Find the maximum speed at which a truck can safely travel without toppling over, on a curve of radius `250m` . The height of the centre of gravity of the truck above the ground is `1.5m` and the distance between the wheels is `1.5m` , the truck being horizontal.

Explanation

To find the maximum speed at which a truck can safely travel without toppling over on a curve, we will use the principles of circular motion and moments.

### Given Data:
- Radius of curvature, \( r = 250 \, m \)
- Height of center of gravity, \( h = 1.5 \, m \)
- Distance between wheels, \( d = 1.5 \, m \)
- Acceleration due to gravity, \( g = 10 \, m/s^2 \)

### Step-by-Step Solution:

1. **Understanding Forces and Moments:**
When the truck is turning, the centripetal force required to keep it moving in a circle is provided by the frictional force acting on the inner tire. If the truck is to remain upright and not topple over, the moments about the edge of the outer tire must be balanced.

2. **Setting Up the Moment Equation:**
Let \( N_o \) be the normal force on the outer tire, and \( N_i \) be the normal force on the inner tire. When the truck is on the verge of toppling, the normal force \( N_i \) approaches zero. Thus, we can focus on the normal force \( N_o \) and the frictional force.

The moment about the outer tire (point O) due to the normal force \( N_o \) is:
\[ \text{Moment due to } N_o = N_o \times \frac{d}{2} = N_o \times \frac{1.5}{2} = \frac{1.5 N_o}{2} \]

The moment due to the frictional force (which provides the centripetal force) is:
\[ \text{Moment due to friction} = \text{Frictional force} \times h = \left(\frac{mv^2}{r}\right) \times 1.5 \]

3. **Equating Moments:**
At the point of toppling:
\[ N_o \times \frac{1.5}{2} = \frac{mv^2}{r} \times 1.5 \]

Rearranging gives:
\[ N_o = \frac{mv^2}{r} \times 2 \]

4. **Finding Normal Force:**
The normal force \( N_o \) can also be expressed in terms of weight:
\[ N_o = mg \]

Setting the two expressions for \( N_o \) equal:
\[ mg = \frac{mv^2}{r} \times 2 \]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g = \frac{v^2}{r} \times 2 \]

5. **Solving for Maximum Speed \( v \):**
Rearranging gives:
\[ v^2 = \frac{gr}{2} \]
\[ v = \sqrt{\frac{gr}{2}} \]

6. **Substituting Values:**
Now substituting the values of \( g \) and \( r \):
\[ v = \sqrt{\frac{10 \times 250}{2}} = \sqrt{\frac{2500}{2}} = \sqrt{1250} \]
\[ v \approx 35.36 \, m/s \]

### Conclusion:
The maximum speed at which the truck can safely travel without toppling over is approximately \( 35 \, m/s \).