The width b into the paper is 3 m. Neglect the weight of the gate. Solution: The horizontal component of water force is. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the bottle. Solution: First, from the manometer, compute the gage pressure at section AA in the Fig.
For the position shown, determine a the hydrostatic force on the gate per meter of width into the paper ; and b its line of action. Does the force pass through point O?
Solution: The horizontal hydrostatic force is based on vertical projection: Fig. Reference to a good handbook will give you the geometric properties of a circular segment, and you may compute that the segment area is 3.
This force passes, as expected, right through point O. Determine the vertical hydrostatic force on circular-arc section AB and its line of action. Solution: Assume unit depth into the paper. Determine the horizontal and vertical hydrostatic.
What is the force in each bolt required to hold the dome down? Solution: Assuming no leakage, the hydrostatic force required equals the weight of missing water, that is, the water in a 4-m-diameter cylinder, 6 m high, minus the hemisphere and the small pipe:. If the end caps are neglected, compute the force in each bolt.
Solution: Consider a cm width of upper cylinder, as seen below. Derive an analytic expression for the hydrodynamic force F on the shell and its line of action. The width into the paper is b. Which force is larger? Solution: It looks deceiving, since the bulging panel on the right has more water nearby, but these two forces are the same, except for their direction. The left-side figure is the same as Example 2.
The right-side figure has the same vertical force, but it is down. Both vertical forces equal the weight of water inside, or displaced by, the half-cylinder AB. Their horizontal forces equal the force on the projected plane AB. If the panel is 4 m wide into the paper, estimate the resultant hydrostatic force of the water on the panel.
The vertical component equals the weight of water above the gate, which is the sum of the rectangular piece above BC, and the curvy triangular piece of water just above arc BC—see figure at right. The curvytriangle calculation is messy and is not shown here. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force.
Determine a the horizontal and b the vertical hydrostatic forces on the hemisphere, in lbf. Compute the hydrostatic force on the conical surface ABC. Estimate the gage pressure of the air if a the hydrostatic force on panel AB is 48 kN;.
The wedding-cake shape of the box has nothing to do with the problem. Find a the hydrostatic force on the bottom; and b the force on a side panel. What is the specific gravity of fluid X? Solution: The block is sketched below. What is its weight in newtons? Solution: The can weight simply equals the weight of the displaced water neglecting the air above :. Was it gold?
When released into the U. Standard Atmosphere, at what altitude will it settle? Solution: The altitude can be determined by calculating the air density to provide the proper buoyancy and then using Table A. Solution: Refer back to Prob. If so, how many cubic centimeters will be exposed? Therefore the sphere is floating exactly half in and half out of the water. Therefore it floats in gasoline.
The volume exposed is. What diameter spherical balloon will just support the weight? Neglect the size of the hot-air inlet vent. Determine a the string tension; and b the specific gravity of the wood. Solution: The rod weight acts at the middle, 2. What is the specific gravity of the rod material? Note that the spar weight is based on the specific weight of fresh water, The steel force acts right through A.
You could also add in the weight of the air inside the sphere, but that is only 49 N. We see that the buoyancy exceeds the weight by 19 kN, or more than 2 tons.
The sphere would float nicely. Estimate a the tension in the mooring line, and b the height in the standard atmosphere to which the balloon will rise if the mooring line is cut.
They estimate the weight of the box and treasure in air at lbf. Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pressure.
The empty balloon weighs lbf. The box is 2 ft wide, 5 ft long, and 18 in high. The box volume is 2ft 5ft 1. Thus the balloon must develop a net upward force of 1. The air weight in the balloon is negligible, but we can compute it anyway. Clean these two equations up:.
Solution: The total beam volume is 3. Sum moments about point A:. A uniform heavy sphere tied to the left corner causes the beam to float exactly on its diagonal. Sum moments about the left corner, point C:. What is the ratio of the distances a and b for this condition? Then the vertical force balance on the block is. If the total weight of barge and cargo is tons, what is the draft H of the barge when floating in seawater? In terms of seawater, this total volume would be equivalent to The density of the hydrogen in the balloon is thus.
