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 $m_b= 21.12kg$

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 $L = \sqrt{9^2 + 14^2}$

                                                     $= 14.56 m$

 The angle  $\alpha = tan^{-1} (\frac{5}{14} )$s

                        $= 19.65^o$

The angle $\beta = tan^{-1}(\frac{4}{14} )$

                      $= 15.95^o$

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

                  $T * sin (\alpha + \beta ) * L -[\frac{m_bg }{2} +mg ]* L cos \beta = 0$

                $T * sin (\alpha + \beta ) * L = [\frac{m_bg }{2} +mg ]* L cos \beta$

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

              $m_b = \frac{2 *[\frac{T sin(\alpha +\beta ) }{cos \beta } -mg]}{g}$

  Substitution the values

            $m_b = \frac{2 *[\frac{3206 * sin (19.65 + 15.95)}{cos 15.95} - 187.5*9.8]}{9.8}$

                 $m_b= 21.12kg$