When you immerse an object like a rubber ducky in water, the object experiences an upward force equal to the weight of the water it displaced. This rule is called Archimedes' principle. Basically, this implies that objects less dense than water will float, while objects that are more dense than water will sink. You can apply Archimedes' principle and some simple math to solving many different kinds of density and buoyancy problems.
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Determine the density of the fluid in which the object is submerged. Typically, this information will be given to you in the problem unless the fluid is water, in which case it has a density of 1 gram per cubic centimeter at room temperature.

Determine the dimensions of the object and use these to set up an equation for its volume.
For example, suppose you are given the following problem: a flatbottomed, rectangular barge is floating in the water. It weighs 500 kilograms when empty; it is 10 meters long, 5 meters wide and 2 meters high from bottom to gunwale. How much cargo could you add to the boat before the water level rose above the gunwale? (Assume the water has the same density as pure water.)
The first thing you will notice is that the barge is rectangular  consequently, its volume must be length x width x height. The volume of the submerged portion will, therefore, be length x width x height submerged also. This can be abbreviated submerged volume = LWH, where H is height.

Multiply the density of the fluid by the volume of the submerged part of the object to find the mass of water displaced, then multiply the mass of water displaced by 9.81 m/s^2, the standard gravity.
Example continued: you know the submerged volume is as follows: Vs = LWH,
therefore, multiply this by the density of water. The density of water is roughly 1 gram per cubic centimeter. There are 100^3 cubic centimeters in a cubic meter, so the density of water is 100^3 grams per cubic meter or (100^3)/1000 kg per cubic meter = 1000 kg per cubic meter. The new equation, then, is as follows: mass of water displaced = LWH x 1000 kg/m^3; and weight of water displaced = LWH x 1000 kg/m^3 x 9.81 m/s^2.

Solve for any unknowns in your equation.
Example continued: in this example, you want to find out what the buoyant force will be when the height of the submerged part of the boat equals the boat's height. To do so, you plug L, W and H into the equation you found earlier, as follows: weight of water displaced = (10 m) x (5 m) x (2 m) x (1000 kg/m^3) x (9.81 m/s^2) = 981000 Newtons.
You now know that the mass of the boat at this point will be equal to this weight divided by the standard gravity constant, 9.81 m/s^2, so if 981000 N is divided by 9.81 m/s^2, you obtain the following: mass = 100,000 kg.
This is the maximum mass of the boat before its gunwale will be underwater. The boat has a mass of 500 kg when empty, so the maximum mass of the cargo will be 100,000  500 = 99,500 kilograms.

Remember two useful equations that you used to solve this problem. You can use these equations to solve other problems like this as well: weight of an object or displaced fluid = mass x 9.81 m/s^2; and buoyant force on object = weight of water it displaced = volume of water displaced x density of water = volume of object submerged x density of water.
Notice that as more and more of the object becomes submerged, the buoyant force on it increases. Once the buoyant force on it equals its weight, it will cease to sink or rise and remain level. If the object always weighs more than the water it displaces, however, it will sink straight to the bottom.
References
 "Physics for Scientists and Engineers"; Richard Wolfson, et al.; 1999
 University of Winnipeg: Buoyant Forces
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