anonymous
  • anonymous
On a carnival ride the static friction between a person and the walls of the ride keep them from falling. The normal force is perpendicular to the vertical. Given that the engineers of the ride do not want the normal force to be more than 1.5 times any persons weight, what is the minimum coefficient of static friction needed to keep riders from falling? parameters below...
Physics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
radius: 6.7m
anonymous
  • anonymous
my thinking is that bounding the normal force to 1.5 times a persons weight means that the centripetal acceleration must not exceed 1.5 (as F=ma). If this is so then, it is my reasoning, that the static friction coefficient would have to be something of the order of 300%
JamesJ
  • JamesJ
Well, the downward force of gravity on the person is \[ F_g = -mg \] On the other hand, the force due to friction is \[ F_f = \mu N \] In balance, \( F_f + F_g = 0 \). You have an upper bound on N. Use that this equation to solve for \( \mu \).

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anonymous
  • anonymous
u(1.5)-9.81=0, u(1.5)=9.81, u=6.54, (but u cant be greater than one!?)
JamesJ
  • JamesJ
Careful. You know that \( N \leq 1.5mg \). Hence as \(-F_g = F_f \), we have that \[ mg = \mu N \leq 1.5 \mu mg \] Now you take it from there.
anonymous
  • anonymous
why is N < 1.5mg, why is gravity involved in the horizontal direction
JamesJ
  • JamesJ
You are given that "the engineers of the ride do not want the normal force to be more than 1.5 times any persons weight"
anonymous
  • anonymous
AH!, you know what i did, i was thinking 1.5 times a persons mass!, and to me that ment that acceleration had to be 1.5, and i was thinking well duh this is impossible. Weight is a force!!! duh, lol
JamesJ
  • JamesJ
Weight is mg, yes
JamesJ
  • JamesJ
what is important here however is not the direction of the weight vector, but the magnitude of N relative to the number mg.

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