If so, b how many cubic centimeters are exposed, and c how high will a spherical cap protrude above the surface?
EES is recommended for the solution. Solution: From Table A. The sphere floats in oil. Knowing that h is small, of order 5 cm, you could guess your way to the answer. Or you could use EES and get the answer directly. Estimate the gage pressure of the air in the tank.
Solution: In order to apply the hydrostatic relation for the air pressure calculation, the density of Fluid X must be found. The buoyancy principle is thus first applied. Let the block have volume V. Neglect the buoyancy of the air on the upper part of the block. Then 0. Solution: A vertical force balance provides a relation for h as a function of S and L,. Is it stable? Solution: The distance h is determined by. The metacenter position is determined by Eq. Solution: As in Prob.
Solution: A vertical force balance gives 0. Use Eq. This float position is thus slightly unstable. The cylinder would turn over. It is piled so high with gravel that its center of gravity is 3 ft above the waterline, as shown. Solution: Example 2. Is this a stable position? Solution: Let r be the radius at the surface and let z be the exposed height. The centroid G is at 0.
The center of buoyancy B is at the centroid of a frustrum of a submerged cone: 0. The base radius is R and the cone height is H, as shown. This is plotted below. Which position is more stable? Assume large body width into the paper. Solution: The calculations are similar to the floating cone of Prob. Let the triangle be L by L by L. List the basic results. Equation 2. Then Eq. Its center of gravity G is exactly at the surface. Solution: Equation 2. How full can the glass be before it spills?
Solution: First, how high is the container? Well, 1 fluid oz. Thus the glass should be filled to no more than 3. There are many ways to compute p A.
For example, we can go straight down on the left side, using only gravity:. Let the tank be pulled along a horizontal road in rigid-body motion. Find the acceleration and direction for which a a constant-pressure surface extends from the top of the front end to the bottom of the back end; and b the top of the back end is at a pressure 0.
The elliptical shape is immaterial, only the 2-m height. Thus, again, the isobar must slope upward through B but not necessarily pass through point C.
This angle must satisfy Eq. The top surface is horizontal. Determine the rigidbody accelerations for which the water at opposite top corner B will cavitate, for a horizontal, and b vertical motion. Solution: From Table A-5 the vapor pressure of the water is Pa. Assuming rigid-body motion, estimate the maximum water depth to avoid spilling.
Which is the best way to align the tank? It is 1 m wide into the paper. Inside is a cm balloon filled with helium at kPa.
Will it lean to the left or to the right? Solution: The acceleration sets up pressure isobars which slant down and to the right, in both the water and in the helium.
This means there will be a buoyancy force on the balloon up and to the right, as shown at right. It must be balanced by a string tension down and to the left. Can you explain this interesting result? In a frame of reference moving with the child, which way will the balloon tilt, forward or backward? The helium-filled balloon is not in contact with any part of the car seats, ceiling, etc. All the windows in the car are closed. When the traffic light turns green, the car accelerates forward.
In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Conduct a scientific experiment to see if your predictions in parts a and b are correct. If not, explain. Solution: a Only the child and balloon accelerate, not the surrounding air.
This is not rigid-body fluid motion. The balloon will tilt backward due to air drag. This is rigid-body motion. The balloon will tilt forward, as in Prob. Our predictions were correct. What uniform acceleration a x will cause the pressure at point C to be atmospheric?
The fluid is water. Find the central rigid-body rotation rate for which a onethird of the water will spill out; and b the bottom center of the can will be exposed.
The bowls are to be 8 inches deep. Solution: The molten plastic viscosity is a red herring, ignore. The appropriate final rotating surface shape is a paraboloid of radius 7 inches and depth 8 inches. Thus, from Fig. Solution: Let h o be the height of the free surface at the centerline. Then, from Eq. At what point in the tube will this happen? When spinning around DC, the free surface comes down from point A to a position below point D, as shown.
Therefore the fluid pressure is lowest at point D Ans. Thus the drawing is wildly distorted and the dashed line falls far below point C! The solution is correct, however. Solution: a If pressures are equal at B and C, they must lie on a constant-pressure paraboloid surface as sketched in the figure.
