How do you calculate the density of an irregularly shaped object that floats in water?

Answer 1

If an object floats in water, we can immediately conclude that it is less dense than the water. So, we've already gained a bit of information. But can we learn more? Yes. We can further "ballpark" our estimate of the object's density through additional observation and deduction. About how much of the object is submerged? If, say, 75 percent of the object is under water, we can then say that its relative density -- that is, its specific gravity -- is about 0.75. In other words, it has a density of 0.75 grams per milliliter or, equivalently, 0.75 grams per cubic centimeter. (Note that the density of water is 1.00 gram per milliliter.) But can we do better? I think so. If we measure the volume of water displaced by the object when it is placed into the container of water, we can calculate the weight of the object, because its weight will be equal to the weight of the water it displaces. If the floating object displaces, say, 100 milliliters of water, then we know it weighs 100 grams, because, as noted above, the density of water is one gram per milliliter. But we're not done. To calculate an object's density, we must know its volume as well as its mass. From the measurement above, we know the object's weight , but we don't know its volume, mainly because of its irregular shape. But if we carefully push the object completely under water, it will displace an amount of water equal to its volume. Let's say that when we submerge the object fully, it displaces 130 milliliters of water. We therefore conclude that its volume is 130 milliliters, which is equal to 130 cubic centimeters. Since the object weighs 100 grams and has a volume of 130 cubic centimeters, its density is 100 grams/130 cubic centimeters = 0.769 g/cm3.