When objects are submerged in a fluid one of three outcomes is expected:
 the object will float with a certain percentage of its volume above the surface of the fluid,
 the object will float in neutral equilibrium with all of its volume entirely submerged, or
 the object will sink and come to rest on the bottom of the container.
These outcomes are based on hydrostatic pressure and Archimedes' Principle. Hydrostatic Pressure and Archimedes' Principle
Hydrostatic pressure is the pressure, or force per unit area (P = F/A), an object experiences as a result of the weight of the water column above its position. It can be calculated with the formula, P = ρgh where

ρ is the density of water (1 gram/cm^{3} = 1000 kg/m^{3}),

g is the gravitational field strength (980 dynes/gram = 9.8 N/kg), and

h is the height of the water column (in cm or in meters).


In the diagram shown below we can see that symmetry results in the cancellation of the forces acting on the four lateral faces of the block since the pressures in those regions are equal. However, the forces acting on the top of the block are smaller than the forces acting on its bottom. We will now show that this force differential is the buoyant force.
Since the area of the top and bottom faces are equal we can factor out A from our expression.
Our derivation has shown us that ΔF, or the buoyant force supplied by the water, is equal to the weight of the displaced water. This statement is known as Archimedes' Principle.

Refer to the following information for the next four questions.
Suppose a specimen, when weighted in air, has a weight of 1.75 N. When the specimen is placed on the end of a spring scale and completely submerged in a beaker of water, it weighs 1.58 N.

Experimentation shows that a submerged, nonporous object will displace an amount of fluid equal to its volume. Thus if the volume of water in a graduated cylinder rises from 74.5 ml to 78.0 ml with the addition of an object that sinks to the bottom of the cylinder, then we know that the submerged object has a volume of V = 78.0  74.5 or 3.5 cm^{3}. Note that we can change the unit from ml to cm^{3} since 1 ml of water occupies 1 cm^{3} of space.
Even if an object floats the buoyant force is still equal to the weight of the water that it displaces. It is just not necessary for the object to displace a volume of water equal to its entire volume. In fact, it turns out that the ratio of the relative volumes is equal to the ratio of the relative densities.
Which of our original three outcomes occur depends on the relative densities of the liquid and the submerged object.
If the density of the object is less than the density of the fluid, the object will rise, or float. If the density of the object is greater than the density of the fluid, the object will sink.
If the density of the object is exactly equal to the density of the fluid, then the object, when submerged, will be in neutral equilibrium. This is the desired result for scuba divers when they add weights to their belts so that they can swim at any desired depth without having to fight hydrostatic forces that are trying to force them to rise or sink. It is also the ultimate goal of our lab.

Refer to the following information for the next two questions.
Suppose a floating block of wood, having a volume of 50 cm^{3}, displaces only 30 cm^{3} of water.

Equipment
 one dry film canister
 20 dry pennies
 one triple beam balance
 24 books
 2 metersticks
 hangers for canister
 1000 ml glass beaker with water
 pipette
 towels
Procedure
Your first task is to use Archimedes' Principle to determine the volume of the film canister.
 Zero your triple beam balance.
 Mass the dry film canister with 20 dry pennies inside.
 Fill the beaker with approximately 500 ml of water.
 Carefully place your triple beam balance on top of the two metersticks so that they spans the gap between two equal stacks of books which are high enough for the beaker to sit under the balance's pan. Secure the balance so that it does not slip. Zero the balance one more time.
 Gently suspend the wire hangers from the bottom of your triple beam balance. Ask for help if you are uncertain how to attach the wires.
 Submerge the bottom "wire basket" in the beaker of water and find its "tare mass."
 After making sure that the lid is snapped on securely, place your loaded film canister in the "wire basket" and completely submerge it into the water. Read the canister's "apparent mass." Remember to subtract the wire basket's tare mass.


Neutral Equilibrum
Your second task is to exactly match the density of the canister to that of water. You may remove and add pennies, and fine tune the mass by using the pipette to slowly add water to the inside of the film canister.
Once you are certain that your canister is ready, bring it up to the front of the room so it can timed in the fish tank. Each group is allowed two trials to release their canister so that it neither breaks the surface or sinks to the bottom of the tank in less than 3 seconds. You are NOT allowed to change its mass between trials. Be careful of air bubbles!


When you are done, leave your film canisters and wet pennies at the front on the room next to the fish tank. Return to your lab stations, gently remove the wire hangers from your triple beam balance, and carefully place the balance and the wires on the surface of the table so that the next period's group can begin.
There is no written lab report to turn in to the oneway box. However, make sure that you submit the lab's online form and get your lab sheets signed/dated for your notebooks.