# Buoyant Force In Liquids

## Physics lesson, subject of buoyancy force. Archimedes’ Principle. Floating and sinking objects in a liquid. The formula for buoyant force. Subject expression and solved questions.

The materials may be in solid, liquid or gaseous phase.

There is no bond that binding the moleculaes each other, in gas phase. In gas phase, molecules moves freely.

In solid phase there are strong bonds that binding the molecules each other. The molecules of the solid objects does only vibration. Therefore, solid materials have a certain shape.

In liquid phase, the bonds that binding the molecules each other is weak. That is, they are moves at a limited area. The liquid molecules not fly and not binding firmly.

This bond structure imparts fluid property to matter.

Because of this properties the liquids take the shape of the container that they are in.

When a solid materials is put on surface of a liquid, the solid material is exert a force downward to the liquid. Whereas liquid matter, too exert an upward force to the solid matter (Law of action and reaction).

In the interaction of this opposing forces, the force that the greater density material applies is greater.

If the density of the solid is greater than the density of the liquid, the solid sinks to the bottom of the liquid.

If the density of the liquid is greater than the density of the solid, the solid floats on the liquid.

If the density of the liquid is equal the density of the solid, the solid not move upward or downward. It stays where you put in the liquid. However, usually all volume of the solid is in the liquid.

**Archimedes’ Principle**

Any liquid applies an upward force to any object that fully or partially in the liquid. The magnitude of this force applied is equal the weight of the liquid displaced by the object.

The upward force applied by a liquşd to an object is called the buoyant force.

Also Archimedes’ law applies for gases, too.

According to this principle, ,if we can find amount of liquid that displaced by the object, we can find the buoyant force.

**Calculation Of the Buoyancy Force**

The mathematical formula of Archimedes’ law is given in the following.

F = V_{s}. ρ_{l} . g

F: Buoyant force

V_{s}: The volume of the part of the object in the liquid

ρ_{l}: Density of the liquid

g: Acceleration of gravity

**Floating Objects On Liquid**

In this case, density of the body is smaller than density of the liquid.

Weight of the liquid that displaced by object is equal to weight of the object. The amount of liquid that displaced by the object is equal to the volume of the object’s part in the liquid.

V_{s} |
| |||||

V |

V_{s}: Volume of the sinking part of the object

V: Total volume of the object

ρ_{object}: Density of the object

ρ_{liquid}: Density of the liquid

For the object in the above figure,

1 |
| |||||

4 |

ρ_{l} = 4ρ_{o}

**Sinking Objects**

In this case, the density of the object is greater than the density of the liquid. The volume of liquid that displaced by the object is equal to volume of the object.

The weight of the liquid that displaced by the object is smaller than the weight of the object. The weight of the liquid that displaced by the object is equal to the buoyant force.

**The Objects that Stays Hanging in Liquid**

In this case, the density of the object is equal to the density of the liquid.

The weight of the object = The buoyancy force

The density of the object = the density of the liquid

The volume of the displaced liquid = the volume of the object

**Example:**

Volume of the segments of the X, Y, and Z objects are equal.

These objects are in balance in the liquid.

Compare density of these objects.

**Solution:**

Suppose that density of the X liquid is ρ.

For X object.

V_{s} |
| |||||

V |

2 |
| |||||

3 |

ρx = | 2ρ | |

3 |

For Y object,

1 |
| |||||

2 |

ρ_{Y} = | ρ | |

2 |

For z object,

3 |
| |||||

3 |

ρ_{z} = ρ

According to this, relationship exists between density of the X, Y and Z object as follows.

ρ_{z} > ρ_{Y} > ρ_{x}

**Example:**

In the figure above, the segments of the X are equal. The X object is in balance. The volume of the segments of the X object are 200 cm3.

The buoyance force that the liquid applies to upon the X object is 1200 N.

How many g/cm^{3} the density of the X object.

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

**Solution:**

The buoyant force that the liquid applies to upon the X object is found as follows.

F = V_{s}. ρ_{l}. g

Let we use this formula.

1200 = 0,2 . ρ_{l} . 10

ρ_{l} = 600 kg/m^{3}

Force of 1200 N applied to the object X is equal the weight of the object X.

W_{x} = V_{x} . ρ_{x} . g

1200 = 0,6 . ρ_{x} . 10

ρ_{x} = 200 kg/m^{3}

Practically, If 1/ 3 of the volume of the object is in the liquid, the density of the object is 1 / 3 of the density of the liquid.

**Buoyant Force Questions with its Solutions**

**Pascal's Principle And Hydraulic Systems**

RISE KNOWLEDGE

20/08/2018

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