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

To determine the possible speeds at which the vehicle neither slips down nor skids up on the banked turn, we need to consider the forces acting on the vehicle.

When a vehicle goes around a banked turn, there are two main forces at play:

1. The gravitational force (mg) acting vertically downwards.
2. The frictional force (F) acting horizontally towards the center of the turn.

To prevent slipping down, the gravitational force must be balanced by the component of the frictional force acting opposite to gravity (upwards). To prevent skidding up, the component of the frictional force acting towards the center of the turn must be strong enough to provide the necessary centripetal force.

Let's analyze these forces in more detail:

1. Balancing gravitational force:
The component of the frictional force acting opposite to gravity can be calculated using the formula:
F_up = mg * tan(θ)

where
m = mass of the vehicle
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the banking

2. Providing centripetal force:
The centripetal force required to keep the vehicle on the turn can be calculated using the formula:
F_centripetal = m * v^2 / r

where
v = velocity of the vehicle
r = radius of the turn

To find the possible speeds where neither slipping down nor skidding up occurs, we equate the two forces:

mg * tan(θ) = m * v^2 / r

Next, we can solve this equation for velocity (v):

v = √(g * r * tan(θ))

Substituting the given values:
g = 9.8 m/s^2
r = 20 m
θ = angle of banking (which is not given in the question)

Without the value of the banking angle (θ), we cannot determine the exact speeds. To find the range of speeds, you would need to provide the banking angle.

To determine the possible speeds of the vehicle so that it neither slips down nor skids up on the banked turn, we need to consider the forces acting on the vehicle.

1. The gravitational force (mg) acting vertically downwards.
2. The normal force (N) acting perpendicular to the banked turn surface.
3. The centripetal force (Fc) acting horizontally towards the center of the turn.

The maximum value of the static friction force (fs) can be expressed as fs = μsN, where μs is the coefficient of static friction between the road and the tires.

Since the vehicle is on the verge of slipping, the static friction force must be equal to the horizontal component of the gravitational force (mg * sinθ) which provides the necessary centripetal force to keep the vehicle on the turn.

Now, let's calculate the maximum and minimum speeds at which the vehicle can complete the turn without slipping down or skidding up:

1. Minimum Speed (Vmin):
The minimum speed occurs when the static friction force is at its maximum.
fs = μsN = mg * sinθ
μsN = mg * sinθ
μs * (mg * cosθ) = mg * sinθ (since N = mg * cosθ)
μs * cosθ = sinθ
μs = sinθ / cosθ
Vmin² / r = g * sinθ / cosθ (since Fc = fs = m * Vmin² / r)
Vmin² = r * g * sinθ / cosθ
Vmin = √(r * g * sinθ / cosθ)

2. Maximum Speed (Vmax):
The maximum speed occurs when the static friction force is at its minimum.
fs = μsN = mg * sinθ
μsN = mg * sinθ
μs * (mg * cosθ) = mg * sinθ (since N = mg * cosθ)
μs * cosθ = sinθ
μs = sinθ / cosθ
Vmax² / r = g * sinθ / cosθ (since Fc = fs = m * Vmax² / r)
Vmax² = r * g * sinθ / cosθ
Vmax = √(r * g * sinθ / cosθ)

Now substitute the given values into the formulas:
- radius (r) = 20m
- coefficient of static friction (μs) = 0.4
- acceleration due to gravity (g) = 9.8m/s²
- angle of inclination (θ) = tan^(-1)(1/20) = 2.86°

Vmin = √(20 * 9.8 * sin(2.86°) / cos(2.86°))
Vmax = √(20 * 9.8 * sin(2.86°) / cos(2.86°))

Calculating the values:
Vmin ≈ 15.6 m/s
Vmax ≈ 15.6 m/s

Therefore, the possible speeds of the vehicle so that it neither slips down nor skids up are approximately 15.6 m/s.