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

2. 2 Limiting friction If the block remains at rest, f = F A The frictional force to be overcome before it moves is called limiting friction. Frictional force f Applied force F A

3. 3 Limiting friction Frictional force f Applied force F A F A / Nf / N 00 11 22 33 43 53 Limiting friction = 3 N

4. 4 Limiting friction Frictional force f Applied force F A Limiting friction = 3 N F A / N f / N 0 1 2 3 4 5 1 2 3 At rest Moving Static Kinetic friction

5. 5 Limiting friction F A / N f / N 0 1 2 3 4 5 1 2 3 At rest Moving Static Kinetic 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.

6. 6 From experiment, limiting friction f is slightly greater than kinetic friction f’. i.e. f > 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. Friction f Applied force F Static frictionKinetic friction f’ f For simplicity, take f = f’

7. 7 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 F A Normal reaction N For simplicity, take f = f’ =  N and  =  ’

8. 8 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  =  ’ RubberConcrete (Dry)  0.6 – 0.85 RubberConcrete (Wet)  0.45 – 0.75 Frictional force does not depend on the area of contact of the surfaces.

9. 9 A block of mass 2 kg starts from rest, sliding down a rough inclined plane making an angle of 30 o 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 at 2 8 = 0(4) + ½ a(4) 2 a = 1 ms -2 mg sin 30 o – f = ma (2)(10)sin 30 o – f = (2)(1) f = 8 N R = mg cos 30 o = 17.32 N By f =  R 8 =  (17.32)  = 0.462 30 o 8 m R mg f a

10. 10 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 60 o = 250 x 2 cos 60 o S = 72.17 Newton f = S = 72.17 Newton The ladder is about to slip, f =  N 72.17 =  x 250  = 0.289 60 o 250 N 4 m smooth S N f A

11. 11 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 .