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Intro Pully Question

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    The figure (attached (5-56))shows a box of mass m2 = 2.60 kg on a frictionless plane inclined at angle θ = 33°. It is connected by a cord of negligible mass to a box of mass m1 = 4.00 kg on a horizontal frictionless surface. The pulley is frictionless and massless.

    (a) If the magnitude of horizontal force is 2.3 N, what is the tension in the connecting cord?

    (b) What is the largest value the magnitude of may have without the cord becoming slack?


    2. Relevant equations
    Newton's Laws


    3. The attempt at a solution

    Unfortunately this is the part of physics where I became lost in high school and im having a difficult time putting everything together.

    So the first thing that I did was set up axes for each. I set m1 as a normal x and y coordinate system, and m2 with a tilted x-y coordinate system (attached FBD).
    So first I attempted to find w2 because that should help me find tension.

    [tex]\sum F_x=m_2*a_x[/tex]
    [tex]T-w_{x2}=m_2*a_x[/tex] \\w in the x direction of mass 2
    [tex]w_{x2}=-2.6*sin(33)[/tex]
    I should use this to solve for [tex]a_x[/tex]?

    moving on to m1
    [tex]\sum F_x=m_1*a_x[/tex]
    [tex]T+F=m_1*a_x[/tex]
    Do I plug in for a_x because if I do, then I have a T on both sides, in a place that wouldnt be easy to solve?

    And even where I am at I believe I have made a huge mistake. I am having problems understanding this. Please help me.
     

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  2. jcsd
  3. Oct 19, 2007 #2

    Doc Al

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    Staff: Mentor

    Careful with signs. Which way are the masses accelerating?

    You need to use the equations for both masses to solve for the acceleration.

    Good.
    Once you correct the sign error and rearrange the first equation, it may be easier to see how to eliminate T.
     
  4. Oct 19, 2007 #3
    Thank you for the quick response.
    I'm not sure why I put that negative sign there, but even after fixing that I'm still stuck.
    [tex]w_{x2}=2.6*sin(33)[/tex]

    and
    [tex]w_{x2}-T=m_2*a_x[/tex]
    [tex]a_x=\frac{w_{x2}-T}{m_2}[/tex]

    but after plugging in
    [tex]T=m_1\frac{w_{x2}-T}{m_2}-F[/tex]
    making it difficult to solve for T. Am I not understanding something?
    [edits for minor mistakes]
     
  5. Oct 19, 2007 #4

    Doc Al

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    Staff: Mentor

    Stop right there! Now combine this equation with the equation for m1 to eliminate T. (Write them one after the other and compare them.)
     
  6. Oct 19, 2007 #5
    Thank you so much for helping me. I feel like I am close, but I am getting a negative tension (and the magnitude of it is not correct).

    [tex]w_{x2}-m_2*a_x+F=m_1*a_x[/tex]
    [tex]a_x=\frac {w_{x2}+F}{m_1+m_2}[/tex]

    [tex]T=w_{x2}-m_2*a_x[/tex]
    but I'm getting -.0479
    why?
     
  7. Oct 19, 2007 #6

    Doc Al

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    Staff: Mentor

    This is correct. What did you get for the acceleration?

    This is also correct. So if you plug your answer for acceleration into this expression, you should be good to go.

    The equations are fine. Just do the arithmetic over. (If it's not an arithmetic error, then you are inputing a wrong value somewhere.)
     
  8. Oct 19, 2007 #7
    a_x = (2.6*sin(33)+2.3) / (4+2.6)
    a_x = 0.563

    T = 2.6*sin(33) - (2.6(.563))
    T = -.0477

    [physics makes me feel ignorant :( ]
     
  9. Oct 19, 2007 #8

    Doc Al

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    Staff: Mentor

    You left something out: That should be mg sin(33), not m sin(33).


    Me too! :wink:
     
  10. Oct 19, 2007 #9
    Thank you so much!
    I think I need to do some practice problems similar to this, because I am having a very hard time with this. (I'm sure I can find plenty on this board :) ).
     
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