Bricks Underwater 2
Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The buoyant force on each can be calculated with the formula Fb = ρfVdg. None of these change significantly over a reasonable depth so the buoyant force remains constant. But what if this depth difference is enormous, say in the ocean? For each of the variables in this formula (plus the buoyant force itself), estimate if it...
Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The buoyant force on each can be calculated with the formula Fb = ρfVdg. None of these change significantly over a reasonable depth so the buoyant force remains constant. But what if this depth difference is enormous, say in the ocean? For each of the variables in this formula (plus the buoyant force itself), estimate if it...
A) increases a bit
A) increases a bit
B) remains exactly the same
B) remains exactly the same
C) decreases a bit
C) decreases a bit
Solution:
Solution:
Density of water increases a bit with depth. Liquids are generally assumed to be incompressible, but an increase in pressure will decrease the volume of a given mass of water, thus increasing its density. Another cause of increased density with depth is that the water generally gets colder as you go deeper in the ocean. This is called a thermocline.
Density of water increases a bit with depth. Liquids are generally assumed to be incompressible, but an increase in pressure will decrease the volume of a given mass of water, thus increasing its density. Another cause of increased density with depth is that the water generally gets colder as you go deeper in the ocean. This is called a thermocline.
Volume displaced will decrease because the brick itself will compress with increasing pressure. This effect will likely be larger than the change for density of water.
Volume displaced will decrease because the brick itself will compress with increasing pressure. This effect will likely be larger than the change for density of water.
The variable g will actually increase with depth. If the Earth was uniform density, then g would decrease with depth. But since the Earth's density increases with depth, there is a maximum g (about 10.5 m/s^2) near the boundary of the mantle and core.
The variable g will actually increase with depth. If the Earth was uniform density, then g would decrease with depth. But since the Earth's density increases with depth, there is a maximum g (about 10.5 m/s^2) near the boundary of the mantle and core.
For submarines (and probably the bricks), the buoyant force generally decreases with depth because the decrease in volume displaced is a greater percentage than the changes in density or g (submarines are more compressible than water). For this reason, a submerged submarine is generally in unstable equilibrium.
For submarines (and probably the bricks), the buoyant force generally decreases with depth because the decrease in volume displaced is a greater percentage than the changes in density or g (submarines are more compressible than water). For this reason, a submerged submarine is generally in unstable equilibrium.
It should also be noted that since g increases with depth, the weight force increases with depth. So with a decreasing buoyant force and an increasing weight force, the force needed to hold the brick in place increases.
It should also be noted that since g increases with depth, the weight force increases with depth. So with a decreasing buoyant force and an increasing weight force, the force needed to hold the brick in place increases.