The focus of the mirror is to be 4 m from the mirror, measured along the centerline. The focal point F is far above the mirror itself. All tubes are long and have equal small diameters. Solution: a The free-surface during rotation is visualized as the dashed line in Fig. The total displacement between outer and center menisci is, from Eq.
The center meniscus. If atmospheric pressure is How deep is the instrument? The top edge of the gate is 2 m below the surface. What is the hydrostatic force on the gate?
What is the specific gravity of the sphere? How high above the water surface does the wooden end of the rod protrude? We can then neglect movement of the reservoir level. If the reservoir is not large, its level will move, as in the figure. Tube height h is measured from the zero-pressure level, as shown.
Write an exact Expression for p 1gage as a function of h, d, D, and gravity g. Solution: Let H be the downward movement of the reservoir. But volumes of liquid must balance:. That is,. Nevertheless, the U-tube is still to be used to measure the pressure in the air tank. Calculate h in cm, ignoring surface tension and air density effects. Dynamics, riding the merry-go-round with his son, has brought along his U-tube manometer.
You never know when a manometer might come in handy. As shown in Fig. The manometer center is 5. Determine the height difference h in two ways: a approximately, by assuming rigid body. How good is the approximation? Apply this to the left leg z 1 and right leg z 2. The glass is cylindrical, twice as tall as it is wide, and filled to the brim.
He wants to know what percent of the cola he should drink before the ride begins, so that none of it spills during the big drop, in which the roller coaster achieves 0. Make the calculation for him, neglecting sloshing and assuming that the glass is vertical at all times. Solution: We have both horizontal and ver-tical acceleration. So the student should drink the cola until its rest position is 0.
The percentage drop in liquid level and therefore liquid volume is. Since p z increases with z at a greater than linear rate, the center of pressure will always be a little lower than predicted by linear theory Eq.
Integrate Eq. We will study loss coefficients in Chap. Example 2. What is the air pressure in the closed chamber B? Here we must account for reservoir volume changes. If d is not small, this is a considerable difference, with surprisingly large error. What is the pressure difference between points 1 and 2 in the pipe?
Assume the reservoir is very large. Of course, you could also go straight down to the bottom of the tube and then across and up. Find the gage pressure, in Pa, in the air gap in the tank. Neglect surface tension. Assume no change in the liquid densities.
The mercury in the left vertical leg will drop Determine the pressure at point A in pounds per square foot. Normal levels, however, are 2. The manometer uses mercury and air as fluids. Determine the total pressure drop and also the part due to friction only. Which part does the manometer read? Is it higher or lower than Patmosphere?
Estimate the specific gravity of fluid X. The pump stops when it can no longer raise the water pressure. Combining these two gives a quadratic equation for H: 0. Solution: a The vertical elevation of the water surface in the slanted tube is 1. Stay with BG units. Some teachers 6 ft say it is more instructive to calculate these by direct integration of the pressure forces. Find the water force on the panel and its line of action.
From Eq. The center of pressure is thus 3. The three bottom shapes and the fluids are the same. This is called the hydrostatic paradox. Explain why it is true and sketch a freebody of each of the liquid columns. In b side pressures are horizontal. In c upward side pressure helps reduce a heavy W.
For what water depth h is this condition reached? Neglecting atmospheric pressure, find the resultant hydrostatic force on panel BC, a from a single formula; b by computing horizontal and vertical forces separately, in the spirit of curved surfaces. What water level h will dislodge the gate? Neglect Fig. The CG would be For what water depth h will the force at point B be zero? The centroid of Ans. This weight acts downward at the CG of the full gate as shown not the CG of the submerged portion.
Thus, W is 7. For Ans. If the water level is high enough, the gate will open. Compute the depth h for which this happens. The forces on AB and BC are shown in the freebody at right. This solution is independent of both the water density and the gate width b into the paper. Determine the required force P for equilibrium. Thus it is 3. Find the hydrostatic force on surface AB and its moment about C.
