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
