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Two blocks on a frictionless horizontal surface connected by a massless string

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Two blocks on a frictionless horizontal surface are connected together by a massless string. A second massless string is attached to the block of mass m2, as shown. Suppose you pull on this second string with a horizontal force directed to the right.

    Determine the ratio of the tensions for the two strings.

    NOTE: THE PICTURE SHOWS ONE BLOCK LARGER THAN THE OTHER. HOWEVER, IN THIS QUESTION WE ACTUALLY DO NOT KNOW WHICH BLOCK IS LARGER.

    2. Relevant equations

    Let T= tension

    use ratio: T1/T2

    3. The attempt at a solution

    So, would it just simply be (T(m2onm1))/(T(Fonm2) - T(m1onm2))

    where m2onm1 is the subscript of the 1st T, Fonm2 is the subscript of the 2nd T, and m1onm2 is the subscript on the 3rd T.
    (These are the action reaction forces I obtained by completing a free body diagram- Tm2onm1 with Tm1onm2. The T(Fonm2) is the force the second string applies on block m2).

    Or is the question asking more than that? Are they asking, what does T1/T2 equal ? I am not sure if I understood the question correctly..

    Please help!
    Thank-you!
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2009 #2

    Doc Al

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    Staff: Mentor

    They want the exact ratio T1/T2, not just generalities. Your answer should be in terms of m1 & m2.

    Apply Newton's 2nd law! (Twice.)
     
  4. Feb 5, 2009 #3
    Hi,

    well I tried what you suggested, and I got: T1/T2= (0.5)(m1/m2)

    where m1 is the mass of the first block,
    m2 is the mass of the 2nd block,
    T1 is the tension from the first massless string,
    T2 is the tension from the second massless string on block m2


    Is that correct? Or am I still not even understanding the question correctly?
     
  5. Feb 5, 2009 #4

    Doc Al

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    Staff: Mentor

    Please describe how you got that result.

    (It's not quite right, but if you show your work perhaps we can see where you went wrong.)
     
  6. Feb 5, 2009 #5
    OK, here it goes:

    n= normal force
    w= gravitational force


    After drawing the blocks' free body diagrams, I obtained:

    For m1:

    Fy: n - w =0
    n=mg
    Fx: T(m2onm1)=m1a1

    For m2:

    Fy: n - w=0
    n=mg
    Fx: T(Fonm2) - T(m1onm2)=m2a2

    a1=a2=a

    T(m2onm1)=T(m1onm2)=T

    a= T/m1

    a= (T(Fonm2) - T)/m2

    putting them together i get:

    (T/m1) = ((T(Fonm2) - T)/m2)

    multiply both sides by m2:

    ((m2/m1)T) = (T(Fonm2) - T)

    divide both sides by T:

    (m2/m1) = ((T(Fonm2)/T) - T/T)

    (m2/m1) = ((T(Fonm2)/T) - 1)

    ((T(Fonm2)/T) = ((m2/m1) + 1)


    For some reason, now I am getting ((m2/m1) + 1). Although I really dont think that is correct either. Especially since this time, for the ratio, I got the second tension (2nd massless rope " T(Fonm2)") over the 1st tension (from 1st massless rope "T"). Shouldn't it be the other way around? Or does that not matter?

    Thank-you for helping, btw!
     
  7. Feb 5, 2009 #6

    Doc Al

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    Staff: Mentor

    Excellent.
    You found T2/T1. If you want T1/T2, just invert it.

    Here's how I'd do this one.

    First I'd consider both masses together:
    T2 = (m1 + m2)a

    Then m1 by itself:
    T1 = (m1)a

    Then take any ratio you like:
    T2/T1 = (m1 + m2)/m1
    T1/T2 = m1/(m1 + m2)
     
  8. Feb 5, 2009 #7
    OK, thanks a lot!
     
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