Question

A car moving at 36 km/h negotiates a 140 m -radius banked turn designed for 60 km/h .

What coefficient of friction is needed to keep the car on the road?

Express your answer using two significant figures.

Answer #1

Here.

for speed, v = 60 km/hr

v = 16.7 m/s

Now, let the angle of banking is theta

as v = sqrt(g *R * tan(theta))

16.7^2 = 9.8 * 140 * tan(theta)

theta = 11.44 degree

Now, for v = 36 km/hr = 10 m/s

let the coefficient of friction needed is u to keep it from sliding downwards

v = sqrt(gR * (tan(theta) - u)/(1 + u * tan(theta)))

10^2 = 9.8 * 140 * (tan(11.44 degree) - u)/(1 + u * tan(11.44 degree))

solving for u

**u = 0.127**

**the coefficient of friction needed is 0.13**

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