<< Chapter < Page | Chapter >> Page > |
where ${F}_{\text{B}}$ is the buoyant force and ${w}_{\text{fl}}$ is the weight of the fluid displaced by the object. Archimedes’ principle is valid in general, for any object in any fluid, whether partially or totally submerged.
According to this principle the buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is
where ${F}_{\text{B}}$ is the buoyant force and ${w}_{\text{fl}}$ is the weight of the fluid displaced by the object.
Humm … High-tech body swimsuits were introduced in 2008 in preparation for the Beijing Olympics. One concern (and international rule) was that these suits should not provide any buoyancy advantage. How do you think that this rule could be verified?
The density of aluminum foil is 2.7 times the density of water. Take a piece of foil, roll it up into a ball and drop it into water. Does it sink? Why or why not? Can you make it sink?
Drop a lump of clay in water. It will sink. Then mold the lump of clay into the shape of a boat, and it will float. Because of its shape, the boat displaces more water than the lump and experiences a greater buoyant force. The same is true of steel ships.
(a) Calculate the buoyant force on 10,000 metric tons $(1\text{.}\text{00}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{kg})$ of solid steel completely submerged in water, and compare this with the steel’s weight. (b) What is the maximum buoyant force that water could exert on this same steel if it were shaped into a boat that could displace $1\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}$ of water?
Strategy for (a)
To find the buoyant force, we must find the weight of water displaced. We can do this by using the densities of water and steel given in [link] . We note that, since the steel is completely submerged, its volume and the water’s volume are the same. Once we know the volume of water, we can find its mass and weight.
Solution for (a)
First, we use the definition of density $\rho =\frac{m}{V}$ to find the steel’s volume, and then we substitute values for mass and density. This gives
Because the steel is completely submerged, this is also the volume of water displaced, ${V}_{\text{w}}$ . We can now find the mass of water displaced from the relationship between its volume and density, both of which are known. This gives
By Archimedes’ principle, the weight of water displaced is ${m}_{\text{w}}g$ , so the buoyant force is
The steel’s weight is ${m}_{\text{w}}g=9\text{.}\text{80}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}$ , which is much greater than the buoyant force, so the steel will remain submerged. Note that the buoyant force is rounded to two digits because the density of steel is given to only two digits.
Strategy for (b)
Here we are given the maximum volume of water the steel boat can displace. The buoyant force is the weight of this volume of water.
Solution for (b)
The mass of water displaced is found from its relationship to density and volume, both of which are known. That is,
Notification Switch
Would you like to follow the 'Physics of the world around us' conversation and receive update notifications?