A turn of radius 20 m is banked for the vehicles going at a speed of 36km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Answer

A, B, D

Explanation

Banking angle, `theta=tan^(-1)((v^(2))/(Rg))`
`36Km//h=10m//s`
`theta=tan^(-1)((100)/(20xx9.8))=27^(@)`
Angle of repose,
`theta_(r)=tan^(-1)(mu)=tan^(-1)(0.4)=21.8^(@)`
Sinne `thetagttheta_(r)` , vehicle cannot remain in the given position with `v=0` . At rest it will slide down. To find minimum speed, so that vehicle does not slip down, maixmum friction will act up the plane.
To find maximum speed, so that the vehicle does not skid up, maximum friction will act down the plane.
Minimum Speed

Equation of motion are,
`Ncostheta+muNsintheta=mg` ...(i)
`Nsintheta-muNcostheta=(m)/(R)v_(max)^(2)`...(ii)
Solving these two equations, we get
`v_(min)=4.2m//s`
Maximum Speed ltbr. Equations of motion are, `Ncostheta-muNsintheta=mg` ...(iii)
`Nsintheta+muNcostheta=(m)/(R)v_(max)^(2)` ...(iv)

Solving two equations, we have
`v_(max)=15m//s`