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Explain this to me

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A 55kg mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration 1.0m/s^2 .

    Find the tension in the rope.

    my answer : 484N

    the way I got is by doing this : 55(9.8-1.0).

    I did this because my friend told me this but am confused on
    why I have to subtract g from acceleration. Please explain in details.
     
  2. jcsd
  3. Mar 31, 2009 #2

    LowlyPion

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    If there was no retarding force then the acceleration from gravity would have been unaffected. So for acceleration to be just 1, then there must be a counteracting force that relates to 9.8 - 1 ... in order to end with 1 as your acceleration.
     
  4. Mar 31, 2009 #3
    This statement confuses me :
    When the mass is released, it falls with acceleration 1.0m/s^2 .

    Its falling down at 1 meters per second per second. This makes me think that
    with gravity counted, it falls at 1 m/s/s.
     
  5. Mar 31, 2009 #4

    LowlyPion

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    Right it does mean that.

    It means that gravity is accelerating it at 9.8 and the drum is decelerating it at 8.8. The 8.8 then is the Tension in the cable.

    Otherwise if it was free falling it would be slack.
     
  6. Mar 31, 2009 #5
    Why should it not be decelerating at 1m/s/s instead of accelerating at 1m/s/s.
    It seems like if if its accelerating then we should add 1.0 instead of subtract.
     
  7. Mar 31, 2009 #6

    tiny-tim

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    Ftotal = ma …

    and Ftotal = mg - T :wink:
     
  8. Mar 31, 2009 #7
    Ya but why is tension m*8.8 instead of m*10.8. After all it is falling 1.0m/s/s with the
    gravity? It just seems counter intuitive to me.
     
  9. Mar 31, 2009 #8

    LowlyPion

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    No. It's not falling 1m/s2 faster than gravity.

    It's falling at just 1m/s2

    If it was static what would the Tension be? m*g right? And it's not moving.

    So as TinyTim pointed out the tension will be m*g less the net acceleration of 1 m/s2
     
  10. Mar 31, 2009 #9
    So the net acceleration is 1m/s/s.

    If so then why isn't m*a = m*(1). Isn't a the net acceleration?
     
  11. Apr 1, 2009 #10

    tiny-tim

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    Yes …
    ma = mg - T,

    so m*(1) = mg - T, and T = mg - m*(1).

    No? :smile:
     
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