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Homework Help: Ratio of weight at top of a ferris wheel to bottom

  1. Mar 28, 2009 #1
    http://img13.imageshack.us/img13/4697/masteringphysicsq1gwk.jpg [Broken]

    I got the first two, but i'm trying to figure out what to do for the others,

    do i take figure out the net force at the top and divide it by the weight force on ground?

    e.g i figured out at top the net force = 600N (just an example) from mv^2/r-mg=fnetAttop
    and on ground was 90N

    would i put in the awnser 600:90, or 600/90
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 28, 2009 #2


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    Those are found from the equations of the other problem. Just divide them by mg.

    (mg - mv2/r)/mg) for the top and similarly as before for the bottom.
  4. Mar 29, 2009 #3
    I'm not given mass or anything else to help solve for it,
    the question is looking for a ratio for mass at the top of the curve to mass of someone standing
  5. Mar 29, 2009 #4
    can someone help please? i've no idea how to approach the problem
    all i know is, at the top the ratio of weight will be less than on the ground
    and at the bottom it will be greater

    because the centripetal force will also be going with the normal force at the bottom which makes the guy feel like the ground is pushing up harder
    and at the top the normal force is much less making him feel like he is lighter
  6. Mar 29, 2009 #5


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    well if you use the equations, you'd see that you didn't need the mass as it cancels out.
  7. Mar 29, 2009 #6

    normal weight = m*g
    apparent weight at top = mg-(m*v^2/r)

    so the ratio is?
    (mg-(mv^2/r)) / mg?
    which makes it m*(g-a)/mg

    i can cancel the mass out, but how am i suppose to put that in a ratio of weight at the top of the ride to the bottom?
  8. Mar 29, 2009 #7
    OH i didn't realise it was just a number ratio, i thought i'd have to put in symbols or something

    i'm such a ****ing failure, if i got this question in a test, i'd definitely fail

    0.9 for the top ratio
    1.09 for the bottom ratio.
  9. Mar 29, 2009 #8


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    Don't despair. That's the reason for homework. So when it comes exam time you don't make the unthinking errors and you can demonstrate proficiency.

    Edison tried thousands of filaments before settling on tungsten. He didn't consider them failures, only steps to the right answer pointed out by all the things that didn't work.
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