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This is the problem, I need to find forces in members 6-8 and 3-8. I have made a horizontal cut cutting across 3 members but I find out that I am only able to formulate 2 independent equations, one of the equations is dependent on the rest. I am not sure why I am facing this issue. I am interested in finding the reason behind this issue and how to tackle this. What am I missing?

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From global equilibrium, solve $R_1$.

$\Sigma M_3 = 0$

$12R_1 - 0.707F_{68}*6 - 0.707F_{68}*6 -50*6 = 0$, solve $F_{68}$

$\Sigma F_x = 0$

$0.707F_{68} + 0.707F_{37} + F_{34} = 0$ -----(1), $F_{68}$ = known constant

$\Sigma F_y = 0$

$0.707F_{68} - F_{38} + 0.707F_{37} - R_1 + 50 = 0$ -----(2), $F_{68}$ & $R_1$ = known constant

$\Sigma M_1 = 0$

$12F_{38} - 0.707F_{37}*12 - 50*6$ = 0$

$12F_{38} - 8.484F_{37} - 300 = 0$ -----(3)

$F_{37} = \dfrac{12F_{38} - 300}{8.484}$ -----(3')

PLug (3') into (2) and solving for $F_{38}$

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  • $\begingroup$ Wouldn't that result in 4 unknowns and only 3 equations? I am interested in the reasons behind why the approach I used did not work and how to learn to look for better approaches. $\endgroup$ Feb 22 at 16:48
  • $\begingroup$ You didn't even bother to give it a try. First, solve support reactions, then make the cut and sum moment about joint 3, what does it result? $\endgroup$
    – r13
    Feb 22 at 17:25
  • $\begingroup$ I was able to get the value of force in member 6-8. $\endgroup$ Feb 22 at 17:42
  • $\begingroup$ Could you please explain what you are looking for while making these cuts? What should I avoid to getting myself into a tight corner like I did in my approach? $\endgroup$ Feb 23 at 5:51
  • $\begingroup$ Sorry for the previous misses, check my math. $\endgroup$
    – r13
    Feb 23 at 19:42

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