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Neuton's second law

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Two objects of unequal mass are hung vertically over a frictionless pulley of negigible mass. Determine vector a1, vector a2, and the tentions. m1<m2


    2. Relevant equations

    ΣFy=t-m1g=m1ay
    ΣFy=m2g-T=m2ay

    -m1g+m2g=m1ay+m2ay

    ay=m2-m1/(m1+m2)g

    3. The attempt at a solution

    ay=(m2-m1)/(m1+m2)h

    ay=(15-5kg)/(15kg+5kg)(-9.8m/s) (Would gravity be negative?)

    =4.9m/s^2

    Is a1= a2? that is, a1=-4.9m/s^2 and a2=-4.9m/s^2

    T=m1(g+ay)=(2m1m2)/(m1+m2)g

    T=5kg(9.8m/s^2+9.8m/s^2+4.9m/s^2)
    =2(5kg)(15kg)(9.8m/s^2)/(5kg+15kg)

    T=73.5kg

    Does this look right?

    Would you also get the same answer if you did vector a =change in velocity/change in time?

    Thank you very much
     
  2. jcsd
  3. Feb 29, 2008 #2

    kdv

    User Avatar

    You should make really clear what axis you are using. Is it clear to you what are the directions you are using? It's a bit confusing because you use two opposite directions for the two objects. But you can do that as long as everything is clear to you.
    well, you made two mistakes which by luck cancelled out, leaving you with the correct answer. You should have used +9.8 for g. And you made a mistake in going from the first to the second line since you dropped the minus sign.
    Are you talking about [tex] (a__y)_1 [/tex] and [tex] (a_y)_2 [/tex]? To answer your question, you must make clear what your directions are for your two axis.
    And why do you now have a minus sign when there was no minus sign on the previous line?
    A tension is a force so it's not in kg.
     
  4. Feb 29, 2008 #3
    Thank you very much

    It's along the y-axis.

    I ended up getting kg(kg)(m/s^2)/(kg). Would that be kgm/s^2?

    Is ay the same thing as vector a1? Vector a1 and a2 are equal, right? even though m1<m2, would the tention and gravity be the same?

    Does it look correct now?

    Thank you
     
  5. Feb 29, 2008 #4

    kdv

    User Avatar

    Yes, and that's the same as a Newton.
    There are three things involved and people keep mixing them up which causes a lot of confusion. You may talk about the vectors [tex] \vec{a}_1, \vec{a}_2 [/tex], you may talk about their components [tex] (a_y)_1, (a_y)_2 [/tex] and you may talk about their magnitudes (usually written as [tex] a_1, a_2 [/tex]).

    The two acceleration vectors are clearly not equal! But the two magnitudes are equal. As for the y components, it depends on the choice of y axis for each object.
    It's not clear because you jump from positive to negative to positive values of the y acceelration.
     
  6. Feb 29, 2008 #5
    Thank you very much

    Regards
     
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