Physics Session Opener A recent space shuttle accident occurred because of failure of heat protecting devices. How was this heat generated ? Session Objectives Session Objective 1. Friction & frictional force 2. Bodies connected in frictional surface 3. Kinetic friction 4. Static friction 5. Angle of friction 6. Coefficient of friction 7. Laws of friction Friction and Frictional Force Whenever an object moves or tends to move, contact surfaces of object resist the force tending to generate motion. This property of the surface is friction The resistive force is the friction force Bodies Connected in Frictional Surface Consider the experiment : 1. F applied. ‘A’ does not move. An equal and opposite force f acts A F MA f MB 2. ‘A’ still at rest on increased F. f adjusts to match F Bodies Connected in Frictional Surface r 3. F increased further A has acceleration. f has a maximum limiting value 4. F is reduced A moves uniformly f during the motion is lower than at rest. FF F F F ff f F MB f F f F f f f Bodies Connected in Frictional Surface Friction retards motion Friction force (f) is zero if no external force ( F ) exists f adjusts to stay equal and opposite to F f has a limiting value moving friction force is less than friction force at rest. Bodies Connected in Frictional Surface Friction force is a contact force, independent of area of contact. f F Friction force is a non-conservative force as mechanical energy changes to heat during friction. That is why the space shuttle overheated and exploded Kinetic Friction appears when two objects in contact have relative motion. Directed opposite the direction of motion fk is a constant. fk on B may start the motion of B. F F fk A B fk Static Friction F has component along contact surface, but there is no motion. f acts along the surface and opposite the force component f = Fcos till limiting friction. At limiting friction f(=fs) is constant fs > fk F N Fcos fs mg Class Exercise Class Exercise - 1 Blocks A and B are pressed against a smooth wall and are in equilibrium by a horizontal force F as shown. Then friction force due to A on B (a) upward (b) downward (c) dependent on relative mass of A and B (d) system cannot be in equilibrium Solution A B F Wall As wall is smooth, A and B will keep slipping down with acceleration g and cannot be in equilibrium. Hence answer is (d). Coefficients of Friction Limiting static friction fs :cons tan t fs sN N Kinetic friction fk :cons tan t fk kN N are coefficient of friction s k Coefficients of Friction Graphically f fs,max sN Limiting friction fs F fk k N F 0 Static region Kinetic region Laws of Friction 1. Bodies slip over each other fk kN 2. Direction of kinetic friction is opposite to the velocity 3. Bodies do not slip over each other fs sN s N is called the limiting friction. 4. fk or fs do not depend upon the area of contact. Class Exercise Class Exercise - 4 An object of mass m is pressed against a wall with a force F and is in equilibrium. The coefficient of friction between the wall and the object is . Then F must be equal to mg (a) mg (b) (c) mg (d) cannot be found m W a ll F Solution As equilibrium exists, wall presses against the object by a force of F (equal and opposite to force exerted by object against the wall) f F At equilibrium f mg f F F F mg F mg Class Exercise - 5 Three blocks of masses ma,mb and mc are arranged on a smooth horizontal surface as shown. Surface between ma and mb is smooth and the coefficient of friction between mb and mc is .Then the minimum force F required to keep mb from sliding down is: ma mb mc mbg ma mb g a b mc mc c d ma mb mc g ma mb mc mbg mc F m b m c m a S m o o ths u rfa c e Solution Let the force F give an acceleration a to the system F ma mb mc a To give an acceleration a to mc, mb presses against mc with force mca, so mc gives a normal reaction mc a (to left) on mb. mbg mca for mb not slipping down mbg ma mb mc mbg a F mc mc m b m c a F m S m o o thsu rfa ce a Class Exercise - 6 A block of mass M is kept in a lift. Coefficient of friction between the block and the lift is . The force required to initiate horizontal motion of the block is maximum when (a) Lift is moving up with constant acceleration (b) Lift is moving down with constant acceleration (c) Lift is stationary (d) Lift is in free fall Solution Condition of motion of M: F N F is largest when N is maximum N is maximum when lift moves up with constant acceleration a [N = m(g + a)]. So F is maximum at that condition. N F f W Class Exercise - 8 A 70 Kg box is pulled along a horizontal surface by a 400 N force at an angle of 30° above horizontal. If the coefficient of friction is 0.50, what is the acceleration of the box? (g = 10 m/s2) Solution Vertical: N = mg – F sin30 Horizontal: F cos 30 – f = ma F cos30 N ma F cos30 sin30 mg ma Fsin 3 0.5 400 0.5 70 10 70 a 2 2 F N 30° f Solving for a: a = 1.37 m/s2 mg Fcos Angle of Friction f along surface (surface property) N normal to surface Angle of friction f tan N R N fs,max Static > kinetic Class Exercise Class Exercise - 2 The angle between the resultant contact force and the normal reaction force exerted by a body on the other when they are one on top of the other is f. Then, what is the coefficient of friction, (a) = tan f (b) > tan f (c) < tan f (d) f and are not related Solution : f tan f N Hence answer is (a) Class Exercise - 3 A pushing force F is applied to a body of weight W, placed on a horizontal table, at an angle . The angle of friction is f . The magnitude of F to initiate motion in the body is W sin f a. cos f W cos f b. cos f F W tan f c. cos f W sin f d. sin f w Solution Vertical: N W F sin Horizontal: F cos f (a = 0 at initiation of motion) f N tan f W F sin tan f F cos W F sin tan f sin f F cos sin . W tan f cos f cos cos f sin sin f W sin f F cos f W sin f F cos f cos f F sin N F cos f F cos f W sin f W Angle of Repose Object at rest f increased . Object slips N fs f is angle of repose f tan f N s f f mg Angle of Repose f reduced At f f : Angle of repose (kinetic friction) k k , object moves uniformly. N fk fk tan fk N fk fs f f mg Class Exercise Class Exercise - 9 A block ‘A’ slides from rest down an incline with a 30° inclination with horizontal. It covers 3 m in 5 seconds. What is the value of coefficient of kinetic friction? (g = 10 m/s2) A 30° Solution Along y: N = mg cos Along x : mg sin – f = ma g sin cos a x ut 1 2 1 2 at at 2 2 N y 2x 1 gsin 2 . t cos .g 2 10 2 3 0.53 . 25 3 10 2 f m gsin x m g m gcos Class Exercise - 10 A block of mass 2kg lies on a rough inclined plane of inclination 30° to the horizontal. Coefficient of friction between the block and the plane is 0.75. What minimum force will make the block move up the incline? (g = 10 m/s2) M 30° Solution Along y: N = mg cos F is up the incline (along x) F – (mg sin + f) = ma To start the motion a is put equal to zero Fmin mg sin f y mg sin mg cos mg sin cos 1 3 3 2 10 . 22.98 N 2 4 2 F x N mg cos mg sin mg Thank you