Let's break down the problem step by step: ### Problem Summary - **Curved gate equation:** \( y = 3 \left(\frac{x}{3}\right)^{3/2} \) - **Gate is supported by a string passing over a pulley** - **Water is filled up to a height:** \( h = 4.5 \) m - **String is connected to the gate at \( h = 4.7 \) m** - **Gravity:** \( g = 10 \text{ m/s}^2 \) - **Width of gate:** \( b = 20 \) m ### What is required? **Calculate the tension developed in the string.** --- ## Step-by-Step Solution ### 1. **Find Hydrostatic Force on the Gate** The hydrostatic force on a submerged curved surface is: \[ F = \gamma \cdot A \cdot \bar{h} \] Where: - \( \gamma \) = unit weight of water = \( \rho g = 1000 \times 10 = 10,000 \text{ N/m}^3 \) - \( A \) = area of the curved surface - \( \bar{h} \) = depth of centroid of the area from the surface #### **But** (since the gate is curved), we calculate the total vertical force as the weight of the water above the gate. **Area under the curve from \( x = 0 \) to \( x = 3 \):** \[ y = 3 \left(\frac{x}{3}\right)^{3/2} \] So for \( x = 3 \), \( y = 3(1)^{3/2} = 3 \). #### **Volume of Water Above the Gate:** Since the width is given, the volume is: \[ V = \text{area under the curve} \times \text{width} \] \[ \text{Area} = \int_{0}^{3} y \, dx = \int_{0}^{3} 3 \left(\frac{x}{3}\right)^{3/2} dx \] Let \( u = \frac{x}{3} \implies x = 3u, dx = 3 du \), when \( x = 0, u = 0 \), when \( x = 3, u = 1 \): \[ = \int_{u=0}^{1} 3 u^{3/2} \cdot 3 du = 9 \int_{0}^{1} u^{3/2} du = 9 \left[ \frac{u^{5/2}}{5/2} \right]_{0}^{1} = 9 \left( \frac{2}{5} \right) \left[ u^{5/2} \right]_{0}^{1} = \frac{18}{5} (1 - 0) = 3.6 \, \text{m}^2 \] \[ \text{Volume} = 3.6 \times 20 = 72 \text{ m}^3 \] \[ \text{Weight of water} = V \times \gamma = 72 \times 10000 = 720,000 \text{ N} \] --- ### 2. **Calculate Tension in the String** - The tension in the string must balance the moment produced by the hydrostatic force (weight of water) about the hinge at the bottom. #### **Moment due to Water Weight about the Hinge** - The centroid of the area will be at a certain distance from the bottom (let's call this \( x_c \)). - The tension acts at the top, which is at height \( h \). #### **But since the tension supports the gate at the top, the moment about the hinge:** \[ T \times h = W \times d \] Where: - \( T \) = Tension in string - \( h \) = height from hinge to the point of application of tension (4.7 m) - \( W \) = weight of water (720,000 N) - \( d \) = distance from hinge to centroid of water volume #### **Find \( d \) (Centroid Height)** For a function \( y(x) \): \[ \bar{y} = \frac{1}{A} \int y \cdot dA \] For this case, since \( y \) is vertical: \[ \bar{y} = \frac{\int_{0}^{3} y \cdot y \, dx}{\int_{0}^{3} y \, dx} = \frac{\int_{0}^{3} y^2 dx}{\text{Area}} \] But \( y = 3 \left( \frac{x}{3} \right)^{3/2} \): \[ y^2 = 9 \left( \frac{x}{3} \right)^3 = 9 \left( \frac{x^3}{27} \right) = \frac{x^3}{3} \] So, \[ \int_{0}^{3} y^2 dx = \int_{0}^{3} \frac{x^3}{3} dx = \frac{1}{3} \left[ \frac{x^4}{4} \right]_{0}^{3} = \frac{1}{3} \cdot \frac{3^4}{4} = \frac{1}{3} \cdot \frac{81}{4} = \frac{27}{4} \] So, \[ \bar{y} = \frac{27/4}{3.6} = \frac{27}{4 \times 3.6} = \frac{27}{14.4} \approx 1.875 \text{ m} \] --- ### 3. **Calculate Tension** \[ T \times h = W \times d \] \[ T = \frac{W \cdot d}{h} = \frac{720,000 \times 1.875}{4.7} \approx \frac{1,350,000}{4.7} \approx 2872 \text{ kN} \] --- ### 4. **Check the Answer Choices** - 4368 kN - 3289 kN - 3623 kN - 2883 kN #### Closest value is **2883 kN**. --- ## **Final Answer** **The tension developed in the string is approximately:** \[ \boxed{2883 \text{ kN}} \] --- If you need the detailed calculus steps or a diagram, let me know!