# Concept of Pressure

## The physics lesson, subject of pressure. The factors that pressure is depend on. The formula for pressure. The pressure of an object that on the inclined plane.

**What Is Pressure?**

The pressure is defined as the following.

The pressure is the force acting on the unit surface.

The pressure unit is Pascal (Pa) according to the International System of Units (SI).

1 Pa is a force of magnitude 1 Newton acting on a 1 m^{2}.

1 Pa = | 1 N | |

1 m^{2} |

According to the above formula, if the force exerted to the surface is kept constant and the surface area is increased, the pressure is reduced. If the force exerted to the surface is kept constant and the surface area is reduced, the pressure is increases.

If the surface area is kept constant and the exerted force is increased, the pressure is increases. If the surface area is kept constant and the exerted force is reduced, the pressure is decreases.

**Pressure In Solids**

The solids apply a force to the object they are its on. By dividing this force into the surface area, pressure is found.

**Example:**

The base Radius of the barrel which seen in the figüre is 50 cm, and its mass of 50 kg.

How much pressure is applied by the barrel to the floor.

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

**Solution:**

The bottom surface area of the barrel.

S = π•r^{2}

S = 3.14 • 0.5^{2}

S = 0.785 m^{2}

Now, let we find the force exerted to the floor by the barrel. This force is equal to the weight of the barrel.

F = 100 • 10

F = 1000 N

Let we find the pressure.

P = | F | |

A |

P = | 1000 | |

0.785 |

P = 1273.88 Pa

**Example:**

The pressure exerted to the floor by the barrel that seen in the figure is 1200 Pa.

The mass of the barrel is 1200 kg.

How many cm the radius of the barrel.

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

**Solution:**

P = | F | |

A |

F = 1200 • 10

F = 12000

A = π•r^{2}

A = 3.14 • r^{2}

1200 = | 12000 | |

3.14•r^{2} |

3.14•r^{2} = 10

r^{2} = 3.18

r = 1.78 m

r = 178 cm

**Example:**

There is two rectangular prisms in the figure the mass of the bottom X body is 400 kg. Its the base length is 160 cm, the base width is 50 cm.

The mass of the upper Y body is 300 kg. Its the length of the bottom surface is 80 cm, width is 50 cm.

The pressure exerted to the X body by the Y body is P1, and the pressure exerted to the floor by the X body is P2.

Calculate the P1 and P2 pressures.

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

**Solution:**

The pressure that the upper object applies to the lower object.

P1 = | Fy | |

Ay |

Fy = m•g

Fy = 300•10 = 3000 N

Ay = 0.8 • 0.5

Ay = 0.4 m^{2}

P1 = | 3000 | |

0.4 |

P1 = 7500 Pa

The pressure that the X object applies to the floor.

The pressure that X object applied to the floor is found by dividing the sum of the weight of the object X and the object Y into the surface area of X.

P2 = | Fx + Fy | |

Ax |

Fx = 400 • 10 = 4000 N

Ay = 1.6 • 0.5

Ay = 0,8 m^{2}

P2 = | 4000 + 3000 | |

0.8 |

P2 = | 7000 | |

0.8 |

P2 = 8750 Pa

**Pressure On the Inclined Plane**

The pressure of the solids that on the inclined plane is calculated by taking the normal force of the inclined plane.

The normal force is the force acting perpendicular to the surface of the inclined plane.

If the angle between the horizontal surface and the inclined plane is α, the force applied by the inclined surface is,

F = m•g•cosα

M: is mass

G: acceleration of gravity

If the bottom surface area the object is A,

P = | m•g•cosα | |

A |

**Example:**

The angle between the inclined surface in the figüre and horizontal surface is 33 degrees.

The mass of the K object that on the inclined plane is 60 kg, the base surface is 0.4 m^{2}.

Determine the pressure the K object applies to the inclined surface.

(g = 10 m/s^{2}, cos33 = 0.84)

**Solution:**

The weight of the K object.

G = m•g

G = 60 • 10

G = 600 N

The force that it weight applies to the inclined plane.

F = 600 • cos33°

F = 600 • 0.84

F = 504 N

P = | F | |

A |

P = | 504 | |

0.4 |

P = 1260 Pa

RISE KNOWLEDGE

01/08/2018

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