Question

What is angle of banking derive the expression for angle of banking on a curved road with certain coefficient of friction

Answer 1

Acting forces on the body while talking on the curve:  

(1) Normal force (N) - - Vertical for plane  

Component along x-axis: N cosθ

Component along  y-axis: N sinθ  

(2) Centripetal force (Fnet)  = mv2/r  - - towards the center of the curve (horizontal)  

(3) friction force (F) = μs N - Increase the plane in the direction of the bottom  

Component along x-axis: μs N cosθ

Component along y-axis: μs N sinθ

(4) body weight = mg - - Vertical bottom direction x –  

The sum of forces with axis:  

ΣFx = mv2/r    

N sinθ + μs Ncosθ = mv2/r  

N (sinθ + μs cosθ) =mv2/r. . . . . . (1)  

The sum of forces along x- axis:  

ΣFy = 0  

N cosθ - μsNsinθ - mg = 0  

N(cosθ - μs sinθ) = mg. . . . . . (2)  

with division (1) (2):

(sinθ + μs cosθ)/(cosθ-μs sinθ)  = V2/rg

v =√rg ((sinθ + μs cosθ )/( cosθ-μs sinθ)) ------------ -  

where μs = coefficient of static friction  

θ = angle of banking