Examples of hydrostatics
The hydrostatic is the study pertaining to fluid mechanics, which analyzes the fluid balance, or in other words, study fluids at rest. It is the cohesion that fluids present that allows their molecules to be always very close.
The principle of hydrostatics , also known as hydraulics, indicates that, given a liquid, the pressure difference between two of its points is equal to the product of the specific weight of the liquid calculated by the difference in the levels.
This is the formula that represents such a principle:
There are two principles for calculating hydrostatics and its pressure:
Pascal’s principle . Point out that a liquid at rest exerts pressure at all points.
Archimedes principle . Indicates that increasing depth produces greater pressure that allows objects to float in the liquid.
In thermodynamic physics, it is said that hydrostatic pressure is that exerted by the same fluid on its weight when it is at rest. Among the concepts linked to this area is atmospheric pressure , which is that which the atmosphere exerts on the fluid.
Fluids, unlike liquids, have the special ability to change shape by increasing their volume due to increased stress , but their mass remains the same. While liquids have a constant volume according to temperature.
Another aspect to take into account about liquids is viscosity . It is the resistance that fluids exert against gradual deformations generated by two types of stresses: shear , which occurs tangentially to a given plane, and traction stress , which is caused by two internal forces placed in the direction otherwise, causing the fluid to stretch.
Hydrostatics is measured by the density (p), gravity (g) and depth (h) of the fluid, and not by its mass or volume.
In this way the formula to calculate the hydrostatic pressure is:
Examples of hydrostatics
Here are some examples of hydrostatic troubleshooting:
- A diver descends 10 meters deep into the sea. What is the pressure it is supporting, if the density of seawater is 1025 kg / m 3 ?
h = 10 m
dmar = 1025 kg / m 3
g = 9.8 m / s 2
According to the fundamental principle of hydrostatics, the pressure at a point in a fluid with density d, located at a depth h, is calculated using the following expression:
P = d⋅h⋅g
Substituting the data provided in the problem in the equation, the diver will be supporting:
P = 1025 kg / m 3 ⋅10 m⋅9.8 m / s2⇒P = 100450 Pa
- The hydraulic lift of a garage works by means of a hydraulic press connected to a water intake from the urban network that reaches the machine with a pressure of 5,105N / m². If the radius of the piston is 20 cm and the efficiency is 90%, determine the value in tons of the maximum load that the elevator can lift. According to Pascal’s principle:
p 1 = p 2
which for a hydraulic press becomes:
F 1 / S 1 = F 2 / S 2
In this case, the data that would correspond to the small piston of the press is provided in the form of pressure, with this we have:
p 1 = F 2 / S 2 or F 2 = p 1 / S 2
Since S2 = π.R² = 0.126 m²
F 2 = 5.10 5 .N / m² 0.126 m² = 6.3.10 4 N
As the efficiency is 90% (η = 0.9) the effective value of the maximum load will be:
Maximum F = 0.9.6.3.10 4 N = 5.67.10 4 N
One metric ton is equivalent to the weight of a body of mass 1000 kg, that is:
1t = 1000 kg. 9.8 m / s² = 9.8.10³ N ⇒ F maximum (t) = 5.67.10 4 N / 9.8.10³ N ≈ 5.8 t