CHAPTER 4 SEC 8Applications involving Friction

What is Friction? Friction is a FORCE that opposes or

impedes the motion of an object. Friction is caused by microscopic bumps

between solid objects in contact. Friction can exist from sliding objects or

rolling objects (though rolling friction is less than sliding).

Friction depends on the types of materials AND on the normal force (or weight) of the object.

Types of Friction Friction between solid objects sliding

against each other (or rolling) is called Kinetic Friction. (Greek for moving)

Friction between solid objects can exist parallel to the surface of them even when they are not moving. This is Static Friction.

Each type of material changes the amount of Friction present, so we have a coefficient of friction, μk or μs, for kinetic and static coefficients.

Friction depends very little on surface area.

Friction Coefficients

The friction force, Ffr, is always perpendicular to the normal force, FN.

In calculating kinetic friction: μ depends on whether the object is

wet or dry, what type of finish is on them, but NOT on speed of the objects sliding.

Ffr kFN

Static Friction Static friction is a force that exists

between objects that are in contact, but NOT moving when a force is applied.

Eventually with a hard enough push, the object will move and kinetic friction takes over as you exceed the MAXIMUM static friction.

Fmax = μsFN and since static friction can vary from 0 to max, we write

μs is generally more than μk as its harder to start an object than

keep it moving.

Ffr sFN

Example 1

A 10.0kg box rests on the floor. μs = 0.4 and μk =0.3

Determine the force of friction, Ffr, acting on the box if the horizontal applied force, FA = 20N. If FA = 40N. Draw a free body diagram and label all forces.

Solution for Example 1 In the vertical direction there is no

motion, so the net force, Σfy = ma = 0 yields FN – mg = 0.

In all cases, the normal force, FN = mg = (10.0kg)(9.80m/s/s)= 98 N.

The force of static friction will oppose any applied force up to the maximum of

Ffr = μsFN = (0.40)(98N)=39N If FA = 20 N, the box won’t move so Ffr

= -20N to balance the applied force.

Continued

If FA = 40N, which is more than the maximum static friction force, the box will accelerate and we have kinetic friction, Ffr = μkFN

Ffr = (0.30)(98N) = 29 N. ΣFx= FA + Ffr = 40 N + (-29N) = 11N,

so the box will accelerate at a = Fnet / mass.

a = 11N / 10.0 kg = 1.1 m/s/s in a direction of the applied force. (positive horizontal).

FAFf

Fg

FN

More Examples

Look at the Example 4-16 on page 99. (two boxes on a pulley)

Now look at the Example 4-17 on page 100. (skier on a hill)

Recall how to obtain components of weight when not perpendicular to the surface?

FN is always perpendicular to the surface and Ffr is always parallel to the surface here.

Your turn to Practice

Please do Chapter 4 Review p 107 #s 38 and 39

Please do Chapter 4 Review p 108 #s 40, 42, 43, 44, 46.