The floor under water is usually pretty light despite having hundreds of pounds of water above it, how is this possible?
In: Physics
Dirt, sand, rocks and organic matter is all porous so it lets water through. So the water that pushes down on the sediments is going straight through it creating pressure under it. This pushes up with an equal force that it is pushed down and therefore it does not get packed. It takes a lot of sediments and rock in order to be relatively watertight so the weight of the water is distributed over a lot of rock under the lake. So eventually it will compress and pack the material but not at the top.
The water typically goes much deeper than the bottom of the lake. So the rock and sediment at the bottom is just sitting on top of a mix of rock, sediment, and water. So the water is pushing on all sides of the material – not just on top – so the top layer is just suspended in the water, so it is easy for a fish or something to kick up sand or mud at the bottom of a lake.
Now, that material still has greater density than the water, which is why it sinks to the bottom. And as more layers build up on top of it, then those layers of other material start to pack it down. Eventually, those layers can end up deep underground (like 2,000 feet), and over a few million years the pressure of the ground above turns it into stone.
Lakes themselves are generally some of the briefest living geographical features. They may appear and disappear over 100,000 years, while it takes tens of millions of years to wear down a mountain.
Sediment is individual grains that touch each other but also with space between them, which in this case is 100% filled with water. Water pressure when there is no flow (hydrostatic) pushes equally in all directions (is isotropic), meaning the upward stress (force per unit area) from the water is the same as the downward stress at every location. The effective vertical stress that soil is under at the bottom of the lake is equal to the stress caused by the weight of the overburden minus the stress caused by pore pressure. In hydrostatic, drained (no excess pore water pressure) conditions, the pore pressure in any situation is always equal to the depth of the water multiplied by the density of water (1Mg/m³ – and 1Mg = 1000kg) multiplied by acceleration due to gravity (we geotechnical engineers tend to take 10m/s² as the value for this, even though it is not strictly accurate). The overburden pressure in general is calculated by taking the thickness of the overburden, multiplied by the density, multiplied by acceleration due to gravity. In this case the overburden is just water, so the density is 1Mg/m³. Therefore in a 20m deep lake, for example, we have an overburden pressure of 20x1x10=200kPa (kPa = kilopascal, units of stress. 10Pa = 10 Newtons per square meter. 10 Newtons = 1kg x acceleration due to gravity 10m/s²) and a pore pressure of 20x1x10=200kPa. The effective stress on the bed of the lake is 0, so the sediment is not compacted.
Lakes aren’t usually like a cup holding water. They are more like colander for spaghetti noodles with the water moving through them. A hole in the ground that reveals the water under the dirt and rock.
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