FRICTION Friction 1 Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages.   Disadvantages ---- Power loss, wear and tear etc. Advantages ---- Brakes, traction for vehicles etc. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 2 F (Friction) P W Hills & Vales Magnified Surface N Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 3 Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented.  When P = 0, F = 0  block under equilibrium When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.   www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 4 When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction. Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.   www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 5 Sliding friction friction experienced when a body slides over another surface. Dynamic Friction Rolling friction  friction experienced by a body when it rolls over a surface.   www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 6 F  N Fmax = N Fmax (limiting friction) P W  N R Where Fmax = Limiting Friction N= Normal Reaction between the contact surfaces  =Coefficient of friction   = Fmax N Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Angle  is known as Angle of limiting Friction. Angle of Friction Friction 7 The angle between N & R depends on the value of F. This angle θ, between the resultant R and the normal reaction N is termed as angle of friction. As F increases, θ also increases and will reach to a maximum value of  when F is Fmax (limiting friction) W P Fmax (limiting friction)  R N i.e. tan = (Fmax )/N =  Angle  is known as Angle of limiting Friction. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 8 Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending. Angle of repose   When granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Significance of Angle of repose: Friction 9 Significance of Angle of repose: The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose. Angle of repose is numerically equal to Angle of limiting friction www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Laws of dry friction to the normal reaction between the two surfaces.  1. The magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces. (Experimentally proved)     2. The force of friction is independent of the area of contact between the two surfaces.  3. For low velocities the total amount of friction that can be developed is practically independent of velocity. It is less than the frictional force corresponding to impending motion. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

FRICTION IN BELT/ ROPE DRIVES The transmission of power by means of belts or rope drives is possible only because of friction between the wheels and the belt. Tension in the belt is more on the side it is pulled and less on the other side. Accordingly they are called as tight side and slack side.  T2 (Tight side) Pull T1 (Slack side) W www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

RELATIONSHIP BETWEEN TIGHTSIDE AND SLACKSIDE FORCES IN A ROPE Friction 12 A load W is being pulled by a force P over a fixed drum. Let the force on tight side be T2 and on slack side be T1. (T2>T1 because of frictional force between drum and the rope). Let  be the angle of contact in radians between rope and the drum. Consider an elemental length of rope as shown. Let T be the force on slack side and T+dT on tight side. There will be normal reaction N on the rope in the radial direction and frictional force F= N in the tangential direction.  F d/2 T T+dT N F d w T1 T2 P www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

 Forces in radial direction = 0 Friction 13  Forces in radial direction = 0 N-T Sin d/2 – (T+dT)Sin d/2 = 0 { Sin d/2 = d/2 as d is small }  N-T d/2- (T+dT) d/2 = 0 i.e. N = ( T+dT/2) d ------(1) We know that F = N  F =  ( T+dT/2) d-----(2)  Forces in tangential direction = 0 (T+dT) Cos d/2 = F + T Cos d/2 { Cos d/2 = 1 as d is small }  T + dT = F + T i.e. dT = F------(3) From (2) & (3) dT =  ( T+dT/2) d Neglecting small quantity of higher order, dT = T d dT/T =  d www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

 = Angle of contact in radians Friction 14 Integrating both sides, T2  dT/T =   d T1 0 T2  (log T) = () T1 0 Log (T2/T1) =   T2/ T1 = e where T2 = Force on tightside T1 = Force on slackside  = Angle of contact in radians www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

1 ) For the block shown in fig., determine the smallest EXERCISE PROBLEMS Friction15   1 )   For the block shown in fig., determine the smallest force P required  a)    to start the block up the plane b)    to prevent the block moving down the plane. Take μ = 0.20 [Ans.: (a) Pmin = 59.2N (b) Pmin = 23.7N θ = 11.3o] P  100N 25 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 16 2) A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If μ between the block and the plane is 0.35, determine the unknown force P for impending motion (a) to the right (b) to the left [Ans.: (a) P = 132.8N (b) P = 1252N]  30 800N 2000N P www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 17  3) Determine value of angle θ to cause the motion of 500N block to impend down the plane, if μ for all contact surfaces is 0.30. [Ans.: θ = 28.4°] 200N 500N  = ? www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 18  4) In Figure, μ between rope and the fixed drum and between all contact surfaces is 0.20. Determine the minimum weight W to prevent the downward motion of the 1000N body. [Ans. : T1 = 0.76W, T2 = 1.424W, W = 253N] W 1000N 3 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS 4

Friction 19  5) A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends. [Ans.: x = 3.5m] 100N 200N X = ? 2 m A B 60 30 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 20  6) Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o. [Ans.: P = 328.42N] P 1600N B 20 A www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Friction 21 7) Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction,  = 15o for all contact surfaces. [Ans.: P = 1192N] A 20 P wedge www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N] Friction 22 8) A ladder of weight 200N, 6m long is supported as shown in fig. If μ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slipping b) the reactions at A & B [Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]   1000N B 5m  www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS A

Friction 23   9) An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. μ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor? [Ans.: FA = 52N] Smooth wall B 12m www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS A 5m

[Ans.: (a) x = 2.92m (b) Wboy = 458N] Friction 24 A ladder of length 5m weighing 500N is placed at 45o against a vertical wall. μ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder. [Ans.: (a) x = 2.92m (b) Wboy = 458N] www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS