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Homework Help: Why is y-y0 positive?

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    In a stage production of Peter Pan, the 63 kg
    actress playing Peter has to fly in vertically,
    and to be in time with the music, she must
    be lowered, starting from rest, a distance of
    2.6 m in 2.2 s. Backstage, a smooth surface
    sloped at 58 supports a counterweight of mass
    m. What is the mass of the counterweight that
    must be used? The pulley is of negligible mass
    and is frictionless. The acceleration of gravity
    is 9.81 m/s2 .

    2. Relevant equations
    y-y0 = Vot + 0.5at^2
    m1a = T - m1gsin 58
    m2a = m2g - T

    3. The attempt at a solution
    I already got the answer.
    a= 1.07438 m/s^2
    m1=58.57211916 kg

    My question is, why is y - y0 positive? Should it not be negative since Peter Pan is lowered, making y < y0 which y0 = 2.6?? If I make y - y0 = -2.6, then I do not get correct value of the mass.
  2. jcsd
  3. Sep 24, 2011 #2
    In your equations of motion, you've set the -y direction as positive, thereby you arrived at a positive acceleration.
    Set the acceleration negative, then set y-y0 as the negative value you tried, and see for yourself; That will get you the right answer...
  4. Sep 24, 2011 #3
    My acceleration does come out negative when I set y - y0 = -2.6. However, the value of the mass is wrong if I use a = -1.07438
  5. Sep 24, 2011 #4
    It's not merely enough to change the sign. You have to set a different direction for it, in your force equations.
    T is positive, heading upwards, the Weight is negative, downwards.
    Alter this:
    m1gsin(...) should thus have precedence.
    Last edited: Sep 24, 2011
  6. Sep 24, 2011 #5
    Then would it be entirely wrong to set y - y0 = 2.6 instead of -2.6?
  7. Sep 24, 2011 #6
    Of course not!
    It's only a question of convention, of the selection of axes. In these mechanical problems, and indeed in many such ones in physics, the choice of our reference frame is truly arbitrary...
    You've done everything properly, in my view,
  8. Sep 24, 2011 #7
    oh wow ok thank you very much. I undersand it clearly. Oh may I ask one more question then? is gravity = -9.8 or +9.8?
    because in y-component velocity equation, i do not know which one is correct:
    y - y0 = V0t + .5gt^2
    y - y0 = V0t - 0.5gt^2
  9. Sep 24, 2011 #8


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    Staff Emeritus
    Science Advisor
    Gold Member

    Most people define the constant g = +9.81 m/s2. Therefore, whether the acceleration due to gravity is positive or negative once again depends entirely on your choice of sign convention (just as Daniel was saying before). Gravity points downward (in fact, this is what "down" means -- it's the direction in which gravity points). Therefore if you choose to define a coordinate system in which "downward is the negative y-direction," then in your equation, the acceleration due to gravity would be written as a = -g. In contrast, if you choose to define a coordinate system in which "downward is the positive y-direction," then you'd write the acceleration due to gravity as a = +g. Again, the choice is yours, but you must stick to it consistently.
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