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(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.
Recall that centripetal acceleration has a magnitude a of
a = v ² / R
where
v = tangential speed
R = radius of the curve
so that
a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²
Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have
∑ F = Fs = m a
where
Fs = magnitude of static friction
m = mass of the car
Then
Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N
(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,
N - W = 0
where
N = magnitude of normal force
W = weight
so that
N = W = m g = (950 kg) (9.8 m/s²) = 9310 N
Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:
Fs = µ N
Assuming 35 m/s is the maximum speed the car can travel without skidding, we find
µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58