A car weighs `1800 kg`. The distance between its front and back axles is `1.8 m`. Its centre of gravity is `1.05 m` behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Options

A. 4000 N on each front wheel, 5000 N on each back wheel
B. 5000 N on each front wheel, 4000 N an each back wheel
C. 4500 N on each wheel, 4500 N on each back wheel
D. 3000 N on each front wheel, 6000 N on each back wheel

Explanation

To solve the problem of determining the forces exerted by the level ground on each front wheel and each back wheel of the car, we can follow these steps:

### Step 1: Write down the given data
- Mass of the car (m) = 1800 kg
- Distance between front and back axles (d) = 1.8 m
- Distance from the front axle to the center of gravity (h) = 1.05 m
- Gravitational acceleration (g) = 9.8 m/s²

### Step 2: Calculate the total weight of the car
The weight (W) of the car can be calculated using the formula:
\[ W = m \cdot g \]
\[ W = 1800 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 17640 \, \text{N} \]

### Step 3: Set up the equations for equilibrium
Let \( R_f \) be the force exerted by the ground on the front wheels and \( R_b \) be the force on the back wheels. According to the equilibrium of forces:
\[ R_f + R_b = W \]
\[ R_f + R_b = 17640 \, \text{N} \] (Equation 1)

### Step 4: Set up the torque equilibrium equation
Taking moments about the center of gravity (CG) to find the relationship between \( R_f \) and \( R_b \):
- The distance from the center of gravity to the front axle is \( 1.05 \, \text{m} \).
- The distance from the center of gravity to the back axle is \( 1.8 \, \text{m} - 1.05 \, \text{m} = 0.75 \, \text{m} \).

Setting up the torque equilibrium about the center of gravity:
\[ R_f \cdot 1.05 = R_b \cdot 0.75 \]
From this, we can express \( R_b \) in terms of \( R_f \):
\[ R_b = \frac{R_f \cdot 1.05}{0.75} \]
\[ R_b = 1.4 R_f \] (Equation 2)

### Step 5: Solve the equations simultaneously
Substituting Equation 2 into Equation 1:
\[ R_f + 1.4 R_f = 17640 \]
\[ 2.4 R_f = 17640 \]
\[ R_f = \frac{17640}{2.4} = 7350 \, \text{N} \]

Now substituting \( R_f \) back into Equation 2 to find \( R_b \):
\[ R_b = 1.4 \cdot 7350 = 10290 \, \text{N} \]

### Step 6: Calculate the force on each wheel
Since the car has two front wheels and two back wheels:
- Force on each front wheel:
\[ \text{Force on each front wheel} = \frac{R_f}{2} = \frac{7350}{2} = 3675 \, \text{N} \]

- Force on each back wheel:
\[ \text{Force on each back wheel} = \frac{R_b}{2} = \frac{10290}{2} = 5145 \, \text{N} \]

### Final Results
- Force on each front wheel = **3675 N**
- Force on each back wheel = **5145 N**