# Mechanics - Banked Slope

I have an example in a textbook that goes like this:

A car travels around a bend which has radius 100 m and is banked at an angle of 20° the horizontal. The car is travelling at a speed of 30 ms-. What is the least possible value of the coefficient of friction if the car does not slip up the slope?

Their solution is to resolve vertically and horizontally to find an two expressions for reaction force and the coefficent of friction.

let F = uR where R is the reaction force and u is the coefficient of friction.

Resolve vertically

Rcos20 - Fsin20 - mg = 0

Rcos20 - uRsin20 = mg

R(cos20-usin20) = mg


Resolve Horizontally

Rsin20 + Fcos20 = mv^2/r
Rsin20 + uRcos20 = m30^2/100
R(sin20 + ucos20) = 9m


Solving dividing these two equations to remove R and m and then solving gives u = 0.416

I would have attempted the problem a different way by resolving parallel and perpendicular to the slope.

Resolve perpendicular

R - mgcos20 = 0 => R = mgcos20


Resolve parallel

F - mgsin20 = 0 => F = mgsin20 => uR = mgsin20


Solving this gives

uR/R = sin20/cos20 => u = tan20 = 0.364


I don't understand where in my method I have gone wrong and any help would be much appreciated.

The acceleration is not zero along neither the parallel axis nor the perpendicular one. You have to decompose the centripetal acceleration which is horizontal, all according to Newton's second law $$\Sigma \vec F_i=m\vec a$$ :
$$R - mg\cos20º = \dfrac{mv^2}r\sin 20º$$, perpendicular
$$F + mg\sin20º = \dfrac{mv^2}r\cos 20º$$ parallel (note the plus sign, as the bank makes the gravity to help to turn)