## Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between

Question

Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between the tires and the roadway. The radius of the curve is R=50m.

Part A

If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?

Part B

What is the minimum speed the automobile can have before sliding down the banking?

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3 years 2021-08-26T12:09:53+00:00 1 Answers 311 views 0

a) v = 20.9 m/s

b) v = 8.46 m/s

Explanation:

Given:-

– The coefficient of static friction is us = 0.30

– The coefficient of static friction is uk= 0.25

– The radius of the curve R = 50m

– The bank Angle β = 25

Find:-

a) If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?

b) What is the minimum speed the automobile can have before sliding down the banking?

Solution:-

– We will investigate the sliding-up case first. Develop a FBD as given in (attachment).

– Use Newton’s second law of motion vertical to slope of bank where the car is in equilibrium:

Sum ( F_n ) = 0

N*cos(β) – m*g – Ff*sin(β) = 0

Where,            Frictional Force Ff = us*N

N (cos(β) – us*sin(β)) = mg   … Eq 1

– Use Newton’s second law of motion horizontal to slope of bank where the car is accelerating:

Sum ( F_h ) = m*a

Ff*cos(β) + Nsin(β) = m*v^2 / R

N (us*cos (β) + sin (β) ) = m*v^2 / R  …. Eq 2

– Divide the two equations:

v^2 / gR = [ us*cos (β) + sin (β) ] / [ cos (β) – us*sin (β) ]

v^2 = [ 0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) – 0.25*sin (25) ]

v = 20.9 m/s

– For the slip down case. We have, friction force Ff reversed hence us = -us. Then the v can be given as:

v^2 / gR = [ -us*cos (β) + sin (β) ] / [ cos (β) + us*sin (β) ]

v^2 = [ -0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) + 0.25*sin (25) ]

v = 8.46 m/s