law of friction
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Law of friction
Properties of frictional force1. act along the surface between two bodiessurface between two bodies2. act in a direction so as to oppose relative oppose relative motionmotion of the surfaces.
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Limiting friction
If the block remains at rest,f = FA
The frictional force to be overcome before it moves is called limiting friction. Frictional
force f
Applied force FA
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Limiting friction
Frictional force f
Applied force FA
FA / N f / N
0 0
1 1
2 2
3 3
4 3
5 3Limiting friction = 3 NLimiting friction = 3 N
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Limiting friction
Frictional force f
Applied force FA
Limiting friction = 3 NLimiting friction = 3 NFA / N
f / N
0 1 2 3 4 5
1233
At rest Moving
Static Kineticfriction friction
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Limiting friction
FA / N
f / N
0 1 2 3 4 5
1233
At rest Moving
Static Kineticfriction friction
•Kinetic / Dynamic friction is the frictional force acting on an object when it is moving.
•Static friction is the frictional force when the object is stationary.
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From experiment, limiting friction f is slightly greater than kinetic friction f’. i.e. f > f’
Suppose f = 3.1 N, f’ = 3NA 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.
Friction f
Applied force F
Static friction Kinetic friction
f’f
For simplicity, take f = f’
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Law of friction
1. The limiting frictional force f is directly proportional to the normal reaction N exerted by the surface. i.e. f ∝ N or f = N where 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 ’ = ’N where is a constant called
coefficient of kinetic friction.
Frictional force f
Applied force FA
Normal reaction N
For simplicity, take f = f’ = N and = ’
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If the surface is smooth, = 0 ⇒ f = 0 N Coefficient of friction depends on the two
contacting materials.
For simplicity, take f = f’ = N and = ’
Rubber Concrete (Dry) 0.6 – 0.85
Rubber Concrete (Wet) 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: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 = 8 N
R = mg cos 30o = 17.32 N By f = R
8 = (17.32) = 0.462
30o
8 m
R
mg
f a
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For the following situation, find the least coefficient of friction between the ground and the ladder.
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 = N72.17 = x 250 = 0.289
60o
250 N
4 m
smoo
th
S
N
f A
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How to find coefficient of limiting friction? p 22 The coefficient of limiting friction, , can be found by
placing the block on a surface and tilting the latter to an angle at which the block is just about to slip.
Along the plane: f = mg sin --- (1)
Perpendicular to the plane: N = mg cos --- (2)
Since the block is just about to slip, f = N --- (3)
Sub (3) into (1):N = mg sin --- (4)
(4)/(2): = tan
mg
f
N
Hence, the coefficient of limiting friction is just tan .
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