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.
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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