Friction is unavoidable in our day-to-day lives. Man keeps trying so many techniques to reduce friction in any process, but he also realises that without friction none of his work would have been possible. Life would have been impossible without the aid of friction. Yet friction is termed as an evil. Without friction one can't walk, write, sit on a chair without slipping off or for that matter, even properly hold on to his or her morning cup of tea.

Friction is the retarding force coming into play when two bodies are in contact with each other.

Friction formula in general is given by

_{n} is the magnitude of normal force.

The Normal force = Weight of the given body. Hence Normal force F_{n} is given by

g is the gravity.

Friction formula is used to calculate the friction between the any two given bodies.

Friction formula in general is given by

Where,

F_{f} is the magnitude of friction,

F

$\mu$ is the coefficient of friction and

FThe Normal force = Weight of the given body. Hence Normal force F

Where,

m is the mass of the body and

m is the mass of the body and

Friction formula is used to calculate the friction between the any two given bodies.

Related Calculators | |

Friction Calculator | Calculate Kinetic Friction |

Calculate Static Friction | Friction Loss Calculator |

Below are problems on friction which helps to know where the formula can be used:

### Solved Examples

**Question 1: **A boy is pulling box of mass 10 Kg. What is the normal force acting and also find the frictional force if the coefficient of friction $\mu$ is 0.3?

** Solution: **

Given: Mass, m = 10 Kg,

Normal force, F_{n} =?

Normal force is given by F_{n} = mg

= 10 Kg $\times$ 9.8 m/s^{2}

= 98 N.

The Frictional force is given by F_{f }= $\mu$ F_{n}

= 0.3 $\times$ 98 N.

= 29.4 N.

**Question 2: **Nancy is of mass 40 Kg is sliding on the ice. If the coefficient of friction acting is 0.45. Find the frictional force acting between her and ice layer?

** Solution: **

Given: Mass m = 40 Kg,

Coefficient of friction $\mu$ = 0.3,

The Normal force is given by F_{n} = mg

= 40 Kg $\times$ 9.8 m/s^{2}

= 392 N.

The Frictional force is given by F_{f} = $\mu$ F_{n}

= 0.45 $\times$ 392 N

= 176.4 N.

Given: Mass, m = 10 Kg,

Normal force, F

Normal force is given by F

= 10 Kg $\times$ 9.8 m/s

= 98 N.

The Frictional force is given by F

= 0.3 $\times$ 98 N.

= 29.4 N.

Given: Mass m = 40 Kg,

Coefficient of friction $\mu$ = 0.3,

The Normal force is given by F

= 40 Kg $\times$ 9.8 m/s

= 392 N.

The Frictional force is given by F

= 0.45 $\times$ 392 N

= 176.4 N.

More topics in Friction Formula | |

Kinetic Friction Formula | Static Friction Formula |

Friction Loss Formula | Friction Force Formula |