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...
Stacey Warren - Expert brainly.com
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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%
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 \( \mu \).