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1800-102-2727Have you ever thought how garage mechanics manage to transport unfathomably big cars or bicycles? They use hydraulic lifters, which can move objects that are too heavy for us to lift with just our hands but weigh several thousand kg. A piston applies pressure to the fluid, which is then evenly delivered. The way hydraulic brakes work is as follows. Have you ever seen a hydraulic press being used to bend or pierce metal objects at a steel mill? To evenly convey pressure to a die that compresses the metal that comes into contact, it is constructed of a piston and an oil-like fluid. Let’s learn more about this concept here!
Table of Contents
Pascal’s law states that the pressure applied to an incompressible, enclosed fluid is transmitted undiminished to every portion of the fluid, as well as to the walls of the container. Take the following example of a liquid filled in a container up to a height h upon which a lid of weight W and area of cross section A. A small hole is made at the top of the container. The liquid has a density . Now the liquid is acted upon by three different pressures
Fig shows three pressures experienced by the liquid
Now according to Pascal’s law, the pressure exerted at the bottom of the container can be written as
${P}_{bottom}={P}_{0}+\rho gh+\frac{W}{A}$
Tools and machinery classified as hydraulic employ fluid power to operate. Through tiny tubes and hoses, a significant quantity of fluid is transported in these machines. Here, fluid is moved throughout the machine to hydraulic cylinders and motors, where it is pressurised and then moved through tubes and control valves to the end effectors.
The hydraulic lift works on the principle of Pascal’s law. It consists of two pistons with different areas of cross sections A_{1} and A_{2}. The heavy object to be lifted is placed on the piston with bigger cross sectional area A_{2}. A force F1 is exerted on the piston having area A_{1}. Now,the pressure exerted on the first piston is ratio of the force to its cross sectional area .i.e.,${P}_{1}=\frac{{F}_{1}}{{A}_{1}}$
This pressure is equally transmitted in all directions towards the second piston where the car to be lifted is placed on. If P2 is the pressure on the second piston, then ${P}_{2}=\frac{{F}_{2}}{{A}_{2}}$
By pascal’s law, ${P}_{1}={P}_{2}\Rightarrow \frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}};{F}_{2}=\frac{{A}_{2}}{{A}_{1}}\times {F}_{1}.Since{A}_{2}>>{A}_{1};{F}_{2}{F}_{1}$
What does this mean? The hydraulic lift is a force multiplier. Whatever force we apply at the piston of the shorter cross section area gets multiplied; this enables us to lift heavy objects like trucks placed on the bigger piston.
Fig shows hydraulic lift having two pistons of areas of cross section A_{1} and A_{2 }(A_{2 }>>A_{1})
Hydraulic brakes also work on the principle of Pascal’s law. It consists of a master cylinder fitted with a pipeline that contains the fluid. When the driver manning the vehicle depresses the pedal, the oil flows into the second cylinder which presses the brake pads (also known as brake callipers) against the wheel.
Fig shows working of a hydraulic brake system. The master cylinder transmits pressure evenly so that it reaches the calliper
Q. A hydraulic machine has two pistons with diameters of 4 cm and 40 cm. On the smaller piston, a 100 kg wt weight is placed. What is the force applied to the larger piston?
A. Area of the smaller piston ${A}_{1}=\pi {{r}_{1}}^{2}=\pi {\left(2\times {10}^{-2}\right)}^{2}=\pi \left(4\times {10}^{-4}\right){m}^{2}$
Area of the larger piston ${A}_{2}=\pi {{r}_{2}}^{2}=\pi {\left(20\times {10}^{-2}\right)}^{2}=\pi \left(4\times {10}^{-2}\right){m}^{2}$
According to Pascal’s law;
Q. Two pistons of cross sectional areas 5 cm2 and 2 cm2 are fitted to both ends of a hydraulic lift. When a force of 20 N is applied on the smaller piston, how much force is experienced by the larger piston?
A. Here A_{2}=5 cm^{2}
${A}_{1}=2{cm}^{2};{F}_{1}=20N$
${F}_{2}=\frac{{A}_{2}}{{A}_{1}}\times {F}_{1}\Rightarrow \frac{5}{2}\times 20=50N$
Q. Water is filled in airtight container up to a height of 10 cm. What is the force experienced by the base of the container? (Take g=10 ms-2)
A. As we know that, total pressure = hg
Total pressure$=h\rho g=0.1\times 1000\times 10=1000Pa$
Q. A cylindrical container is filled with water up to height of 10 m upon which a lid of mass 750 Kg is kept. Find the force on the bottom surface if area of cross section is 2 m2?
A.
${P}_{bottom}={P}_{0}+h\rho g+\frac{mg}{A}$
${P}_{bottom}=1\times {10}^{5}+10\times 1000\times 10+\frac{750\times 10}{2}$
${P}_{bottom}=2\times {10}^{5}+3750$
${P}_{bottom}=203.75KPa$
$Force=Pressure\times Area$
$Force=203.75\times {10}^{3}\times 2$
$Force=4.075\times {10}^{5}N$
Q. Why is Pascal's law not applicable for compressible fluid?
A. Density of compressible fluid varies because of applied pressure, so Pascal’s law is not valid for compressible fluid.
Q. Is Pascal’s law valid for accelerated fluid?
A. Yes, Pascal’s law is applicable for accelerated fluid.
Q. A U-tube manometer has a liquid filled with density up to a height h. What is the pressure exerted by the liquid on the bottom of the manometer?
A. The pressure exerted by the liquid on the bottom of the manometer is P=hg.
Q. Will pressure at the same level always be the same?
A. Not always, pressure at the same level will be the same only if fluid is incompressible and static.