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Tension and Newton's Second Law

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data

    A cable is being used to lift a 350 kg piano into a helicopter. The cable can exert a maximum force of 5000 N without breaking. What is the maximum acceleration that the cable can give to the piano?

    2. Relevant equations

    w=mg


    3. The attempt at a solution

    w=?
    a=?
    m=350 kg
    Tension=5000 N


    w=mg
    w=350kg(9.8 m/s2
    w=3430 N

    Fnet=ma
    w-Fcable=ma
    a=w-fcable /m
    a=3430 N-5000 N / 350 kg

    -1570/350 n/kg = -4.49 m/s2

    Is this answer correct? The negative sign makes sense to me because when we used 9.8 as a positive number, we decided that down was the negative direction. I'm not really sure how quickly a piano can be lifted, but 4.49 seems a bit slow. Can anyone critique?

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2014 #2

    lightgrav

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    gravity pulls mass down. using 9.8 N/kg as positive is calling _down_ as positive. so up is negative.
     
  4. Jan 27, 2014 #3
    Correct, I interchanged my words accidentally. Thanks! So is my answer correct?
     
  5. Jan 28, 2014 #4

    collinsmark

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    It looks good to me (ignoring any sign/direction nuances). :approve:
     
  6. Jan 28, 2014 #5
    Thanks! The sign in the end is correct, right?
     
  7. Jan 28, 2014 #6

    lightgrav

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    it can be accelerated upward almost half as quickly as it would fall .
    ... it is more common (traditional among Engineers) to call up as positive.
     
  8. Jan 28, 2014 #7

    collinsmark

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    It is correct if you use the positive is down convention.

    In general, there is no "correct" convention to use (negative being up or positive being up). It's your choice (unless your instructor tells you otherwise).

    The general idea is that you pick a convention at the beginning, and then stick with it for consistency throughout the problem.

    In this particular case, you started by treating the force of gravity mg = 3430 N as positive. Since the force of gravity is down, you assigned down as positive at that point.

    Had you instead intended on down being negative, you would append on negative sign on your gravitational force. [Edit: And also treat the 5000 N force from the cable as positive instead of negative.]

    The important part is not which convention you use, but rather consistency. Once you pick a convention you must stick with it consistently though the final answer.
     
  9. Jan 28, 2014 #8
    Thank you both for your answers! Thanks for the explanation, collinsmark. I think my question was really if I was consistent. I thought so, but I wanted to be sure I didn't miss anything. Thanks again! :)
     
  10. Jan 28, 2014 #9

    lightgrav

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    it helps to show the positive direction on your diagram "+→" , or "←+" .
     
  11. Jan 28, 2014 #10
    I did, but I still don't really understand where the confusion is. Is the negative sign in my answer okay?

    Thanks!
     
  12. Jan 28, 2014 #11

    collinsmark

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    Yes, it's correct. But to clear up any confusion, write on your paper somewhere that positive is down (or negative is up). You can do this with an arrow and '+' sign as lightgrav suggests.

    Then for your final answer, don't include the negative sign, but instead say "4.49 m/s2 up." (In other words, give the magnitude and direction of the acceleration. 4.49 m/s2 is the magnitude and "up" is the direction.) :wink:
     
  13. Jan 28, 2014 #12
    Thank you! That seems to be a much simpler way.
     
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