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Weight on libra

  1. Jul 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A man that has an ordinarily weight of 780 N stands on a libra on the equator.

    2. Relevant equations
    What is the weight measured by the libra?

    3. The attempt at a solution
    I thought it was simple, so I just did:
    P = m * g
    Since it gives me P I can simply do:
    m = P / g = 780 / 9.81 = 79.51 kg
    But looking at the solution on the book it says that it's 0.
    Did I do something wrong?
     
  2. jcsd
  3. Jul 27, 2015 #2

    BvU

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    To me it seems you do the right thing. A libra compares two masses and ordinarily we use 9.81 as gravitational acceleration.
    Could the book be off ? Misprint or something ?

    Otherwise it's a corny interpretation of "weight measured by the libra" ... after all it doesn't literally measure weight
     
  4. Jul 27, 2015 #3

    HallsofIvy

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    A "Libra" is a "scale"? Where does he "ordinarily weigh 780 N"? If he "ordinarily" weighs himself on the equator, then his weight on the equator would, of course, be 780 N! You calculate the person's mass but I don't see how that is relevant to the question. You say the books answer is "0" (are there no units) so perhaps I am misunderstanding "libra".

    A person's true 'weight', the force the earth exerts on a person, is 9.81 (approximately) times the mass of the person. What the person weighs is that force minus the mass of the person times the centripetal force on the person due to the rotation of the earth. That will be 0 at the poles and maximum at the equator.
     
  5. Jul 27, 2015 #4
    With "libra" I mean the object. Like the sign.
    Where? Wherever on the earth but the equator, it seems.
    Now that I think about it actually calculating the mass doesn't bring me anywhere...
    It is 0 on the book. Without anything after.
    Anyway this was the third question the the entire problem I didn't thought you could have needed everything since I thought every question was just by itself.
    Anyway, in a previous question on the book I calculated the angular velocity so, doing the calculations in the second thing you said we have the centripetal force:
    Fcentripetal = m * ω2 * RT
    The ω I calculated was because the question before asked what was the ω if the centripetal force was the same as the gravitational force at the equator on the same mass.
    So we have:
    P - Fcentripetal = 780 - [79.5 * (1.24 * 10-3)2 * 6.371 * 106] = 2
    It's not 0 but it is near I think.
     
  6. Jul 27, 2015 #5

    BvU

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    Confusing. What is the complete question + context ?
    Seems to me your original post forgot to mention the earth isn't rotating with the same angular speed that we are used to "ordinarily" :rolleyes:
     
  7. Jul 27, 2015 #6
    Yeah, you are right. Let me write everything.
    "What's the value of the angular velocity of rotation of the Earth so that the centrifugal force at the equator is the same as the gravitational force on the mass on the same spot?
    If the Earth turn at this speed, how long would it be a day? If a man that weighs ordinarily 780 N would stand on foot on a libra situated on the equator, what would the weight measured by the libra be?"
    I'm sorry if something doesn't seem clear but I'm translating the exercise from my language to English language.
     
  8. Jul 27, 2015 #7

    BvU

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    That's somewhat different than what I gathered from post #1 :smile:

    If centrifugal force cancels gravitation, a scale (a force measurement, with a spring or something) would show 0 since there is no force left over.
    A libra (I interpret that as: a balance) wouldn't work either: nothing is pulling the scales down, so any two masses can be glued to the scales and seem to be "in balance".

    On such a fast-rotating planet, wouldn't it be risky to jump ? :rolleyes:
     
  9. Jul 27, 2015 #8
    Hahahahaha! Now I got it!
    Okay, okay. Next times I'll write always the full the exercise.
    Sorry for not being clear from the beginning!
     
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