Question

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

(a) 5 m/s and 

(b) 15 m/s. Assume that friction is sufficient to prevent slipping. (g = 10 m/s²)

Answer 1

At the correct speed for which the road is banked, the self-adjusting static friction acting on the vehicle is zero. When the speed decreases, the vehicle has a tendency to slip downwards, thus the static friction acts upwards. When the speed increases, the vehicle has a tendency to slip upwards, thus static friction acts downwards.

(a) the turn is banked for speed v = 10 m/s

Therefore

Now, as the speed is decreased, force of friction f acts upward.

using the equaitons \(\sum f_x\) = \(\frac{mv^2}{r}\) and \(\sum f_y\) = 0 we obtain

N sin\(\theta\) - f cos\(\theta\) = \(\frac{mv^2}{r}\)  ............. (i)

and N cos\(\theta\) + fsin\(\theta\) = mg ................ (ii)

Now, by substituting, \(\theta\) = tan^-1 (1/2), v = 5 m/s, m = 200 kg and r = 20m, in the above equations,

we obtain f = 300 \(\sqrt{5} N\) (outward)

(b) In the second case, force of friction f will act downward.

Using \(\sum f_x = \frac{mv^2}{r}\) and \(\sum f_y = 0\) we obtain

Nsinθ + f cos rθ = … (iii)

Ncosθ - f sin mgθ = … (iv)

Substituting, θ = tan^-1(1/2), v = 15 m/s, m = 200 kg and r = 20 m, in the above equations, we obtain f = 500√5 N = (downward).