First of all, we must determine the amount of centripetal force enabling the car to make this turn without slipping:
#F_c=(mv^2)/r#
We have #v=25# m/s and #r=80# m. While we do not have a value for #m#, we will continue, as it will cancel out of the problem later.
This centripetal force must be provided by the component of friction and of the normal force provided by the banked roadway that act toward the centre of the circle.
The diagram below shows us these forces (using #N# for the normal force and #f# for friction), along with the horizontal and vertical components of each force. The centripetal force will consist of the sum of horizontal components, namely #Nsintheta# and #fcostheta#
So, at this point, we have the centripetal force as
#F_c=Nsintheta + fcostheta = (mv^2)/r#
Further, the force of friction, as always, is #f=muN#, and examination of the diagram shows
#Ncostheta=mg#, so that
#N=(mg)/costheta#
Substituting all this into the equation fr #F_c# above, we obtain
#(mgsintheta)/costheta+(mumg)/costheta=(mv^2)/r#
Now, we can cancel #m# (which we don't know but which does not factor into the problem),
#g sintheta/costheta+(mug)/costheta=(v^2)/r#
Now we rearrange and solve for #mu#. First, multiply each term by #costheta#
#g sintheta+mug=(v^2costheta)/r#
Divide through by #g#, and take the first term on the left over to the right side of the equation:
#mu = (v^2costheta)/(gr) - sintheta#
#mu = (25^2cos10°)/(9.8*80) - sin10°#
#mu = (25^2*cos10°)/(9.8*80) - sin10°#
#mu=0.785-0.174=0.611#
