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P3.2.3 Hydraulics and Pressure A hydraulic system uses pressure in a liquid to transfer force from one place to another to do mechanical work. You must.

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Presentation on theme: "P3.2.3 Hydraulics and Pressure A hydraulic system uses pressure in a liquid to transfer force from one place to another to do mechanical work. You must."— Presentation transcript:

1 P3.2.3 Hydraulics and Pressure A hydraulic system uses pressure in a liquid to transfer force from one place to another to do mechanical work. You must assume that a)The liquid is INCOMPRESSIBLE : it is not elastic: you can’t squash it. b)Pressure acts in all directions within the liquid (Note: similar systems use air pressure: they are described as PNEUMATIC but air is compressible) ‘Candidates should understand that this means that a force exerted at one point on a liquid will be transmitted to other points in the liquid.’ The pressure in different parts of a hydraulic system is given by: Pressure = Force or P = F/A Area P is measured in Pascals (Pa), F is the force in Newtons, N A is the cross-sectional area in metres squared, m 2

2 P3.2.3 Hydraulics Simple examples of the pressure idea include the difference in pressure when the SAME FORCE is applied through different AREAS. Try these examples yourself (you may need a calculator): 1) An office safe has a weight of 500N. If the area of the base is 1.25 square metres, what is the pressure on the floor of the office? 2) A physics teacher has a weight of 800N. If his feet have an area of 0.025 square metres each, what pressure does he exert on the ground? (Remember, he has two feet!)

3 P3.2.3 Hydraulics The use of different cross-sectional areas on the effort and load side of a hydraulic system enables the system to be used as a force multiplier. Example calculation: If the FORCE IN is 20N and the area of cylinder A is 0.02m 2 then the PRESSURE in the HYDRAULIC LIQUID is P = F/A = 20/0.02 = 1000 Pa The pressure all the way to cylinder B is THE SAME. If the area of cylinder B is 0.08m 2 then the FORCE OUT = P x A = 1000 x 0.08 = 80N So the system has multiplied the force by a factor of 4

4 P3.2.3 Hydraulics The use of different cross-sectional areas on the effort and load side of a hydraulic system enables the system to be used as a force multiplier. This principle is used in the braking systems of vehicles. http://www.darvill.clara.net/enforcemot/pressure.htm

5 P3.2.3 Hydraulics Hydraulic disc brakes use the pressure of the fluid (light blue) to push a piston towards the brake rotor. The piston pushes brake pads onto the disc so friction acts between the pads and the rotor or disc. http://www.darvill.clara.net/enforcemot/pressure.htm

6 P3.2.3 Hydraulics The use of different cross-sectional areas on the effort and load side of a hydraulic system enables the system to be used as a force multiplier. http://www.darvill.clara.net/enforcemot/pressure.htm On a disk brake, the fluid from the master cylinder is forced into a caliper where it presses against a piston. The piston, in-turn, squeezes two brake pads against the disk (rotor), which is attached to the wheel, forcing it to slow down or stop.disk brakerotor This process is similar to a bicycle brake where two rubber pads rub against the wheel rim creating friction.

7 P3.2.3 Hydraulics The use of different cross-sectional areas on the effort and load side of a hydraulic system enables the system to be used as a force multiplier. http://www.darvill.clara.net/enforcemot/pressure.htm

8 P3.2.3 Hydraulics The use of different cross-sectional areas on the effort and load side of a hydraulic system enables the system to be used as a force multiplier. http://www.darvill.clara.net/enforcemot/pressure.htm


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