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...

Question

anonymous · Answer

radius: 6.7m

anonymous · Answer

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 · Answer

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 $.

anonymous · Answer

u(1.5)-9.81=0, u(1.5)=9.81, u=6.54, (but u cant be greater than one!?)

jamesj · Answer

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 · Answer

why is N < 1.5mg, why is gravity involved in the horizontal direction

jamesj · Answer

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 · Answer

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 · Answer

Weight is mg, yes

jamesj · Answer

what is important here however is not the direction of the weight vector, but the magnitude of N relative to the number mg.