Finding the Unknown Angle in Equilibrium for Symmetrical Bars Under Load

[Karstedt]

Homework Statement

Trying to find angle u of bars (symmetrical) with ground at equilibrium under load.
Two bars, negligible weight of length .4m
Spring at midpoint k = 3kN/m, unstretched when bars at 45 deg
F = 150N, 5 deg
W of block 10kg
Slope of ground under block 5 deg
coef friction = .2

Homework Equations

Fx = 0 = Ax + 150sin5 - Nsin5 - .2Ncos5
Fy = 0 = Ay - 150cos5 + Ncos5 - .2Nsin5
Ma = 0 = 150cos5*.4cosu + 150sin5*.4sinu - Ncos5*.8cosu + .2Nsin5*.8cosu

The Attempt at a Solution

4 unknowns, Ax, Ay, u, N... this course doesn't include indeterminate structures so I tried anyway. Since I could not get an answer via the standard eq equations I tried to eliminate a variable first in the whole structure then by doing a FBD of the block. I got N in terms of u but it still doesn't work anything out. It's incredibly messy using the block FBD and I re-did it several times because I kept finding errors but I eventually got to a quadratic of tan(u) that spit out ~56.1 deg. which is the closest thing to a reasonable value I've gotten, but that value doesn't seem to work when I run with it. I tried to find a trig identity that would simplify things, but can not work anything out.

 

[NascentOxygen]

That 45°, is it relative to the other bar or relative to the vertical?

 

[Karstedt]

It's relative to the horizontal actually... but at 45 that's the same as vertical. The angles of both bars relative to the horizontal ground.