A small square plank of oak floats in a beaker half full of water. The piece of

Question

A small square plank of oak floats in a beaker half full of water. The piece of oak is 10 cm on a side and 2 cm thick and floats on its side as shown in Figure P10.80

The Center of Mass with respect to the waters surface is .500 cm


A very light vegetable oil is poured slowly into the beaker, so that the oil floats on the water without mixing and until the oak plank is a few centimeters below the surface of the oil. If the density of the oil is 614 kg/m3, calculate the new position of the center of mass of the plank with respect to the water’s surface

Explanation / Answer

The buoyant force on the plank is equal to the weight of whatever it is displacing. This is called the archimedes principle. You can calculate the weight of the piece of oak by calculating the weight of the water it displaces. Buoyant force = weight of water displaced = volume of oak under water * density of water = weight of the oak Using the same principle Buoyant force = weight of water displaced + weight of oil displaced = volume of oak in water * density of water + volume of oak in oil * density of oil = weight of oak If you provide Fig. P10.80 I can help you more.

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