Download 1.2 Law of Friction

Law of friction
Properties of frictional force
1.
2.
act along the surface between two bodies
act in a direction so as to oppose relative
motion of the surfaces.
1
Limiting friction
Applied
force FA


Frictional
force f
If the block remains at rest,
f = FA
The frictional force to be
overcome before it moves
is called limiting friction.
2
Limiting friction
Applied
force FA
Frictional
force f
Limiting friction = 3 N
FA / N
f/N
0
0
1
1
2
2
3
3
4
3
5
3
3
Limiting friction
Applied
force FA
f/N
At rest
Moving
3
Frictional
force f
2
1
0
Limiting friction = 3 N
1 2 3 4 5
Static
friction
FA / N
Kinetic
friction
4
Limiting friction
f/N
At rest
Moving
•Kinetic / Dynamic friction is
the frictional force acting on
an object when it is moving.
3
2
1
0
1 2 3 4 5
Static
friction
FA / N
•Static friction is the frictional
force when the object is
stationary.
Kinetic
friction
5

Friction f
Static friction
Kinetic friction
From experiment, limiting
friction f is slightly greater
than kinetic friction f’.
i.e. f > f’
f
f’

Applied force F
Suppose f = 3.1 N, f’ = 3N
A force of 3.1 N is required to
make the block move.
But once the block is moving,
a force of 3 N only is required
to keep the block moving.
For simplicity, take f = f’
6
Law of friction
Normal
reaction N
Applied
force FA
Frictional force f
1.
2.
The limiting frictional force f is directly proportional to
the normal reaction N exerted by the surface.
i.e. f ∝ N or f = mN where m is a constant called
coefficient of limiting friction.
The kinetic frictional force f is directly proportional to
the normal reaction N exerted by the surface.
i.e. f’ ∝ N or f’ = m’N where m is a constant called
coefficient of kinetic friction.
For simplicity, take f = f’ = mN and m = m’
7
For simplicity, take f = f’ = mN and m = m’



If the surface is smooth, m = 0 ⇒ f = 0 N
Coefficient of friction depends on the two
contacting materials.
Rubber
Concrete (Dry)
m = 0.6 – 0.85
Rubber
Concrete (Wet)
m = 0.45 – 0.75
Frictional force does not depend on the
area of contact of the surfaces.
8
A block of mass 2 kg starts from rest, sliding down a
rough inclined plane making an angle of 30o with the
horizontal. Length of plane is 8 m. It takes 4 seconds to
reach the bottom. Find the coefficient of kinetic friction.





Solution:
t = 4 s, s = 8 m, u = 0 ms-1
By s = ut + ½ at2
8 = 0(4) + ½ a(4)2
a = 1 ms-2
mg sin 30o – f = ma
(2)(10)sin 30o – f = (2)(1)
f=8N
R = mg cos 30o = 17.32 N
By f = mR
8 = m(17.32)
m = 0.462
R
f
a
8m
mg
30o
9
For the following situation, find the least coefficient of
friction between the ground and the ladder.


smooth
S
N


60o
f A
250 N
N = 250 Newton
Take moment about A,
S x 4 sin 60o = 250 x 2 cos 60o
S = 72.17 Newton
f = S = 72.17 Newton
The ladder is about to slip,
f = mN
72.17 = m x 250
m = 0.289
10
How to find coefficient of limiting friction? p 22

The coefficient of limiting friction, m, can be found by
placing the block on a surface and tilting the latter to an
angle q at which the block is just about to slip.

N

f
q


mg

Along the plane:
f = mg sin q
--- (1)
Perpendicular to the plane:
N = mg cos q
--- (2)
Since the block is just about to slip,
f = mN
--- (3)
Sub (3) into (1):
mN = mg sin q
--- (4)
(4)/(2):
m = tan q
Hence, the coefficient of limiting friction is just tan q.
11