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Finding angles of Phi. Very challenging.

  1. Oct 22, 2013 #1
    1) Given: two ropes of negligible weight suspend an object weighing 50N from the ceiling There are two angles of PHI that form from as a measure from the vertical. PHI1 + PHI2= 60°. The tension of the left rope is given to be 80N and the tension of the left rope is given to be 70N. The entire system is stationary.

    2) Find the measures of each angle PHI as well as each complimentary θ.

    3) I tried to use ƩF=0=T1 + T2 + (-Fg)

    ∴ 50N = 80N*cos(60-PHI1) + 70N*cos(60-PHI2)

    I couldn't think of a way to solve this by hand so I tired to use my Ti-84 calculator's equation solver and got PHI values of 10° and 50°. Logically, this doesn't make sense. My teacher decided to just omit the question, but I am still curious as to how it would actually be done.
     

    Attached Files:

  2. jcsd
  3. Oct 22, 2013 #2
    Why do you use ## 60 - \phi_1 ## and ## 60 - \phi_2 ##? Why not simply ##\phi_1## and ##\phi_2##?

    Secondly, the equation you got is for the vertical components of the forces. You should obtain one for the horizontal components. That will give you a system of two linear equation for two unknowns, which you can then solve.
     
  4. Oct 22, 2013 #3

    gneill

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    Staff: Mentor

    It's possible that the given conditions cannot be satisfied and that there is no solution to the problem as given.

    Any solution would have to satisfy the conditions for stasis for horizontal and vertical components separately. So set ##\phi_2 = 60° - \phi_1## and use the horizontal force balance to find all solutions for ##\phi_1##. Check to see if any value you found satisfies the vertical force balance. If yes, huzzah! If no, the problem is broken :smile:
     
  5. Oct 22, 2013 #4
    Which would be the horizontal components equation? I'm confused. I don't want to trouble you any further, but could you possibly provide me with the two equations you are referring to? I'm a little frustrated so my mind is unclear.
     
  6. Oct 22, 2013 #5
    The tensions in the ropes are vectors. Your general equation is ##\newcommand{b}[1]{\boldsymbol{#1}} \b {T_1} + \b {T_2} + \b {W} = 0 ##, where ## \b {T_1}, \ \b {T_2}, \ \b {W} ## are the left rope's tension, the right rope's tension and the weight. Each of those vectors has two components: vertical (y) and horizontal (x), so that one general equation gives you two equations for the components: $$

    T_{1x} + T_{2x} + W_{x} = 0

    \\

    T_{1y} + T_{2y} + W_{y} = 0

    $$ All that you need to do is find out these components, some of which will depend on ##\phi_1## and ##\phi_2##, and solve for ##\phi_1## and ##\phi_2##.
     
  7. Oct 22, 2013 #6
    The problem doesn't have a solution that can satisfy the specified input data. No pair of angles summing up to 60 degrees can even balance the vertical force, let alone match the horizontal force balance.
     
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