Pressure and Buoyancy Problems

Let's state the two working equations we have so far.

The solutions to the problems below can be found at the end of this page. As always, try all the problems before looking at the solutions. It's much easier to understand a solution put before you than to come up with the solution yourself. To develop the skills necessary to solve the problems yourself, you must spend the time doing it.

Problems

1. What is the absolute pressure at the bottom of the Virgin Islands Basin (located between St. Thomas and St. Croix), at a depth of 4000 meters? Express your answer in atmospheres of pressure. What is the gauge pressure? If there are fish at this depth, how would they deal with this pressure? The density of sea water is 1.03 x 10³ kg/m³

2. A water hose is connected to a spigot located at the bottom of a cistern. The cistern is half full with 5 ft of water. The nozzle at the other end of the hose is turned off but is left down by the papaya tree, which is 20 ft below the bottom of the cistern. If the spigot is left open, what is the pressure at the nozzle? Why would it be a good idea to turn off the spigot when you are finished watering the tree?

3. A large part of Holland is below sea level. Earthen dikes keep the sea at bay. There's a Holland legend of a boy who uses his finger to plug a hole in the dike and saves the country side. Assume the hole is located 3.0 meters below the sea level. The hole is the same size as the childs finger, a diameter of about 1 cm. How much force would the child have to exert against the sea pressure in order to keep the sea at bay? Do you think a child could do this?

4. A 10 lb box falls overboard and is floating. The box has the shape of a cube, 1 ft on a side. What is the buoyancy force on the box?

5. The float in a toilet tank is a sphere of diameter 10 cm.

   1) What is the buoyancy force on the float when it is completely submerged? You might need a reminder that the volume of a sphere is V = ⁴/₃p(r)³

   2) Here's a slightly tougher one. If the float must have an upward buoyancy force of 3.0 N to shut off the ballcock valve, what percentage of the float will be submerged?

6. Here's an interesting puzzle to see if you really understand buoyancy and displacement. You are floating in a small dingy in your pool. There's a brick in the boat. You toss the brick out of the boat and into the pool. The brick sinks to the bottom of the pool. Does the water level at the side of the pool rise, stay the same, or decrease?
   

Solutions

Don't look at the answers until you've tried the problems on your own!!

1. Using the SI system of units, P = P_o + rgh = 1.01 x 10⁵ + 1.03 x 10³ x 9.8 x 4000 = 4.0 x 10⁷ Pa. In terms of atmospheres, that would be 3.9 x 10⁷ Pa / 1.01 x 10⁵ Pa/atm = 400 atmospheres! The gauge pressure is P - P_o which is just the rgh term. That would be about 399 atm. If fish lived at that depth, they would not notice the pressure anymore than we notice the 15 psi pressure pushing on us. Organism generally adapt to the pressure around them. The fish take water into their bodies at the ambient pressure so there is no net or gauge pressure difference. However, changing depth can present problems. Many sea mammals, such as sea lions, have developed systems that allow them to dive to extraordinary depths.

2. The nozzle end is 5 + 20 = 25 ft below the water level. We can convert this to meters and apply the static pressure equation in the SI units. But we could also use the fact that 34 ft of fresh water produces a pressure of 1 atmosphere = 14.7 psi. So 25 ft corresponds to 14.7 x 25/34 = 11 psi. Note that this is the gauge pressure, which is appropriate since atmospheric pressure act both on the surface of the water and on the hose. This means there will be a net force of 11 lb pushing outward on every square inch of the hose. It's probably best to turn the spigot off.

3. The gauge pressure would be 1.03 x 10³ x 9.8 x 3.0 = 3.1 x 10⁴ Pa. The force exerted by his "round" finger would be F = PA = 3.1 x 10⁴(p(.01/2)² = 2.4 N. This is about .53 lb ... no problem!

4. The info on the size of the box is not relevant. If the box is floating, then the buoyancy force must be equal to the weight of the box ... = 10 lb! Here's another problem to try. A cubic foot of water weighs about 64 lb. Can you see why the box would float with 10/64 ^th of its volume submerged? This would mean about 1.9 inches below the water.

5. The volume of the float is V = ⁴/₃p(.05)³ = 5.2 x 10^-4 m³. Assuming there is freshwater in your toilet tank, then F_buoyancy = 10³ x 9.8 x 5.2 x 10^-4 = 5.1 N.

   If you need 3.0 N of upward force to shut off the valve and there's 5.1 N of buoyancy force when completely submerged, then you would need 3.0 / 5.1 x 100% = 59% of the float to be submerged.

6. Did you figure this one out? The water level in the pool goes down! Some of our physics major get fooled by this one. While in the boat, the entire weight of the brick is being supported ... ultimately by water displaced by the dingy. Since the brick sinks when out of the boat, it must be more dense than water. Hence, the volume of water displaced is greater than the volume of the brick. But when the brick is tossed into the pool, it displaces only its own volume. OK, try again. What if the object tossed overboard floated?