Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The mass of the boom is [tex]m_b= 21.12kg[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 187.5kg[/tex]
The tension the cable is [tex]T = 3206\ N[/tex]
A sketch of the free body diagram is shown on the second uploaded image
From the second diagram
The length L is evaluated as [tex]L = \sqrt{9^2 + 14^2}[/tex]
[tex]= 14.56 m[/tex]
The angle [tex]\alpha = tan^{-1} (\frac{5}{14} )[/tex]s
[tex]= 19.65^o[/tex]
The angle [tex]\beta = tan^{-1}(\frac{4}{14} )[/tex]
[tex]= 15.95^o[/tex]
At equilibrium the net torque about x is zero and this can be represented mathematically
[tex]T * sin (\alpha + \beta ) * L -[\frac{m_bg }{2} +mg ]* L cos \beta = 0[/tex]
[tex]T * sin (\alpha + \beta ) * L = [\frac{m_bg }{2} +mg ]* L cos \beta[/tex]
Where g is the acceleration due to gravity with a value of [tex]g =9.8 m/s^2[/tex]
Making [tex]m_b[/tex] the mass of the boom the subject of the formula
[tex]m_b = \frac{2 *[\frac{T sin(\alpha +\beta ) }{cos \beta } -mg]}{g}[/tex]
Substitution the values
[tex]m_b = \frac{2 *[\frac{3206 * sin (19.65 + 15.95)}{cos 15.95} - 187.5*9.8]}{9.8}[/tex]
[tex]m_b= 21.12kg[/tex]
