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Unknown angle in equilibrium

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

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

    3. 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.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 16, 2013 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    That 45°, is it relative to the other bar or relative to the vertical?
     
  4. Oct 17, 2013 #3
    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.
     
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