Explain ---**Other Relevant Text (Header)**
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:** Engineering diagram / Cross-sectional view illustrating a hydraulic gate system.
* **Main Elements:**
* A body of water is depicted on the left side, labeled "Water". Its free surface is indicated by an inverted triangle symbol and is annotated with "Open to atmosphere".
* A rigid, L-shaped gate, implicitly labeled "AOB" from the problem statement, is shown. It consists of a vertical section and a horizontal section.
* Point O is marked near the top of the vertical section of the gate and is labeled "Hinge", indicating the pivot point.
* Point A is the corner where the vertical and horizontal sections of the gate meet.
* Point B is the far end of the horizontal section of the gate.
* A textured, grey block represents a rigid support structure, against which point B of the gate rests, and part of which extends underneath the horizontal section of the gate.
* A horizontal arrow labeled "P" points to the left, indicating a horizontal force applied to the right side of the vertical section of the gate, near its bottom.
* **Dimensions:**
* A vertical dimension of "3 m" is indicated from the water surface down to the hinge point O.
* A vertical dimension of "4.4 m" is indicated from the hinge point O down to the corner A of the gate.
* A horizontal dimension of "2 m" is indicated for the length of the horizontal section of the gate (from A to B).
* **Labels and Annotations:** "Open to atmosphere", "Water", "Hinge", "P", "3 m", "4.4 m", "2 m", "O", "A", "B".
* **Relative Position and Direction:** The water is contained on the left side of the gate. The gate is hinged at point O, allowing rotation. Point B is supported by the rigid structure. The force P is applied horizontally from the right to hold the gate closed against the water pressure. The region to the right of the gate is open to the atmosphere.
**Other Relevant Text (Figure Title)**
Figure 1
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答案文本
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We have a hydraulic gate problem involving an L-shaped rigid gate AOB that is hinged at point O and supported at point B. The gate holds back water on the left side, while the right side is exposed to atmosphere. We need to find the minimum horizontal force P required to keep the gate closed. The key dimensions are: 3 meters from the water surface down to hinge O, 4.4 meters from hinge O down to corner A, and 2 meters horizontally from A to B. The gate width is 3 meters, and we neglect the gate's weight and friction at the hinge.
Now let's analyze the hydrostatic pressure distribution on the vertical section of the gate. Water pressure increases linearly with depth according to the formula p equals rho g h, where rho is water density, g is gravitational acceleration, and h is depth. At the water surface, pressure is zero. At hinge O, which is 3 meters below the surface, pressure equals 1000 times 9.81 times 3, which is 29.43 kilopascals. At corner A, which is 7.4 meters below the surface, pressure reaches 72.59 kilopascals. This creates a triangular pressure distribution acting horizontally on the gate surface, with pressure arrows showing the direction and magnitude of the hydrostatic force.
Now we calculate the resultant hydrostatic force acting on the vertical gate section. The total force equals the area under the triangular pressure distribution. Using the formula F equals one-half times rho times g times h squared times width, where h is 7.4 meters, we get F equals one-half times 1000 times 9.81 times 7.4 squared times 3, which equals 805,778 Newtons, or approximately 806 kilonewtons. The center of pressure, where this resultant force acts, is located at one-third of the height from the bottom of the submerged section. This puts it 2.47 meters from point A, or 1.93 meters below hinge O. The green arrow shows the magnitude and location of this resultant force.
Now we apply static equilibrium principles to find the required force P. We analyze moments about the hinge point O since this eliminates the unknown reaction forces at the hinge. The gate is in equilibrium, so the sum of moments about O must equal zero. Three forces act on the gate: the hydrostatic force F of 806 kilonewtons acting at the center of pressure, the applied force P that we need to find, and the reaction force at support B. The hydrostatic force creates a clockwise moment with a moment arm of 1.93 meters below the hinge. The applied force P creates a counterclockwise moment with a moment arm of 4.4 meters below the hinge. For equilibrium, these moments must balance, giving us the equation: F times 1.93 equals P times 4.4.
Now we complete the numerical solution by substituting our calculated values into the moment equilibrium equation. We have F times 1.93 equals P times 4.4. Substituting F equals 806,000 Newtons, we get 806,000 times 1.93 equals P times 4.4, which gives us 1,555,580 equals P times 4.4. Solving for P, we get P equals 1,555,580 divided by 4.4, which equals 353,540 Newtons, or approximately 354 kilonewtons. Let's verify our solution: the units are correct in Newtons, the magnitude is reasonable for a hydraulic gate system, and the force direction is correct to counteract the water pressure. Therefore, the minimum horizontal force required to hold the gate closed against water pressure is 354 kilonewtons.