825 mm 1. If the water rises to 150 mm above the bottom of the door, calculate the hydrostatic force on the door (using standard units), Whr? (Hint
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- A vertical plate is submerged in water and has the shape as seen in the picture. Using the fact that the weight of water is 62.5 lbs/ft³, calculate the hydrostatic force (in Ibs) against the end of the tank. Give your answer correct to the nearest hundred. (Hint: First set up a Riemann sum that approximates the hydrostatic force, which can then be used to obtain an integral that represents the force.) 12 ft 8 ft 20 ft Hydrostatic force = Ibswhat is the total hydrostatic force acting on the tainter sector gate? (in lbs)H-W 2 Q1/ Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle panel at the bottom of the water tank in Fig. Water 2m
- A vertical plate is submerged in water and has the shape as seen in the picture. Using the facts that the density of water is 1000 kg/m³ and acceleration due to gravity is 9.8 m/s², calculate the hydrostatic force (in N) against the end of the tank. Make sure your answer is correct to the nearest thousand. (Hint: First set up a Riemann sum that approximates the hydrostatic force, which can then be used to obtain an integral that represents the force.) 20 m Hydrostatic force =A closed vessel 1 m in diameter is completely filled with water. If the vessel is rotated at 1200 rpm, what increase in pressure will occur at the top of the tank at the circumference? Ans. 1970 kPaA vertical plate is submerged in water and has the indicated shape. 10 m 6 m A triangle is submerged in water pointed up. The top most vertex of the triangle touches the surface of the water. The base of the triangle is 6 m and is parallel to the surface of the water. The height of the triangle is 10 m. Express the hydrostatic force (in N) against one side of the plate as an integral (let the positive direction be downwards) and evaluate it. (Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1,000 kg/m3.)
- The total pressure force of a square plate is 25 kN, placed vertically in water so that the center of the plate is 1.5 m below the free surface. Determine the size of the plate and the depth of center of pressure. The above square plate is taken out of the water which weighs 370.5 N in air and 180.3 N in water of specific gravity 0.99. Find the volume and specific gravity of the square plate. Draw a neat sketch.Q.1: An inclined, circular gate with water on one side is shown in the figure below. Determine the total hydrostatic force acting on the gate and the location of the center of pressure. Water surface 1.5 m 60 Gate Side view Gate Top 1 m view Q.2: In the figure shows a curved surface LM, which is in the form of a quadrant of a circle of radius (2.4 m), immersed in the water. If the width of the gate is unity, calculate the horizontal and vertical components of the total force acting on the curved surface. Free water surface 1.2 m 2.4 m TTTTTTTM Curved surfaceAt what depth below the surface of the oil, relative density 0.83 will produce a pressure of 100 kN/m2. What depth of water is this equivalent to? Depth of oil in meters Answer for part 1 Depth of water is this equivalent to Answer for part 2
- Determine the hydrostatic force exerted by the water on the bottom and sidewall of the pool of width 12 ft. D, 6 ft 8 ft B 20 ft ANSWER: (round it off to the nearest whole number) FBc = || kips, FAB = kips, FDc = kipsA square plate (each side equal to 3m) is submerged vertically in water such that the upper edge of the plate is at depth 1m from the free surface. What will be the total hydrostatic force on the plate? Select the correct response: 145.89 kN 132.44 kN 220.73 kN 88.29 kNThe floodgate of a dam is in the shape of a parabola opening upwards. It is 20 ft tall and 10 ft wide at the top. Calculate the hydrostatic force experienced by the floodgate if the top of the floodgate is 50 ft below the surface of the water. To simplify calculations, use w for the weight density of water in lieu of 62.4 lb/ft Use Insert -> Equation when necessary