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Question

A 1050 kg1050 \mathrm{~kg} car rounds a curve of radius 72 m72 \mathrm{~m} banked at an angle of 1414^{\circ}. If the car is traveling at 85 km/h85 \mathrm{~km} / \mathrm{h}, will a friction force be required? If so, how much and in what direction?

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Answered 1 year ago
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To determine whether the friction is necessary or not consider the figure below. We will assume there is friction and that it points towards the center of the curve. Balancing the forces in vertical direction we have

mg+Ffrsinθ=FNcosθ.(eq-1)mg+F_{fr}\sin\theta=F_N\cos\theta.\qquad\text{(eq-1)}

Requiring that all of the components in the direction towards the center provide for the centripetal force we have

FNsinθ+Ffrcosθ=mv2r.(eq-2)F_{N}\sin\theta+F_{fr}\cos\theta=\frac{mv^2}{r}.\qquad\text{(eq-2)}

From (eq-1) we have that

FNsinθcosθ=mgsinθ+Ffrsin2θ.F_{N}\sin\theta\cos\theta=mg\sin\theta+F_{fr}\sin^2\theta.

Returning this to (eq-2) premultiplied by cosθ\cos\theta we get

mgsinθ+Ffrsin2θ+Ffrcos2θ=mv2rcosθ.mg\sin\theta+F_{fr}\sin^2\theta+F_{fr}\cos^2\theta=\frac{mv^2}{r}\cos\theta.

Using the trigonometric identity sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1 we find

mgsinθ+Ffr=mv2rcosθ,mg\sin\theta+F_{fr}=\frac{mv^2}{r}\cos\theta,

yielding for friction

Ffr=m(v2rcosθgsinθ).(eq-3)F_{fr}=m\left(\frac{v^2}{r}\cos\theta-g\sin\theta\right).\qquad\text{(eq-3)}

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