# Pascal’s Principle and Hydraulic Systems

## The physics lesson, subject of pressure. The definition and applications of the Pascal’s principle. The hydraulic system problems and its solutions.

Solid materials are transmit the force applied to them the same. But, they do not exactly transmit the pressure. The situation in liquids is the oppsite of the solids. The liquids transmit the pressure applied to them exactly. But may be not exactly transmit the force applied to them .

Pascal suggested that the liquids exactly transmit the pressure, do not exactly transmit the force.

Pascal explained "The Pascal’s Principle" which called in his own name.

**Pascal’s Principle**

The liquids in the closed container transmits the pressure applied to them to all portions of the fluid and to the walls of container they are in without any loss .

Using Pascal’s principleit is possible to obtain great forces from small forces. The hydraulic systems, cranes, car lifters are made using Pascal’s principle.

In this section the hydraulic systems will be examined. In the next section the U-tube will be examined.

**Hydraulic Systems**

These system basically have one small and one large cylinder. There is a liquid in these cylinders. There is a piston at the end of the cylinders. The force applied to the small piston is taken as a greater force from the big piston.

Te cause of this is that the liquids transmit the pressure in all directions.

The force applied to the small piston applies a pressure to the liquid. This pressure must be taken the same magnitude from second piston. Since the pressure force is multiplication of the pressure and the surface area, a larger force is obtained from the big piston.

**The Formula for Hydraulic Systems**

Let we find the formula of the hydraulic systems.

Supposei that the pressure applied to the small piston is P1, the pressure taken from the large piston is P2.

In the reason that the liquids transmitted the pressure in all directions equally,

P1 = P2

P1 = | F1 | |

A1 |

P2 = | F2 | |

A2 |

Because of P1 = P2

F1 |
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A1 |

The balance formula of the hydraulic system,

F1 |
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A1 |

**Example:**

In the system in the figure, the pistons are weightless.

The weight of the K object is 9000 N.

What is the magnitude of the F force.

**Solution:**

P1 = P2

P1 = | F1 | |

A |

P2 = | F2 | |

B |

F1 |
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A |

F1 |
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0.002 |

F1 |
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2 |

F1 = 3000 N

**Example:**

In the system in the figure, the pistons weights are insignificant. The surface area of piston in the "A" cylinder is 30 cm^{2}, the surface area of piston in the "B" cylinder is 90 cm^{2 }.

The difference of the liquid heights between the two cylinder is 30 cm^{2 }.

The magnitude of the F1 is 300 N.

The density of the liquid is 1200 kg/m^{3}.

How many kg the mass of the K object.

(g = 10 m/s^{2})

**Solution:**

In this system there is a difference in liquid heights between the two cylinder. The formula of hydraulic system should be arranged according to the point which the liquid height is equal.

In the question, the balance formula will be written according to dashed line.

At this point, the right cylinder pressure is the sum of the liquid pressure of height of 30 cm and the pressure of the K object.

F1 |
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A |

300 |
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30•10^{-4} |

10^{5} = | K | + 3600 |

90•10^{-4} |

100000 – 3600 = | K | |

90•10^{-4} |

96400 • 90•10^{-4} = K

K = 867.6

The weight of the K body is 867,6 N. The mass of the K body is,

m = 867.6/10

m = 86.76 kg

RISE KNOWLEDGE

06/08/2018

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