Resultant Force for circular motion on a banked track

[dumdum000333]

[SOLVED] Resultant Force for circular motion on a banked track

Homework Statement

A car travels 77 m/s around a circular track of radius 71.9
mass of car= 2800 kg
coefficient of friction = .1
angle of track with horizontal = 22 degrees
acceleration of gravity = 9.8 m/s^2
What is the magnitue pf the resultant force on the 2800 kg driver and his car?

Homework Equations

normal force= mgcos(angle)
force of friction = (normal force)(coefficient of friction)

The Attempt at a Solution

i know that the force of friction and normal force do not create the centripetal force necessary to cause uniform circular motion as
Fn= 2800(9.8)cos(22)=25441.9N
Ff= 25441.9(.1)=2544.19N
and the necessary centripetal force= mv^2/r= 2800(77)^2/71.9= 230892.9068N

so i tried making the resultant= the horizontal component of Ff+horizontal component of Fn= cos(22)(2544.19)+sin(22)(25441.9)= 11889.6N which was not the right answer

 

[Mentz114]

What is the magnitude pf the resultant force on the 2800 kg driver and his car?

You may ignore friction. At any moment the force will be the resultant of the centripetal force and gravity. The question isn't about whether the car will stick.

 

[dumdum000333]

never mind, the answer was just centripetal force