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Force with 3 masses

  1. Feb 15, 2008 #1
    A mass m2= 1.5kg rests on a horizontal table. us= .3 and uk = .25. M2 is attached by strings to m1= 2.5kg hanging over oneside of the table and m3= 4.5kg hanging over the other side. The system is initially at rest. Once released what is the acceleration of m2?

    Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T-m1g
    For the force involving m3: m3 is going down so F3=m3g - T

    I do not know what to do from here.

    Any help would be appreciated. Thank you.

    Stephen
     
  2. jcsd
  3. Feb 15, 2008 #2

    Doc Al

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    The tensions in the strings are different, so label them T1 & T3 (or something).

    You found the net force on m1 & m3, so apply Newton's 2nd law to each.

    And don't forget m2. What forces act on it? Apply Newton's 2nd law to it as well.

    You'll get three equations and three unknowns.
     
  4. Feb 15, 2008 #3
    Ok for the force involving m1: m1 < m3 so m1 will be going up, so F1= T1-m1g=m1a
    For the force involving m3: m3 is going down so F3=m3g - T4=m3a

    F2=T3-(fk+T2)=m3a????

    I do not know what to do from here.
    What do I solve for to find the acceleration of m2????
    Any help would be appreciated. Thank you.

    Stephen
     
  5. Feb 15, 2008 #4

    HallsofIvy

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    First you should calculate the static friction- the coefficient of static friction, [itex]\mu_S[/itex] times the weight of M2. (I guessed that us and ks were the coefficients of static and kinetic friction- it would have been a good idea to say that.) None of the masses will move until the net force on M2 is greater than the static friction.

    The net force on M2 is, of course, the difference between the weights of the two other masses.

    If M3 and M1 are such that the net force on M2 is greater than the static friction, you must calculate the kinetic friction and subtract that from the net force to find the total force accelerating M2.
     
  6. Feb 15, 2008 #5

    Doc Al

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    There are only two strings, so there are only two tensions. Let's call them T1 (for the string attached to m1) and T3 (for the string attached to m3).
    Good.
    Good, but let's rewrite it using T3: m3g - T3 = m3a

    I assume this is supposed to be the force equation for m2, so it should be:
    T3 -fk -T1 = m2a

    What's fk?

    To find the acceleration (for all the masses, since they are connected) solve those three equations together. Hint: See if you can combine the equations to eliminate the tensions.
     
  7. Feb 15, 2008 #6

    Doc Al

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    That's only true for the static case; if it accelerates, the tensions will no longer equal the weights of the hanging masses.
     
  8. Feb 15, 2008 #7
    there are 2 strings?
    one from the first hanging mass to the mass2 then from mass2 to mass three?

    So F for mass 1= T1-mg
    F for mass 3= mg-T2

    and for mass 2 F= T2-(k friction) - (T1)

    But if there are 2 strings with four sections of tension: one from first hanging mass to pulley, next from pulley to mass 2, then from mass 2 to other pulley, and lastly from pulley to hanging mass 3, then my origional 3 equations were right.

    Which set of equations do I use??

    And are we not assuming that after they are released the k friction is greater than the s friction? If not the s friction = 4.45N.
    What do I do with this information?

    How do I find the acceleration of m2?? And what do I solve for to help me find the acceleration??

    thank you

    Stephen
     
    Last edited: Feb 15, 2008
  9. Feb 15, 2008 #8

    Doc Al

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    Correct.

    The tension is uniform throughout the string. There are two strings and thus two tensions. String 1 exerts the same tension force on m1 that it does on m2.

    See post #5.

    (Note: I assume that the strings are massless, and that any pulleys are massless and frictionless.)
     
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