Explain ---**Fluid Mechanics Problem Extraction** **Section Title:** FLUID MECHANICS ASSIGNMENT 3 **Question Stem:** The rigid gate, AOB, of the Figure 1 below is hinged at O and rests against a rigid support at B. What minimum horizontal force, P, is required to hold the gate closed if its width is 3 m? Neglect the weight of the gate and friction in the hinge. The back of the gate is exposed to the atmosphere. **Chart/Diagram Description:** * **Type:** Schematic diagram illustrating a fluid mechanics scenario involving a gate, water, and applied force. * **Main Elements:** * **Water Body:** A large blue shaded area representing water, located to the left of a vertical gate segment. The water surface is indicated by an inverted triangle symbol and is labeled "Open to atmosphere". * **Gate AOB:** An L-shaped rigid gate (colored orange), composed of a vertical segment OA and a horizontal segment AB. * Point **O** is a circular hinge located at the top of the vertical gate segment. The text "Hinge" is placed next to it. * Point **A** is the corner where the vertical segment meets the horizontal segment. * Point **B** is the end of the horizontal segment, resting against a rigid support (a concrete-like structure). * **Support Structure:** A textured, light gray, irregular shape representing a rigid structure or ground, providing support for the gate at point B and forming the base of the water reservoir. * **Force P:** A horizontal arrow pointing to the left, applied to the right side of the gate, at the same vertical level as point A. It is labeled "P". * **Dimensions:** * Vertical distance from the water surface to the hinge O: 3 m. * Vertical length of the gate segment OA (from hinge O down to point A): 4.4 m. * Horizontal length of the gate segment AB (from point A to point B): 2 m. * **Labels and Annotations:** * "Open to atmosphere" is located above the water surface. * "3 m" indicates the depth of water above hinge O. * "Water" is written within the water body. * "Hinge" is located near point O. * "4.4 m" indicates the vertical length of the gate segment OA. * "B" and "A" label the respective points on the gate. * "P" labels the applied horizontal force. * "2 m" indicates the horizontal length of the gate segment AB. * "Figure 1" is located below the entire diagram.

视频信息

答案文本 复制

视频字幕 复制

We have a fluid mechanics problem involving an L-shaped gate AOB that controls water flow. The gate is hinged at point O and rests against a rigid support at point B. We need to find the minimum horizontal force P required to hold this gate closed. The system has specific dimensions: the gate width is 3 meters, water depth above the hinge is 3 meters, the vertical segment OA is 4.4 meters long, and the horizontal segment AB is 2 meters long. We'll neglect the weight of the gate and friction at the hinge, and note that the back of the gate is exposed to atmospheric pressure. Now let's analyze the hydrostatic pressure distribution on the vertical gate segment. Hydrostatic pressure increases linearly with depth according to the formula p equals gamma h, where gamma is the specific weight of water and h is the depth below the water surface. At the water surface, pressure is zero. At the hinge point O, which is 3 meters below the surface, the pressure is 3 gamma. At point A, which is 7.4 meters below the surface, the pressure reaches its maximum value of 7.4 gamma. This creates a triangular pressure distribution on the gate, with zero pressure at the top and maximum pressure at the bottom. Note that atmospheric pressure acts on both sides of the gate and therefore cancels out in our analysis. Now we calculate the resultant hydrostatic force acting on the vertical gate segment. The total force is given by F equals gamma h_c A, where h_c is the depth to the centroid of the gate area and A is the gate area. The centroid of the vertical segment is located at its geometric center, which is 3 meters plus 4.4 divided by 2, equals 5.7 meters below the water surface. The gate area is 4.4 meters times 3 meters width, which equals 13.2 square meters. Therefore, the resultant force is gamma times 5.7 times 13.2, which equals 75.24 gamma. This force acts at the center of pressure, not at the centroid. Using the formula for center of pressure, we find it acts at 6.27 meters below the water surface, or 2.87 meters below the hinge point O. Now we apply static equilibrium principles by taking moments about the hinge point O. For the gate to remain in equilibrium, the sum of all moments about point O must equal zero. There are two main forces that create moments about the hinge. First, the hydrostatic force of 75.24 gamma acts horizontally at the center of pressure, which is 2.87 meters below the hinge. This creates a clockwise moment. Second, the applied force P acts horizontally on the horizontal segment of the gate, with a moment arm of 4.4 meters from the hinge, creating a counterclockwise moment. Note that the gate weight is neglected as stated in the problem, and the hinge reactions pass through point O, so they create no moment about the hinge. Now we complete the solution by substituting our calculated values into the moment equilibrium equation. The hydrostatic force creates a clockwise moment equal to 75.24 gamma times 2.87 meters, which equals 215.94 gamma meters. The applied force P creates a counterclockwise moment equal to P times 4.4 meters. Setting these moments equal for equilibrium, we get P times 4.4 equals 215.94 gamma. Solving for P, we divide 215.94 gamma by 4.4, which gives us P equals 49.08 gamma. Using the specific weight of water as 9810 Newtons per cubic meter, the minimum horizontal force P required to hold the gate closed is 481,275 Newtons, or 481.3 kilonewtons.