Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of
a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²
There are two forces acting on the car in this situation:
• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n
• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)
By Newton's second law, the net forces in the perpendicular and parallel directions are
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) = ma
==> sin(θ) = a/g ==> θ = arcsin(a/g) ≈ 17.8°
(Notice that in the paralell case, the positive direction points toward the center of the curve.)
When rounding the curve at 31.8 m/s, the car's radial acceleration changes to
a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²
and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:
(perp.) ∑ F = n - mg cos(θ) = 0
(para.) ∑ F = mg sin(θ) + f = ma
The first equation gives
n = mg cos(θ)
and substituting into the second equation, we get
mg sin(θ) + µmg cos(θ) = ma
==> µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12