# Infinite Solution in Truss Problem by Method of Sections

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?

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 $$F_{38}$$

• 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. Feb 22 at 16:48
• 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?
– r13
Feb 22 at 17:25
• I was able to get the value of force in member 6-8. Feb 22 at 17:42
• 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? Feb 23 at 5:51
• Sorry for the previous misses, check my math.
– r13
Feb 23 at 19:42