A turn of radius `20 m` is banked for the vehicle of mass `200 kg` going at a speed of `10 m//s`. Find the direction and magnitude of frictional force (a) `5 m//s`
(b) `15 m//s`
Assume that friction is sufficient to prevent slipping. `(g=10 m//s^(2))`

Options

A. `f=200sqrt(5)N` (down wards)`
B. `f=500sqrt(5)N` (up wards)`
C. `f=500sqrt(5)N` (down wards)`
D. `f=200sqrt(5)N` (up wards)`

Explanation

To solve the problem of finding the direction and magnitude of the frictional force acting on a vehicle of mass 200 kg moving around a banked turn of radius 20 m at different speeds (5 m/s and 15 m/s), we will follow these steps:

### Step 1: Calculate the banking angle (θ)
The banking angle can be calculated using the formula:
\[ \tan \theta = \frac{v^2}{rg} \]
Where:
- \( v = 10 \, \text{m/s} \) (the speed of the vehicle)
- \( r = 20 \, \text{m} \) (the radius of the turn)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ \tan \theta = \frac{(10)^2}{20 \times 10} = \frac{100}{200} = 0.5 \]
Now, calculating θ:
\[ \theta = \tan^{-1}(0.5) \approx 26.56^\circ \]

### Step 2: Analyze the forces acting on the vehicle at 5 m/s
When the vehicle is moving at a speed less than 10 m/s, the frictional force will act towards the center of the circular path to prevent slipping outward.

- **Forces in the horizontal direction (centripetal force)**:
\[ N \sin \theta - F \cos \theta = \frac{mv^2}{r} \]
- **Forces in the vertical direction**:
\[ N \cos \theta + F \sin \theta = mg \]

### Step 3: Substitute known values for 5 m/s
1. **Calculate the centripetal force**:
\[ \frac{mv^2}{r} = \frac{200 \times (5)^2}{20} = \frac{200 \times 25}{20} = 250 \, \text{N} \]

2. **Set up the equations**:
- Horizontal:
\[ N \sin(26.56^\circ) - F \cos(26.56^\circ) = 250 \]
- Vertical:
\[ N \cos(26.56^\circ) + F \sin(26.56^\circ) = 2000 \, \text{N} \quad (\text{since } mg = 200 \times 10) \]

### Step 4: Solve the equations
Using the trigonometric values:
- \( \sin(26.56^\circ) \approx 0.447 \)
- \( \cos(26.56^\circ) \approx 0.894 \)

Substituting these values into the equations:
1. \( 0.894N - 0.447F = 250 \) (1)
2. \( 0.447N + 0.894F = 2000 \) (2)

From equation (1):
\[ N = \frac{250 + 0.447F}{0.894} \]

Substituting \( N \) into equation (2):
\[ 0.447\left(\frac{250 + 0.447F}{0.894}\right) + 0.894F = 2000 \]
Solving this will yield the value of \( F \) (frictional force).

### Step 5: Calculate the frictional force
After solving the above equations, we find:
\[ F \approx 300 \sqrt{5} \, \text{N} \quad \text{(upward direction)} \]

### Step 6: Analyze the forces acting on the vehicle at 15 m/s
When the vehicle is moving at a speed greater than 10 m/s, the frictional force will act towards the center of the circular path.

1. **Calculate the centripetal force for 15 m/s**:
\[ \frac{mv^2}{r} = \frac{200 \times (15)^2}{20} = \frac{200 \times 225}{20} = 2250 \, \text{N} \]

2. **Set up the equations**:
- Horizontal:
\[ N \sin(26.56^\circ) + F \cos(26.56^\circ) = 2250 \]
- Vertical:
\[ N \cos(26.56^\circ) - F \sin(26.56^\circ) = 2000 \]

### Step 7: Solve the equations for 15 m/s
Using the same trigonometric values as before, we can substitute and solve:
1. \( 0.894N + 0.447F = 2250 \) (1)
2. \( 0.447N - 0.894F = 2000 \) (2)

After solving these equations, we find:
\[ F \approx 500 \sqrt{5} \, \text{N} \quad \text{(downward direction)} \]

### Final Answers:
- For speed 5 m/s: Frictional force = \( 300 \sqrt{5} \, \text{N} \) (upward)
- For speed 15 m/s: Frictional force = \( 500 \sqrt{5} \, \text{N} \) (downward)