First of all, we must determine the amount of centripetal force enabling the car to make this turn without slipping:
F_c=(mv^2)/rFc=mv2r
We have v=25v=25 m/s and r=80r=80 m. While we do not have a value for mm, 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 NN for the normal force and ff for friction), along with the horizontal and vertical components of each force. The centripetal force will consist of the sum of horizontal components, namely NsinthetaNsinθ and fcosthetafcosθ
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So, at this point, we have the centripetal force as
F_c=Nsintheta + fcostheta = (mv^2)/rFc=Nsinθ+fcosθ=mv2r
Further, the force of friction, as always, is f=muNf=μN, and examination of the diagram shows
Ncostheta=mgNcosθ=mg, so that
N=(mg)/costhetaN=mgcosθ
Substituting all this into the equation fr F_cFc above, we obtain
(mgsintheta)/costheta+(mumg)/costheta=(mv^2)/rmgsinθcosθ+μmgcosθ=mv2r
Now, we can cancel mm (which we don't know but which does not factor into the problem),
g sintheta/costheta+(mug)/costheta=(v^2)/rgsinθcosθ+μgcosθ=v2r
Now we rearrange and solve for muμ. First, multiply each term by costhetacosθ
g sintheta+mug=(v^2costheta)/rgsinθ+μg=v2cosθr
Divide through by gg, and take the first term on the left over to the right side of the equation:
mu = (v^2costheta)/(gr) - sinthetaμ=v2cosθgr−sinθ
mu = (25^2cos10°)/(9.8*80) - sin10°
mu = (25^2*cos10°)/(9.8*80) - sin10°
mu=0.785-0.174=0.611
