Coefficient Of Friction (COF) is a measure of the resistance between two surfaces. It can also be defined as the ratio of friction to a normal reaction. For example,

**COF = Friction / Normal Reaction**

Here,

**Friction**– when we push an object across a floor, it moves slower due to an opposing force called friction.**Normal Reaction**– when two bodies are in contact with each other, they exert a force on one another called a normal reaction.

**Properties of the coefficient of friction**

- We use the Greek symbol Mu (μ) to represent the COF.
- It is a unitless and dimensionless quantity.
- Its value depends on the material of the objects.
- An object can only slide if the coefficient of friction is greater than 1. Otherwise, it cannot slide.

## How to find the coefficient of friction?

Let a body be at rest on a horizontal table as shown in the figure below.

When a force **F** is applied on it horizontally, the body just begins to slide along the surface. From the laws of friction, the force of friction **F** is proportional to the normal reaction **R**.

That is,

Friction (F) ∝ Normal Reaction (R)

```
or, F ∝ R
or, F = μR
```

where **μ** is a constant of proportionality called the coefficient of friction between two surfaces in contact.

Therefore,

```
μ = F / R
```

### What are the different types of friction coefficient?

Based on the types of frictions, COF can be divided into two types

- Coefficient of Static Friction
- Coefficient of Kinetic Friction

## 1. Coefficient of Static Friction

We can define the coefficient of static friction as the ratio of limiting friction to normal reaction and its symbol is **μ**** _{s}**.

Hence, the formula of coefficient of static friction is

```
μ
```_{s} = Limiting Friction (F) / Normal Reaction (R)

Here, **Limiting Friction** is the maximum static friction at which one body can move over another body.

## 2. Coefficient of Kinetic Friction

We can define the coefficient of kinetic friction as the ratio of kinetic friction to normal reaction and its symbol is **μ**** _{k}**.

Hence, the formula of coefficient of kinetic friction is

```
μ
```_{k} = Kinetic Friction (F_{k}) / Normal Reaction (R)

Here, Kinetic Friction is the friction between bodies when a body is moving or rolling over another.

### Relation between the coefficient of kinetic and static frictions

Since limiting friction is the maximum friction, it is greater than the kinetic friction. Hence,

```
Limiting Friction (F) > Kinetic Friction (F
```_{k})
or, (F / R ) > (F_{k} / R )
or, μ_{s} > μ_{k}

Hence, the coefficient of limiting friction is always greater than that of kinetic friction.

### 1. How to find the coefficient of kinetic friction on an incline?

Suppose we have an inclined plane that makes 60 degrees with the horizontal plane and a box is sliding down with an acceleration of 5.62 m/s^{2}. What is the coefficient of kinetic friction between the box and the inclined surface?

Here,

- Angle of inclination (A) = 60 degrees
- Acceleration of the box (a) = 5.62 m/s
^{2} - Coefficient of friction (μ
_{k}) = ?

From the above image, the force down the inclined plane is

**Force = mgsinA – Frictional Force**

We know,

- Force = ma
- Frictional Force = μ
_{k}R(normal reaction)

Substituting the value

```
=> ma = mgsinA - μ
```_{k}R

Also, from the image, the value of **R** is **mgCosA**. So, our equation becomes

```
=> ma = mgsinA - μ
```_{k}(mgcosA)

Taking **mg** common

```
=> ma = mg (sinA - μ
```_{k}cosA)

removing **m** from both sides, we get

```
=> a = g (sinA - μ
```_{k}cosA)

substituting the value of **a**, **g** (acceleration due to gravity) and **A**

```
=> 5.62 = 9.8 (sin60 - μ
```_{k}cos60)
=> 5.62 / 9.8 = sin60 - μ_{k}cos60

we know the value of **sin60** is **0.87** and **cos60** is **0.5**

```
=> 0.57 = 0.87 - μ
```_{k}0.5
=> μ_{k}0.5 = 0.87 - 0.57
=> μ_{k} = 0.3 / 0.5
=> μ_{k} = 0.6

Hence, the coefficient of kinetic friction (sliding friction) is **0.6**.

### 2. What is μ_{k} between a block and tabletop?

Suppose a block of mass 6 kg is pushed across the tabletop by a force of 9 Newton. If the block is moving with a constant velocity of 1.8 m/s, find the coefficient of friction between the block and tabletop.

Here,

- Mass of the block (m) = 6 kg
- Force applied (F) = 9 Newton
- Constant speed of block (v) = 1.8 m/s
- μ
_{k}between block and tabletop = ?

As per our diagram, the force applied on the body is given by

```
Force Applied = ma + Frictional Force
F = F
```_{f} + ma

From our question, we know that the block is moving with constant velocity, hence acceleration (a) becomes **0**.

```
=> F = F
```_{f} + m * 0
=> F = F_{f}

We know that the value of frictional force is **μ _{k}R** (reaction force). Substituting the value

```
=> F = μ
```_{k}R
=> F = μ_{k} (mg)
=> μ_{k} = F / (mg)

Substituting the value of **m**, **g**, and **F**.

```
=> μ
```_{k} = 9 / (6 * 9.8)
=> μ_{k} = 9 / 58.8
=> μ_{k} = 0.15.

Hence, the coefficient of friction between the block and tabletop is **0.15**.

## Coefficient of Friction of Common Materials

Materials | Coefficient of static friction | Coefficient of kinetic friction |
---|---|---|

Steel on rod | 0.74 | 0.57 |

Aluminum on steel | 0.61 | 0.47 |

Copper on steel | 0.53 | 0.36 |

Brass on steel | 0.51 | 0.44 |

Copper on cast iron | 1.05 | 0.29 |

Glass on glass | 0.94 | 0.40 |

Copper on glass | 0.68 | 0.53 |

Rubber on concrete (dry) | 1.00 | 0.80 |

Rubber on concrete (wet) | 0.30 | 0.25 |

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