Chapter 4 Sec 8
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Transcript of Chapter 4 Sec 8
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.