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.

Complete Question

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

The mass of the boom is

Explanation:

From the question we are told that

      The mass is

       The tension the cable is  

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

       From the second diagram

  The length L is evaluated as

 The angle  s

The angle

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

Where g is the acceleration due to gravity with a value of

  Making the mass of the boom the subject  of the formula

  Substitution the values

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

the answer is 3.1 his answer is wrong mines is right