Basic theoretical information
Pressure. Pascal’s law. Hydrostatic pressure
The main difference of liquids from solid (elastic) bodies is the ability to easily change their shape. Parts of the liquid can move freely, moving relative to each other. Therefore, the liquid takes the form of a vessel in which it is poured. In a liquid, as in a gaseous medium, solids can be immersed. Unlike gases, liquids are practically incompressible. A body immersed in a liquid or gas is affected by forces distributed over the surface of the body. To describe such distributed forces, a new physical quantity, pressure, is introduced into hydrostatics .
Pressure is defined as the ratio of the modulus of force F , acting perpendicular to the surface, to the area S of this surface:
If the force is directed at a certain angle to the perpendicular to the site , then the pressure created by this force is found by the formula:
In the SI system, pressure is measured in pascals (Pa): 1 Pa = 1 N / m2. Non-systemic units are often used: normal atmospheric pressure (atm) and pressure of one millimeter of mercury (mm Hg):
1 atm = 101325 Pa = 760 mm Hg
Pascal’s law: the pressure exerted on the liquid (or, by the way, gas) is transmitted to any point of this liquid without changes in all directions.
The fluid pressure at the bottom or side walls of the vessel depends on the height of the liquid column above the point at which pressure is measured. Hydrostatic pressure of the liquid column is calculated by the formula:
Please note that the pressure exerted in no way depends on the shape of the vessel, but depends only on the type of liquid (ie its density) and the height of the column of this liquid. The same pressure at depth h in accordance with Pascal’s law also exerts a liquid on the side walls of the vessel.
So, if the hydrostatic problem is about the pressure of a liquid column on a side face at some particular point, then this pressure is given by the previous formula, where h is the distance from this point to the surface of the liquid. But sometimes in hydrostatic problems it is necessary to calculate the average pressure on the entire lateral surface of the vessel. In this case, apply the formula:
In this case, h is the total height of the liquid column in the vessel.
If the fluid is in the cylinder under the piston, then acting on the piston with some external force F , you can create an additional pressure p 0 = F / S in the fluid , where: S is the piston area. Thus, the total pressure in a fluid at a depth h can be written as:
If the piston is removed, the pressure on the surface of the liquid will be equal to atmospheric pressure. If we dive into the water, then the pressure at a certain depth will also consist of two pressures – the pressure of the atmosphere and the pressure of the water column (which is determined by the depth of the dive).
Communicating call receptacles having between them a channel filled with liquid. Observations show that in communicating vessels of any shape a homogeneous liquid is always set at the same level. tasks for communicating vessels are very common in hydrostatics.
Different liquids behave differently even in communicating vessels of the same shape and size. The fact is that in communicating vessels the same pressure should be set at the same height in all parts of the vessel. But if the fluids are different, then the height of the pillars of these fluids must be different in order to create the same pressure. Therefore, dissimilar liquids in communicating vessels may not be installed at the same level.
Algorithm for solving problems of hydrostatics on communicating vessels:
- Make a drawing.
- Choose a horizontal level, below which all the vessels have the same liquid. If there is no such level, then, naturally, for the zero level we select the bottom of the vessels.
- Record the pressure relative to this level in all vessels and equate.
- If necessary, use the incompressibility property of a liquid (the volume of liquid flowing from one vessel is equal to the volume of liquid flowing into another vessel).
- Solve mathematically derived system of equations.
If both vertically arranged cylinders of communicating vessels are closed with pistons, then using external forces applied to the pistons, a large pressure p can be created in the fluid , many times greater than the hydrostatic pressure ρgh at any point in the system. Then we can assume that the same pressure is set in the entire system p (according to Pascal’s law). If the pistons have different areas S 1 and S 2 , then different forces act on them from the fluid side F 1 = pS 1 and F 2 = pS 2. The same in magnitude, but oppositely directed external forces must be applied to the pistons to keep the system in balance. Thus, for a hydraulic press we have the formula:
This ratio follows from the equality of pressures and is performed only in an ideal hydraulic press , i.e. such in which there is no friction. If S 2 >> S 1 , then F 2 >> F 1 . Devices in which these conditions are met are called hydraulic presses (machines, jacks). They allow you to get a significant gain in strength. If the piston in a narrow cylinder is moved down by an external force F 1 to the distance h 1, then the piston in the wide cylinder will move to the distance h 2, which can be found from the relationship:
This ratio follows from the equality of volumes and is carried out in any hydraulic press. This expression is obtained because when the piston is moved, the same volume of fluid moves, that is, how much fluid has gone from one cylinder, as many have come to the second, or V 1 = V 2 . Thus, gain in strength is necessarily accompanied by the same loss in distance. At the same time, the product of force by distance remains unchanged:
The latter formula follows from the equality of work and is performed only for ideal machines in which friction forces do not act. Thus, in the hydraulic press everything happens in full accordance with the “golden rule of mechanics”: how many times do we win in strength, how many times do we lose in distance. In this case, no machine can give a gain in work.
Since the hydraulic press is a mechanism, its work can be characterized by efficiency (efficiency). The efficiency of a hydraulic press in hydrostatic problems is calculated using the following formula:
where: A floor = F 2 h 2 – useful work (work on lifting the load), And cost = F 1 h 1 – the work expended. In most tasks, the efficiency of a hydraulic press is taken as 100%. Efficiency is calculated when it comes to a non-ideal hydraulic press.
We emphasize once again that for a non-ideal hydraulic press, only the ratio that follows from the equality of the volumes of the displaced fluid is fulfilled, and the efficiency is calculated for such presses. The remaining relationships from this section are performed only for an ideal hydraulic press.
Archimedes Law. Body weight in fluid
Because of the pressure difference in the fluid at different levels, a pushing or Archimedean force arises, which is calculated by the formula:
where: V is the volume of the fluid displaced by the body, or the volume of the body part immersed in the fluid, ρ is the density of the fluid into which the body is immersed, and therefore, ρV is the mass of the displaced fluid.
The Archimedean force acting on a body immersed in a liquid (or gas) is equal to the weight of the liquid (or gas) displaced by the body. This statement, called the law of Archimedes , applies to bodies of any form.
In this case, the weight of the body (i.e. the force with which the body acts on the support or suspension) immersed in the liquid decreases. If we assume that the weight of the body at rest in the air is mg , which is exactly what we will do in most tasks (although generally speaking the body also has a very small Archimedes force from the atmosphere, because the body is immersed in gas from the atmosphere), for body weight in liquid, you can easily derive the following important formula:
This formula can be used in solving a large number of problems. It can be remembered. With the help of Archimedes’ law, not only navigation, but also aeronautics is carried out. It follows from Archimedes’ law that if the average density of a body ρ t is greater than the density of a liquid (or gas) ρ (or in another way mg > F A ), the body will sink to the bottom. If ρ t < ρ (or in another mg < F A), the body will float on the surface of the fluid. The volume of the submerged part of the body will be such that the weight of the displaced fluid is equal to the weight of the body. To lift a balloon in the air, its weight must be less than the weight of the displaced air. Therefore, the balloons are filled with light gases (hydrogen, helium) or heated air.
If the body is on the surface of the fluid (floats), then only two forces act on it (Archimedes up and gravities down), which balance each other. If the body is immersed in only one liquid, then by writing Newton’s second law for such a case and performing simple mathematical operations, we can obtain the following expression relating the volumes and densities:
where: V loader is the volume of the submerged part of the body, V is the total volume of the body. With this ratio, most of the problems of swimming bodies are easily solved.