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Atwood's Machine - One mass and one force

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    M9PmJ.jpg

    2. Relevant equations

    ƩFy = may

    3. The attempt at a solution

    I'm unable to scan my work onto the computer, but I've gotten to the part where I this equation:

    ay = [itex]\frac{FA - Fgblock}{mA + mblock}[/itex]

    The problem is now, I have two unknowns ... FA and mA. This is where I'm stuck and I don't know if I missing something from before or ...

    By the way, this is a question from the 2011 OAPT Physics Contest.
     
  2. jcsd
  3. Oct 9, 2011 #2

    gneill

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    You are in a position to draw the FBD for the mass. Write the equation for the acceleration of the mass assuming some tension T is acting in the rope. You're given the acceleration, so T is....?
     
  4. Oct 10, 2011 #3
    This is the work that I have.

    bU5bG.jpg

    I'm stuck here because the m in the equation is for the total mass of the system, but I don't have it. I tried doing something like:

    Rmrtv.jpg

    but you still need to know FA to get the mass, so I don't know what I'm doing wrong :(
     
  5. Oct 10, 2011 #4

    gneill

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    The only mass that's moving is the one that's hanging on the rope. You have its weight.
     
  6. Oct 10, 2011 #5
    Ok, I solved it --> FA = 125N.

    So basically, what I think I learned from this is that not all forces contribute to a mass / have a mass? I'm a little confused because F = ma, and m is mass, so ...

    Normally, when we do normal questions like these, we have one mass on either side of the machine.
     
  7. Oct 10, 2011 #6

    gneill

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    An external force, applied by unspecified means, has no mass associated with it.

    An external force F applied to a given mass M provokes an acceleration A in that mass according to the relationship F = M*A. That's Newton's 2nd law.

    Newton's 3rd law states that as a result of force F being applied to the mass, the mass will respond to that force with an equal and opposite force. This is the so-called reaction force, which is also therefore equal to M*A. This type of force is also what's known as an "inertial force" because it arises out of the fact that the object has inertia and "resists" changes in motion (Newton's 1st law).
     
  8. Oct 10, 2011 #7

    PhanthomJay

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    No, you are forgetting that the block accelerates upward.
     
  9. Oct 10, 2011 #8
    Is the answer wrong or are you just being picky about no direction when I state my force vector?
     
  10. Oct 10, 2011 #9

    PhanthomJay

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    Your answer is incorrect.
     
  11. Oct 10, 2011 #10
    Oops, I think I messed up a sign in my analysis somewhere. I re-did it, hopefully this is the right answer:

    SVEJ3.jpg
     
  12. Oct 10, 2011 #11

    lightgrav

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