## Introduction

Friction exists when one surface is moving across the surface of another object. The direction of the force is opposite the direction of motion of the object. While sometimes the magnitude of friction can be determined from the other forces acting on an object, it can also be determined independently using the formula Ffriction = μ×Fnormal where μ represents a constant called the coefficient of friction. The coefficient of friction value is different for surfaces of different materials. Surfaces that are rough and more likely to induce friction have larger values for μ.

## Coefficient of Friction

The coefficient of friction is a constant that shows the relationship the frictional force and the two surfaces that are in contact with each other. The nature of the material of the surface affects the value for the coefficient of friction, μ. Typically the value for μ varies between 0 and 1, although a value of 0 can only be considered under ideal conditions, that is when surfaces are frictionless.

## The Distance Needed to Come to Rest

Imagine a hockey puck sliding across the ice. As smooth as the ice is, there is friction between the puck and the ice and eventually it will come to rest. The puck is decelerating, so there must be a net force and it is the unbalanced force of friction. Consider the following scenario: A 165 g puck leaves a player's stick with a velocity of 24.6 m/s. The frictional force between the puck and ice is 0.132 N. How long does it take for the puck to come to rest?

The frictional force in a situation like this is equal to the net force because there is no force acting on the puck to move it forward. Yes, the puck was shot from the hockey stick of a player, but once the puck leaves the stick, there is no longer a force on the puck. As a result, Newton's second law can be applied and substitutions for the force and mass can be made:

F = ma

0.132 = 0.165a

Note that when substituting for mass, kilograms must be used. The 165 g puck would be equal to 0.165 kg.

a = 0.800 m/s2

The acceleration can now be used in a kinematic formula to find the distance needed for the puck to come to rest. Recall the puck's initial velocity is 24.6 m/s and its final velocity will be 0 m/s. Since the puck is decelerating, the value for acceleration will be negative when a substitution is made.

v2 = u2 + 2as

02 = 24.62 + 2(-0.800)s

0 = 605 -1.60s

-605 = -1.60s

s = 378 m

## Inclined Plane

Problems involving inclined planes are a frequent occurrence in introductory physics courses. Below is a typical setup for an inclined plane problem. When friction is accounted for in an inclined plane, the either could be accelerating, traveling at a constant velocity, or at rest. In the first example, there is a net force while in the latter two the system remains in equilibrium. First, the equilibrium situation will be considered.

In the diagram above, the frictional force is labeled as f. Note that it is opposite in direction to mgsinθ, a component of the object's weight that is parallel to the surface of the incline. Since the frictional force is also equal to the product of the coefficient of friction and the normal force, the following derivation can be made, beginning with:

mgsinθ = μN

Since the normal force is equal to the component of weight perpendicular to the surface of the incline:

mgsinθ = μmgcosθ

Since mg is on both sides of the equation, they can be canceled out:

sinθ = μcosθ

Dividing each side by cosθ will leave μ on the right:

tanθ = μ

From this two conclusions can be made. First, for an object at equilibrium on an inclined plane, the frictional force is equal to the component of weight parallel to the surface of the object, mgsinθ. Second, the coefficient of friction μ is equal to the tangent on the angle of the incline.

If the object is accelerating down the incline, then mgsinθ is greater than the frictional force. The difference between these two is the net force on the object. Dividing the net force by the object's mass gives its accleration.