Help me in determining the right solution in this problem. I already have an initial solution here but when I compute, my answer does not match to the answer key. I would like to know where did I create a mistake. (By the way, this is not a homework.) Problem

These are my initial equations:

Knowing that member BC is a two force member and it is a duplicate so I arrive to this equation. I dismember the member BC and pin J from the body.

$\sum M_J = 2F_{BC}(\frac{80}{80\sqrt(17)})(60) + 2F_{BC}(\frac{320}{80\sqrt(17)})(440) - 2000(200) = 0$ enter image description here

After I solve the force in member BC, I use that value to solve $F_{Ky}$. I dismember the member HL and member BC from the body.

$\sum M_H = -2000(1700) + F_{Ky}(1100) + 2F_{BC}(\frac{320}{80\sqrt(17)})(280) + 2F_{BC}(\frac{80}{80\sqrt(17)})(1440) = 0$

I got a value of around 2500 N but the answer key states that the answer is around 403 N. I am thinking that the condition that HL is connected to the truck can affect in the computation.

  • $\begingroup$ I don't think you can work from the 2000N which is an applied load to the free end G. Assume this is a cantilever truss simply supported at K & H. So, you shall get the reactions at H & L then work towards K by truss analysis. $\endgroup$
    – r13
    Apr 26 at 1:33
  • $\begingroup$ This might be a difference in nomenclature, but what does it mean when they say that BC is a "duplicate"? $\endgroup$
    – Wasabi
    Apr 26 at 14:41

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