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The problem presents a curve described by the equation \( y^2 = x^3 \), and we are tasked with finding the vertical reaction at point \( O \) for a bar that has a mass per unit length of \( 0.5 \, \text{kg/m} \). The bar spans from \( x = 0 \) to \( x = 1 \, \text{m} \), and the total weight of the bar is the load that causes the reaction at point \( O \).

### Steps to Solve:

1. **Find the differential length of the bar**: The curve is given by \( y^2 = x^3 \). To determine the differential length \( ds \) of the curve in terms of \( dx \), we use the arc length formula.

   - The arc length differential is \( ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx \).

2. **Determine \( \frac{dy}{dx} \)**: From the equation \( y^2 = x^3 \), we can differentiate both sides with respect to \( x \).

   \[
   2y \frac{dy}{dx} = 3x^2
   \]
   \[
   \frac{dy}{dx} = \frac{3x^2}{2y}
   \]

   Since \( y = x^{3/2} \), we substitute into the expression for \( \frac{dy}{dx} \):

   \[
   \frac{dy}{dx} = \frac{3x^2}{2x^{3/2}} = \frac{3}{2} x^{1/2}
   \]

3. **Substitute \( \frac{dy}{dx} \) into the arc length formula**:

   \[
   ds = \sqrt{1 + \left( \frac{3}{2} x^{1/2} \right)^2} dx = \sqrt{1 + \frac{9}{4} x} dx
   \]

4. **Calculate the mass of the bar**: The mass per unit length is \( 0.5 \, \text{kg/m} \), so the mass of a differential element of the bar is \( dm = 0.5 \, ds \).

5. **Determine the vertical force (weight) of the bar**: The weight of the bar acts vertically, and the weight of a differential element is \( dW = dm \cdot g = 0.5 \cdot g \cdot ds \), where \( g \approx 9.81 \, \text{m/s}^2 \).

6. **Integrate to find the total weight**: We need to integrate the weight along the length of the bar to find the total vertical reaction at \( O \). The total weight is:

   \[
   W = \int_0^1 0.5 \cdot 9.81 \cdot \sqrt{1 + \frac{9}{4} x} dx
   \]

Let me calculate this integral to find the total vertical reaction at point \( O \).The total vertical reaction at point \( O \), which corresponds to the weight of the bar, is approximately \( 7.06 \, \text{N} \).

Would you like further clarification or details on any part of this solution?

Here are 5 questions to deepen your understanding:

1. How does the differential arc length \( ds \) change along the curve?
2. What physical significance does the integral represent in this context?
3. How would the reaction change if the mass per unit length was doubled?
4. What would happen if the curve described a different shape, such as \( y = x^2 \)?
5. Can you apply a similar method to determine the horizontal reaction at \( O \)?

**Tip**: For problems involving curved bodies, always start by determining the arc length and its relation to the geometry.

Si la barra tiene una masa por unidad de longitud de 0.5 kg/m, determine la reacción vertical en el soporte fijo O. La curva de la barra está definida por la ecuación y^2 = x^3, y se extiende desde x = 0 hasta x = 1 m.

Determine the vertical reaction at point O for a curved bar with a mass per unit length of 0.5 kg/m. The bar follows the curve y^2 = x^3 and extends from x = 0 to x = 1 m. Learn how to apply calculus to compute arc length and solve for mechanical reactions.

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