m_1 g Cos \theta.(l)/(2 Cos \theta) + m g Cos \theta.(l)/( Cos \theta) -T Cos(90-2 \theta).(l)/(2 Cos \theta) = 0`

By substituting the values, we get

⇒
`(161.5* 9.8* 13)/(2)+72.3* 9.8* 13-T.2 Sin \theta.l =0`

⇒
`10287.55+9211.02-T 2* 0.34* 13=0`

⇒
`T* 8.84=19498.57`

⇒
`T= 2205.72 N`