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Homework Help: 2D Forces Equilibrium Question

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Here is the question from my assignment:

    2. Relevant equations

    3. The attempt at a solution
    I did the free body diagram on the left block. The arrow pointing to the bottom is the Fg, which I will further on refer to as W, the arrow going to the top left corner is the normal force (Fn), and the one going along the cable is the tension (FT).

    From that I was able to get 3 equations for the left side:
    FNL = -FNL * cos (21)i + FNL*sin(21)j
    FTL = FTL*cos(69)i + FTL*sin(69)j
    WL = 0i -9.81*mLj
    Then I wrote the ones for the right side:
    FNR = FNR*cos(60)i+FNR*sin(60)j
    FTR = -FTR*cos(30)i + FTR*sin(30)j
    WR = 0i - 9.81*mRj

    So now I have six equations, resultant of which equals to 0i +0j+0k. I figured the two tensions FTR and FTL had to be equal because it's one string. But I will still have 3 variables even if just use FT to represent both.

    Help please? Thanks
  2. jcsd
  3. Oct 2, 2008 #2


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    Your FBD is good, but in problems of this type, it is much simpler to choose the x axis as the axis along (parallel to) the incline, and the y axis perpendicular to the incline, for each block. This will give you an x component to the weight in your FBD, and the tension force will also be along your chosen x axis. Then apply newton 1 to each block.
  4. Oct 2, 2008 #3
    Thanks Jay. I can do that with the axis on the left side, but how would I transfer them over to the right? I've got to use the same axis orientation for both, don't I?

    So, on the left it would become:
    Fnl = 0i + Fnl*j
    Ftl = Ftl*i+0j
    W = -mL*9.81*cos(21) - mL*sin(21)*9.81

    But what happens on the right side? Does the Tension just have the y-component then?
  5. Oct 2, 2008 #4


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    I don't like this i and j stuff, but I'm old school, I guess. You left out the i and j in your weight breakdown into componets. But anyhow, all the i or x components of the noted forces sum to zero, and all the j or y components sum to zero, per newton 1. Now when you go the block on the right, it's OK to choose your x axis parallel to that right inclne. The tension force will be along the x axis, with the same magnitude as T_l. Use newton 1 again, and solve using the 2 equations thus obtained.
  6. Oct 2, 2008 #5
    YAY Thank you! I think I got it!
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