Question

A crate with a mass of 187.5 kg is suspended from the end of a uniform boom. The upper end of the boom is supported by the tension of 3206 N in a cable attached to the wall. The lower end of the boom pivots at the location marked X on the same wall. Calculate the mass of the boom.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The mass of the boom is
`m_b= 21.12kg`

Step-by-step explanation:

From the question we are told that

The mass is
`m = 187.5kg`

The tension the cable is
`T = 3206\ N`

A sketch of the free body diagram is shown on the second uploaded image

From the second diagram

The length L is evaluated as
`L = √(9^2 + 14^2)`

`= 14.56 m`

The angle
`\alpha = tan^(-1) ((5)/(14) )`s

`= 19.65^o`

The angle
`\beta = tan^(-1)((4)/(14) )`

`= 15.95^o`

At equilibrium the net torque about x is zero and this can be represented mathematically

`T * sin (\alpha + \beta ) * L -[(m_bg )/(2) +mg ]* L cos \beta = 0`

`T * sin (\alpha + \beta ) * L = [(m_bg )/(2) +mg ]* L cos \beta`

Where g is the acceleration due to gravity with a value of
`g =9.8 m/s^2`

Making
`m_b` the mass of the boom the subject of the formula

`m_b = (2 *[(T sin(\alpha +\beta ) )/(cos \beta ) -mg])/(g)`

Substitution the values

`m_b = (2 *[(3206 * sin (19.65 + 15.95))/(cos 15.95) - 187.5*9.8])/(9.8)`

`m_b= 21.12kg`