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?

Answer 1

Answer:

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