Applications involving Friction

 Friction is a FORCE that opposes or impedes the motion of an object.
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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 the normal force (or weight) of the object.
AND on

 Friction between solid objects sliding against each other (or rolling) is called Kinetic Friction . (Greek for moving)
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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,
μ k or μ s
, for kinetic and static coefficients.
Friction depends very little on surface area.

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The friction force, F fr
, is always perpendicular to the normal force, F
N
.
In calculating kinetic friction:
F fr
  k
F
N
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.


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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.
F max
= μ s
F
N and since static friction can vary from
0 to max, we write
μ s is generally more than
F fr
  s
F
N
μ k as its harder to start an object than keep it moving.


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A 10.0kg box rests on the floor. μ s
= 0.4 and μ k
=0.3
Determine the force of friction, F fr
, acting on the box if the horizontal applied force, F
A
=
20N. If F
A
= 40N. Draw a free body diagram and label all forces.

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In the vertical direction there is no motion, so the net force, Σf y
= ma = 0 yields F
N
– mg = 0.
In all cases, the normal force, F
N
(10.0kg)(9.80m/s/s)= 98 N.
= mg =
 The force of static friction will oppose any applied force up to the maximum of
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F fr
= μ s
F
N
= (0.40)(98N)=39N
If F
A
= 20 N, the box won’t move so F balance the applied force.
fr
= -20N to

F
N
F f
F
A
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F g
If F
A
= 40N, which is more than the maximum static friction force, the box will accelerate and we have kinetic friction, F fr
= μ k
F
N
F fr
= (0.30)(98N) = 29 N.
ΣF x
= F
A
+ F fr
= 40 N + (-29N) = 11N, so the box will accelerate at a = F net
/ mass.
a = 11N / 10.0 kg = 1.1 m/s/s in a direction of the applied force. (positive horizontal).

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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)
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Recall how to obtain components of weight when not perpendicular to the surface?
F
N
F fr is always perpendicular to the surface and is always parallel to the surface here.

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Please do Chapter 4 Review p 107 #s 38 and
39
Please do Chapter 4 Review p 108 #s 40, 42,
43, 44, 46.