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Physics Homework Problem

  1. Oct 11, 2005 #1
    I think someone may have previously asked for help on a problem similar to this one:

    There are three blocks connected by a string. m1 is 0.5kg, m2 is 1.7kg.
    m1, and m2 are connected at the top of a surface and m3=2.9kg, is hanging of the end of a pulley. looks kinda like this:

    [ m1 ]----[ m2 ]-----O
    xxxxxxxxxxxxxxxxxxxxx|
    xxxxxxxxxxxxxxxxxxxxx|
    xxxxxxxxxxxxxxxxxxx [m3]

    a) find the acceleration of the masses shown in the figure.

    I realize that the only force acting on m1: F=T= m1a, and I think the force acting on m3: m3g-T=m3a. I am cofused about what to do with m2. Any help is greatly appreciated.

    JB Gibson:confused:
     
  2. jcsd
  3. Oct 11, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that each piece of string--the one between m1 & m2 and the one between m2 & m3--will have a different tension.
     
  4. Oct 11, 2005 #3
    Would I be correct in my assumptions:
    F12=T12=a(m1+m2)
    F23=m3g-T23=m3a

    m3g = a
    m1+m2+m3
     
  5. Oct 11, 2005 #4

    Doc Al

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    I don't understand what you've done. Try this approach:
    (1) Identify the forces on each mass
    (2) Write Newton's 2nd law for each mass

    You'll get three equations that you can solve together. Hint: Assume that the acceleration is "a" to the right and down.
     
  6. Oct 11, 2005 #5
    That's where my confusion begins! How do I derive these equations? I thought I was ok on the first and last equations, but evidently not. The first and last equations are as follows:

    F1=T=m1a
    F3=m3g-T=m3a

    Thaks for all the help!
     
  7. Oct 11, 2005 #6

    Doc Al

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    As I mentioned earlier, the tensions on the strings are different. I'd write these two equations like this:
    [tex]T_1 = m_1 a[/tex]

    [tex]m_3 g - T_2 = m_3 a[/tex]

    Now write the equation for the second mass.
     
  8. Oct 11, 2005 #7
    I'm having a hard time deriving the equation for the second mass. I'm not sure what to do. I know that the gravitational force and the normal force cancels out, and you're left with the tension force and the friction force. How do you apply this information?
     
  9. Oct 11, 2005 #8

    Doc Al

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    Unless the problem states otherwise, assume the surfaces are frictionless. (If friction is involved, you'll need to include friction in the equations for both m1 and m2.)

    So all you need to worry about are the two tension forces that act on m2. Give it a shot.
     
  10. Oct 11, 2005 #9
    Let me get this straight! Here are my equations:

    T1=m1a
    T2=m2a-m1a
    m3g-T2=m3a

    Just a guess for clarification. Thanks again!
     
    Last edited: Oct 12, 2005
  11. Oct 12, 2005 #10

    Doc Al

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    You still haven't got the correct equation for m2. Answer these questions:
    (1) What horizontal forces act on m2?
    (2) What's the net force on m2?
    Now apply Newton's 2nd law ([itex]F_{net} = m a[/itex]) to mass 2.
     
  12. Oct 12, 2005 #11
    To make things a little clearer, I guess I need to know what is the equation for m2. I not asking anyone to do the problem for me or derive any solutions. I just don't see the 2nd equation!
     
  13. Oct 12, 2005 #12
    I think I get it! Correct me if I'm wrong:

    T1=m1a
    T2-T1=m2a
    m3g-T2=m2a

    This would leave me with a final equation of:

    m3g=a(m1+m2+m3)

    and now I would solve for "a"

    Is this correct?
     
    Last edited: Oct 12, 2005
  14. Oct 12, 2005 #13

    Doc Al

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    Now you're cooking!

    (You did have a typo in your third equation.)
     
  15. Oct 12, 2005 #14
    Thanks Doc for all of the help and letting me do the problem. I'm grateful for the help!
     
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