Question

A turn of radius 20m is banked for the vehicle of mass 200kg going at a speed of 10m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed 5m/s

Answer 1

See diagram.
m = 200 kg,    R = 20 m,    Angle of banking = Ф
Speed = v = 10 m/s.

Let f be the friction force and N be the normal force from the ground. When the banking of road is done for speed v, it means that the normal force supplies the centripetal force. Friction plays no part. If the speed is either less or more than v, then it comes into play.

Centripetal force =  N sinФ = m v²/R
Also,     N Cos Ф = mg
=> TanФ = v²/gR = 10²/(10*20) = 1/2
      cosФ = 2/√5    SinФ = 1/√5
      N = mg SecФ = 200*10* √5/2 = 1000 √5 Newtons

When the vehicle is moving at 5 m/s less than the optimum speed,
the centripetal force required is less. But the normal reaction force N is same.
Hence, a friction force f acts upwards (towards outside of the curved road) the incline.

So    N CosФ + f sinФ = m g    => N (CosΦ + μ sinΦ) = m g
        N sinФ -  f cosФ = m u²/R  =>  N (sinΦ - μ CosΦ) = m u²/R
=>   (CosФ + μ SinΦ) * u²/gR  = (sinФ - μ CosΦ)
=>   μ [sinΦ *u²/gR  + cosФ] = [ sinФ - CosФ * u²/gR ]
=>   μ = [sinΦ - cosΦ * u²/gR] / [sinФ * u²/gR  + cosФ]

=>   μ = [1/√5 - 2/√5 * 5²/(10*20) ] / [1/√5 * 5²/(10*20)  + 2/√5]
          = [ 3/(4√5) ] / [ 17/(8√5) ]
          = 6/17

Normal force N = mg/[CosΦ + μ SinΦ] = 200*10/ [2/√5 + 6/17 * 1/√5]
                         = 850√5 N 
Friction force = f = μ N = 300√5 N
        upwards the slope.