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Proving that apparent weight is 5 times actual weight (no numbers)

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Without numbers, identify how you could find the apparent weight of motorcycle to be 5 times that of the actual weight. The motorcycle is going up a loop the loop with a velocity of v.

    2. Relevant equations
    Fnet = mv^2/r
    Fg = mg

    3. The attempt at a solution
    Fnet = mv^2/r
    Fg + Fn = mv^2/r
    Fn = mv^2/r - Fg
    Fn = mv^2/r - mg
    Fn = mv^2/r - mgr/r
    Fn = (mv^2 - mgr)/r

    I have no idea of this is right or if I'm completely misinterpreting the question.
     
  2. jcsd
  3. Mar 4, 2012 #2
    This seems to be an open ended question and does not require you to consider a specific position for the bike.May I suggest that you consider the bike at its bottom most position(that's where its apparent weight will be biggest for a constant speed).What are the two relevant forces on the bike,in what directions do they act and what is an expression for the resultant force?
     
  4. Mar 4, 2012 #3

    tiny-tim

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    hi testme! :smile:
    they're asking for an explanation in words

    start by deciding:

    at which point in the loop is the apparent weight greatest? :wink:
     
  5. Mar 4, 2012 #4
    Thats what I was assuming. The bike was at the bottom, the two forces acting upon it are the normal force (apparent weight) and the gravitational force (actual weight).

    using that I came up with the equation

    Fnet = mv^2/r
    Fg + Fn = mv^2/r
    Fn = mv^2/r - Fg
    Fn = mv^2/r - mg
    Fn = mv^2/r - mgr/r
    Fn = (mv^2 - mgr)/r

    Though I'm not sure if that'd be the right way to find out. Also, that value would have to be 5 times your mass times gravitational acceleration.

    When the questions asks without numbers they mean as in no numbers given, so just theoretically the proccess. That's what my teacher said.
     
  6. Mar 4, 2012 #5
    At the bottom the normal force(apparent weight) acts upwards and the weight acts downwards.
     
  7. Mar 4, 2012 #6

    tiny-tim

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    words only:

    "as soon as it goes into the loop, if will have a centripetal acceleration which can only be supplied by an increased normal force, which (for fixed v) will be proportional to … , which can be made large enough by making … sufficiently … " :wink:
     
  8. Mar 4, 2012 #7
    Fnet = mv^2/r
    -Fg + Fn = mv^2/r
    Fn = mv^2/r + Fg
    Fn = mv^2/r + mg
    Fn = mv^2/r + mgr/r
    Fn = (mv^2 + mgr)/r

    Would that be it then?
     
  9. Mar 4, 2012 #8
    Looks good but remember Fn=5mg
     
  10. Mar 4, 2012 #9
    Would it be fine that we find Fn using that and then we multiply Fg we would find by 5 at the very end?
     
  11. Mar 4, 2012 #10
    Fg is the "true weight" of the bike and that remains constant.What you are finding is an equation giving the necessary speed(v) for a given radius r for the bike to have an "apparent weight"(at the bottom) of 5mg.
     
  12. Mar 4, 2012 #11
    I get that, but what I mean is does it matter if I divide Fn by 5 at the end or multiply Fg by 5? Could I just say 5Fg must equal Fn or Fn/5 must equal Fg for this to be true?
     
  13. Mar 4, 2012 #12
    I'm not sure that I understand you.At the bottom the resultant force is Fn-mg.
    (Fn represents the apparent weight in other words the weight that would be measured,by say,a set of scales over which the bike rides).
    We can write: Fn-mg=mv^2/r
    If the apparent weight is to be 5 times the real weight then Fn=5mg so we can write:
    5mg-mg=4mg=mv^2/r
     
  14. Mar 5, 2012 #13
    Nevermind, I got my answer, thanks for the help!
     
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