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Law of Friction: Surface Between Two Bodies Oppose Relative Motion

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Law of friction

Properties of frictional force


1. act along the surface between two bodies
2. act in a direction so as to oppose relative
motion of the surfaces.
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Limiting friction

Applied
 If the block remains at rest,
force FA f = FA
 The frictional force to be
overcome before it moves
Frictional is called limiting friction.
force f

2
Limiting friction
FA / N f/N

Applied
0 0
force FA 1 1
2 2
Frictional 3 3
force f
4 3

Limiting friction = 3 N 5 3

3
Limiting friction

Applied f/N
force FA At rest Moving
3
2
Frictional 1
force f
0 1 2 3 4 5 FA / N
Limiting friction = 3 N
Static Kinetic
friction friction
4
Limiting friction
f/N
At rest Moving
•Kinetic / Dynamic friction is
3 the frictional force acting on
2 an object when it is moving.
1 •Static friction is the frictional
force when the object is
0 1 2 3 4 5 FA / N stationary.
Static Kinetic
friction friction

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Friction f  From experiment, limiting
friction f is slightly greater
Static friction Kinetic friction than kinetic friction f’.

f i.e. f > f’
f’  Suppose f = 3.1 N, f’ = 3N
A force of 3.1 N is required to
make the block move.
Applied force F But once the block is moving,
a force of 3 N only is required
to keep the block moving.

For simplicity, take f = f’

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Law of friction Normal
reaction N

Applied
force FA
Frictional force f

1. 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.
2. 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’


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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.

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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 R
f a
By s = ut + ½ at2
8 = 0(4) + ½ a(4)2 8m
a = 1 ms-2
 mg sin 30o – f = ma
mg 30o
(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 9
For the following situation, find the least coefficient of
friction between the ground and the ladder.

 N = 250 Newton
 Take moment about A,
S S x 4 sin 60o = 250 x 2 cos 60o
smooth

S = 72.17 Newton
N  f = S = 72.17 Newton
 The ladder is about to slip,
60o f = mN
72.17 = m x 250
f A
m = 0.289
250 N

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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  Along the plane:
f = mg sin q --- (1)
 Perpendicular to the plane:
N = mg cos q --- (2)
f  Since the block is just about to slip,
f = mN --- (3)
q  Sub (3) into (1):
mN = mg sin q --- (4)
mg
 (4)/(2):
m = tan q
Hence, the coefficient of limiting friction is just tan q.
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