Let's break down the problem step by step: --- ## **Given Data** - **Depth of water, \( h \) = 10 m** - **Width of channel, \( w \) = 34 m** - **Gates are inclined at \( 15^\circ \) to the centerline** - **Only upstream side has water** - **We need to find the force exerted on the hinge of Gate A** (magnitude and direction) --- ## **Step 1: Hydrostatic Force on Gate** The hydrostatic force on a vertical surface is: \[ F = \rho g A \overline{h} \] Where: - \( \rho \) = density of water \( \approx 100 \, \text{kg/m}^3 \) - \( g = 9.81 \, \text{m/s}^2 \) - \( A \) = area of gate - \( \overline{h} \) = depth to centroid of area **Area of one gate:** The gates meet at the centerline, so each gate covers half the width: \[ \text{Width of each gate} = \frac{34}{2} = 17\,\text{m} \] \[ A = 17 \times 10 = 170\,\text{m}^2 \] **Depth to centroid:** Centroid of rectangular gate is at half-depth: \( \overline{h} = 5 \, \text{m} \) **Total hydrostatic force on Gate A:** \[ F = 100 \times 9.81 \times 170 \times 5 = 8,348,500 \, \text{N} \] \[ F \approx 8.35 \times 10^6 \, \text{N} \] --- ## **Step 2: Direction of Hydrostatic Force** The force acts **perpendicular** to the gate, at a depth \( \frac{2}{3} \) from the water surface to the base: \[ h_{\text{cp}} = \frac{10}{3} \approx 3.33\,\text{m} \text{ from base} \] So, from the top, \( 10 - 3.33 = 6.67\,\text{m} \) down. --- ## **Step 3: Force Components** Since the gates are at \( 15^\circ \) to the centerline, the hydrostatic force \( F \) acts **normal** to the gate. But for equilibrium, we need to resolve forces **horizontally**. - **Horizontal component:** \( F \cos 15^\circ \) - **Vertical component:** \( F \sin 15^\circ \) --- ## **Step 4: Forces from Gate B** Gate A and B exert forces on each other at their meeting point. This force acts along the line of the gate (i.e., at \( 15^\circ \) to the centerline). --- ## **Step 5: Hinge Force Calculation** Consider equilibrium of Gate A (free-body diagram): - Hydrostatic force acts perpendicular to the gate at center of pressure. - Reaction at the hinge (let's call it \( R \), with components \( R_x \) and \( R_y \)). - Force from Gate B at the meeting point (let's call it \( Q \)), acts along the gate. Sum forces in \( x \) and \( y \) directions and take moments about the hinge to solve for \( R \). --- ### **Sum of Moments about Hinge** \[ F \times \text{distance to hinge} = Q \times \text{distance to hinge} \] But since both gates are identical and symmetrically loaded, the hinge must balance the horizontal hydrostatic pressure. --- #### **Horizontal Component (Main Force)** The main resisting force at the hinge is the horizontal component of the hydrostatic force: \[ F_\text{horizontal} = F \cos(15^\circ) \] \[ F_\text{horizontal} = 8.35 \times 10^6 \times \cos(15^\circ) \] \[ F_\text{horizontal} = 8.35 \times 10^6 \times .9659 \] \[ F_\text{horizontal} \approx 8.07 \times 10^6 \, \text{N} \] --- ### **Direction of Force** - The **direction** is **perpendicular to the gate**, i.e., at \( 15^\circ \) to the channel wall. --- ## **Final Answers** ### **1. Magnitude of Force at the Hinge of Gate A:** \[ \boxed{8.35 \times 10^6 \, \text{N}} \] ### **2. Direction:** \[ \boxed{15^\circ \text{ to the channel wall (perpendicular to the gate surface)}} \] --- ### **Summary Table** | Quantity | Value | |---------------------------|-----------------------------| | Area of gate (A) | \( 170 \, \text{m}^2 \) | | Hydrostatic force (F) | \( 8.35 \times 10^6 \, \text{N} \) | | Direction | \( 15^\circ \) to channel wall (normal to gate) | --- If you want the **exact components at the hinge** (horizontal/vertical), let me know!