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Circular motion: normal force on a loop

  1. Aug 11, 2018 #1
    1. The problem statement, all variables and given/known data
    A 180lb pilot flies a verticle loop with radius 2000 ft at 350 mi/h. With what force does the seat fress upward against him at the bottom of the loop?
    Problem 6.21
    2. Relevant equations
    ##F_c=\frac {mv^2} {r} ##

    3. The attempt at a solution
    There seems to be a discrepancy in the answer key provided? On the first line the equation they came up is ## F= \frac {mv^2} {r} + mg## . But when the number was plugged in, it seems like they use the equation ##F= \frac {mv^2} {gr} + m ## Which equation is correct? My own answer corresponds with the first equation, which is why I get a different numerical answer.

    Thank you!
  2. jcsd
  3. Aug 11, 2018 #2
    Are you familiar with the difference between lb-force and lb-mass? What is the weight (in lb-force) of a body that has a mass of 180 lb-mass?
  4. Aug 11, 2018 #3
    I'm more familiar with the unit with kilogram. But I guess the weight in lb-force would be mass in lb * 32.2 ft/second^2 ? (as opposed to 9.8 m/second^2 for kg). So the weight in lb-force of a body with mass 180 lb is 5796 lb? Still, I haven't quite understood the discrepancy in the answer key.
  5. Aug 11, 2018 #4
    Oh wait I get what you are saying now. 180-lb as indicated in the book's question is the weight (m*a), not the mass (m). I didn't realize that. Thank you!
  6. Aug 12, 2018 #5


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    You really should use lb only for the mass unit and lbf for the force unit. The weight of a body with mass x lb in standard gravity is by definition x lbf, not 32.2x lbf. This is by definition of the pound force unit. There is also a similar kg related unit that sees very little use, kgf.
  7. Aug 12, 2018 #6
    In Imperial units, the weight of 1 lb-mass is 1 lb-force. Crazy, huh? This all resolves itself when we specify that the mass to use in applying Newton's second law using Imperial units is the slug, which is the mass in lb-mass divided by 32.2.
  8. Aug 12, 2018 #7


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    Yes, but be sure you use the correct value for mass in the mv^2/r term!
    Edit: oh already answered, I’m a bit late.
  9. Aug 16, 2018 #8


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    I would convert the whole thing to kg and m/s, then when you get the final force, in newtons, convert that into imperial units of force.
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