Circular motion: normal force on a loop

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Beth N
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Homework Statement


A 180lb pilot flies a vertical 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
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Homework Equations


##F=ma##
##F_c=\frac {mv^2} {r} ##

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!
 

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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.
 
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!
 
Beth N said:
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.
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.
 
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Beth N said:
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.
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.