Calculating Forces on a Banked Curve

[taskev21]

Homework Statement

A curve of radius 152 m is banked at an angle of 12°. An 738-kg car negotiates the curve at 93 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.
(a) the normal force exerted by the pavement on the tires


(b) the frictional force exerted by the pavement on the tires


(c) the minimum coefficient of static friction between the pavement and the tires

Homework Equations

F=ma
F= mv^2 /r

The Attempt at a Solution

I got parts A, and C, but I can't seem to get the force of friction.
I have the centripetal force = 3240.21, so the incline component is 3240.21/cos(12) = 3312.598. Then I add that to the incline component of the car's weight which is 738*9.81*sin(12) = 1505.234. 1505.234 + 3312.598 = 4871.8329 N. so in kiloNewtons it should be 4.871 kN...what is the problem here?

 

[ehild]

The centripetal force is a resultant force, the vector sum of gravity, normal force and friction.

The incline component of the centripetal force is equal to the incline component of gravity + (or -) friction. So the magnitude of friction is equal to the absolute value of the difference between the components of Fcp and of G.

Fr= 3312.598-1505.234 =1807 N

ehild