Could this force tip the dam over? Would fluid seepage under the dam change your argument? As shown in the figure, the line of action of F is 2.
This moment is counterclockwise, hence it cannot tip over the dam. If there were seepage under the dam, the main support force at the bottom of the dam would shift to the left of point C and might indeed cause the dam to tip over. The Fig. What horizontal force P is required at point B for equilibrium?
The gate weight of N is assumed at the centroid of the plate, with moment arm 0. If not, Fig. Neglect the atmosphere.
Because the actual plate force is not vertical. The hinge 15 cm hinge h is 15 cm from the centerline, as shown. Then the moment about the hinge is 0. Just iterate once or twice. What sphere diameter is just right to close the gate?
What is the water depth h which will first cause the gate to open? The hinge at A is 2 ft above the freshwater level. Find h when the gate opens. Solution: Find the force on each panel and set them equal: Fig. Compute a the hydrostatic force of the water on the panel; b its center of pressure; and c the moment of this force about point B. Neglect atmospheric pressure. CD is longer than AB, but its centroid is not as deep.
If you have a great insight, let me know. Neglecting atmospheric pressure, determine the lowest level h for which the gate will open.
Is your result independent of the liquid density? The width into the paper is b. Solution: The critical angle is when the hydrostatic force F causes a clockwise moment equal to the counterclockwise moment of the dam weight W. The negative sign occurs because the sign convention for dF was a downward force.
Determine the horizontal and vertical components of hydrostatic force against the dam and the point CP where the resultant strikes the dam. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs lbf. It is cm wide into the paper. Find a the vertical and b horizontal water forces on the panel.
This is a rectangle, 75 cm by cm, and its centroid is This is in two parts 1 the weight of the rectangular portion above the line AC; and 2 the little curvy piece above the parabola and below line AC.
Recall from Ex. Find the horizontal force P required to hold the gate stationary. The width b into the paper is 3 m. Neglect the weight of the gate. For the position shown, determine a the hydrostatic force on the gate per meter of width into the paper ; and b its line of action. Does the force pass through point O?
This force passes, as expected, right through point O. What is the force in each bolt required to hold the dome down? If the end caps are neglected, compute the force in each bolt.
Derive an analytic expression for the hydrodynamic force F on the shell and its line of action. Which force is larger? Their horizontal forces equal the force on the projected plane AB. If the panel is 4 m wide into the paper, estimate the resultant hydrostatic force of the water on the panel. Compute the a horizontal and 1 m B b vertical water forces on the curved panel AB. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force. Water Determine a the horizontal and b the vertical hydrostatic forces on the hemisphere, in lbf.
Compute the hydrostatic force on the conical surface ABC. Find a the hydrostatic force on the bottom; and b the force on a side panel. What is the specific gravity of fluid X? What is its weight in newtons? Was it gold? When released into the U. Standard Atmosphere, at what altitude will it settle? If so, how many cubic centimeters will be exposed? Therefore the sphere is floating exactly half in and half out of the water.
Therefore it floats in gasoline. What diameter spherical balloon will just support the weight? Determine a the string tension; and b the specific gravity of the wood. What is the specific gravity of the rod material? The rod is neutrally stable for any tilt angle! We see that the buoyancy exceeds the weight by 19 kN, or more than 2 tons.
The sphere would float nicely. Estimate a the tension in the mooring line, and b the height in the standard atmosphere to which the balloon will rise if the mooring line is cut. They estimate the weight of the box and treasure in air at lbf. Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pressure.
The empty balloon weighs lbf. The box is 2 ft wide, 5 ft long, and 18 in high. The box volume is 2ft 5ft 1. Thus the balloon must develop a net upward force of 1. The air weight in the balloon is negligible, but we can compute it anyway. Clean these two equations up: 1. Sum moments about point A: Fig. Then the vertical force balance on 7. What is the ratio of the distances a and b for this condition? If the total weight of barge and cargo is tons, what is the draft H of the barge when floating in seawater?
In terms of seawater, this total volume would be equivalent to If so, b how many cubic centimeters are exposed, and c how high will a spherical cap protrude above the surface? EES is recommended for the solution.
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