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Force- Major Help needed

  1. Oct 12, 2006 #1
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    Suppose that in the same Atwood setup another string is attached to the bottom of m1 and a constant force f is applied, retarding the upward motion of m1. If m1 = 4.00 kg and m2 = 8.00 kg, what value of f will reduce the acceleration of the system by 50%?

    Now I got the equations for these two as

    for M1= T-4g= 4a
    for M2= 8g-T= 8a

    I got a as 3.266

    I got T as 52.264

    But I cant get the answer to the asked question. Please HELP me solve this.
     
  2. jcsd
  3. Oct 12, 2006 #2

    Doc Al

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    Now you have an additional force acting on m1 and a given acceleration. Rewrite your equations accordingly and solve for that force.
     
  4. Oct 12, 2006 #3
    I dont know how to do that, please help
     
  5. Oct 12, 2006 #4

    Doc Al

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    How did you get those first equations? It's the same analysis, only now there is an additional force acting on m1.
     
  6. Oct 12, 2006 #5
    So is it going to be

    T-(4+x)g= (4+x)a

    Please help me, I am so fed up of trying to get this right.
     
  7. Oct 12, 2006 #6
    please help, I am so fed up of this problem.
     
  8. Oct 12, 2006 #7

    OlderDan

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    No. What you have written assumes mass is being added to m1. No mass is being added. There is an additional force acting, but no additional mass. Also, you have been given the acceleration as half what it was without the additional force, so (a) is now known.
     
  9. Oct 12, 2006 #8
    I dont understand still, can you write it in a equation like I did?
     
  10. Oct 12, 2006 #9
    So will it be

    T-4g-F= 4(.5a)

    In that case I got F= 19.596
     
  11. Oct 12, 2006 #10

    OlderDan

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    Did you use the old T, or caclulate the new T? The old T will not work.
     
  12. Oct 13, 2006 #11

    Doc Al

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    Try this. Take your original equations:
    Now add the new force F on M1. The equations become:
    (for M1) T -4g -F = 4a
    (for M2) 8g - T = 8a

    You have two unknowns: T and F. Solve for F. (This time around "a" is not an unknown--it's given as 3.266/2 m/s^2.)
     
  13. Oct 13, 2006 #12
    thank u so much
     
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