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Tension problem - HELP

  1. Feb 4, 2005 #1

    ktd

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    Tension problem - HELP!!

    For some reason, I'm getting myself all confused about this EASY problem:

    A 64.0 kg box hangs from a rope. What is the tension in the rope if:
    (a.) the box is at rest?
    (b.) the box moves up a steady 4.90 m/s^2?
    (c.) the box has Vy=5.50 m/s and is speeding up at 5.00m/s^2?
    (d.) the box has Vy=5.50 m/s and is slowing down at 5.00 m/s^2?

    Why have I been using the F=ma formula and getting things wrong? I'm thinking I'm totally just making things way too hard and my brain is completely frozen!

    Thanks for any help!!
     
  2. jcsd
  3. Feb 4, 2005 #2

    dextercioby

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    Just tell us how u messed things up.

    Daniel.
     
  4. Feb 4, 2005 #3

    ktd

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    (a.) I'm not even sure this is how I should solve this: even though the box is at rest, there is tension on it - the weight of the box itself. wouldn't gravity affect it also? or, would the tension be 0 N? I'm confused about the concept itself.

    For the other parts, I literally plugged in the numbers given to F = ma --> for (c.) F = (64.0kg)(5.00m/s^2)

    I'm so confused.
     
  5. Feb 4, 2005 #4

    dextercioby

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    In every of the 5 cases u need to apply the principles of dynamics ALL OF THEM...For the first problem:what are the forces that act on the wire??

    Daniel.
     
  6. Feb 4, 2005 #5

    ktd

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    Ok so the first part (a.), the forces involved are the tension of the rope (up), the weight of the box (down) and gravity (down)...now what?
     
  7. Feb 4, 2005 #6

    arildno

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    The weight of the box IS the gravity force acting upon the box (i.e, there is no additional gravity force than the weight)
    See if that helps you..:smile:
     
  8. Feb 4, 2005 #7

    dextercioby

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    I remember asking what the forces acting on the WIRE (not on the box) are... :bugeye:

    Daniel.
     
  9. Feb 4, 2005 #8

    ktd

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    The forces on the rope - hmmm. The rope just holds the box, so that would be the only force...?
     
  10. Feb 4, 2005 #9

    dextercioby

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    You didn't say which.Think of the III-rd principle and the fact that forces come in pairs ALWAYS...To get your answer,neglect the gravity force exaerted by the Earth on the rope itself...(and consequently the force acting on the Earth determined by the attraction of the rope).

    Daniel.
     
  11. Feb 4, 2005 #10

    ktd

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    So the gravity force from earth-->rope = force from rope-->earth; they cancel each other out, right?
     
  12. Feb 4, 2005 #11

    rpc

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    in a) the force pulling down is gravity, and since it is at rest, tension must be the equal and opposite force

    F = ma a in this case is gravity .... 9.8
    so, F = mg = Tension
     
  13. Feb 4, 2005 #12

    ktd

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    ok, so get that. now why am I not understanding the other parts? for (b.) I know I'd have to include g in there, but how? if I use the f = ma equation, i get confused
     
  14. Feb 4, 2005 #13

    Doc Al

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    Start by figuring out the acceleration of the box for each case. Find the magnitude and the direction.

    Only then can you apply Newton's 2nd law. Tip: Call up positive, down negative. (Direction matters!) Tension always acts up; the weight of the box always acts down: So the sum of forces = T - mg. Set that equal to ma: T - mg = ma. Since the acceleration is different in each case, so will be the tension.
     
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