1. Branch: Mechanical 3c [b] Guided by: Ankit Sir

2.  Group members: Waghwani Ashish -130450119186 Surti Ankur -130450119 165 Akil Vahulawala -130450119 178 Thakor Hardik -130450119 172 Moshin shiekh - 130450119157

3.  When a body slide or tends to slide on a surface on which it is resting, a resting force opposing the motion is produced at the contact surface. This resisting force is called friction.  In others words, friction is the resistance to motion that occurs when different surfaces are in contact. F P

4.  Friction always act in opposite direction to the movement of the body.  If contact surface is surface is smooth, friction will be less and if surface is rough then friction will be more.  In machine parts like piston, bearing, liner etc. attempts are made to reduce friction to increase life of parts and efficiency of machine.

5. Brakes, traction for vehicles etc.. Disadvantages: Power loss, tear and wear….

6.  The maximum friction force that can be developed at the contact surface, when body is just on the point of moving is called limiting force of friction.

7. Friction Static FrictionDynamic Friction Sliding Friction Rolling Friction

8.  Friction experienced by a body when it is at rest is called static friction.  In case of static friction, P<F There is no motion. Where P=external force F=frictional force

9.  Dynamic friction: Friction experienced by a body, when it is in motion is called dynamic friction. Dynamic friction is always less then static friction.  Sliding friction: Friction experienced by a body when it slides over another body, is called sliding friction.

10.  Rolling friction: Friction experienced by a body, when it rolls over another body, is called rolling friction. e.g. – to move drum by rolling – ball bearing used in machine

11.  The angle between normal reaction and resultant force is called angle of friction or limiting angle of friction.  The value of angle is more for rough surface as compared to smooth surface. R N F W PF

12.  Limiting friction is proportional to the normal friction. F= μ.N μ=F/N  The ratio of limiting friction (F) and normal reaction (N) is called coefficient of friction. tan Φ=F/N Therefore, μ=tan Φ

13.  It is defined as the ratio of he limiting friction force to the normal reaction offered to the body by the surface, provided the body is in impending condition.

14.  It is defined as the ratio of friction force developed while the body is in motion, to the normal reaction developed.

15.  With the increase in angle of the inclined surface, the maximum angle at which, body starts sliding down is called angle of repose. Here we are considering weight (W) which is resting on the inclined at angle α with horizontal. It has two component, Component parallel to plane= W sin α Component perpendicular to plane= W cos α

16.  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.

17. ∴N=W cos α ∴F=W sin α We know that, From 1 and 2, ∴Angle of response=Angle of friction=Φ

18.  It is defined as an imaginary cone formed by revolving the resultant of the normal reaction and limiting frictional force about the normal reaction for one complete revolution.  As long as the resultant lies within or along the periphery of the surface of cone the body will be in rest. As soon as the resultant crosses the periphery of the cone, the body comes in motion.

19.  Static friction force, if body is in equilibrium.  Limiting friction force, when body just start moving.  Kinetic friction force, when body is in motion. Motion Rest Friction 45 Fm Fk Force P0

20.  The friction force always act in a direction, opposite to that in which the body tends to move.  The magnitude of friction force is equal to the external force. ∴ F=P  The ratio of limiting friction and normal reaction is constant. ∴F/N=μ=constant

21.  The friction force does not depends upon the area of contact between the two surfaces.  The friction force depends upon the roughness of the surfaces.

22.  The friction force always act in a direction, opposite to that in which the body is moving.  The ratio of limiting friction and normal reaction is constant and it is known as co- efficient of friction. ∴F/N=μ=constant  For moderate speeds, the friction force remains constant. But, it decreases slightly with the increase of speed.

23. Example: A force required to pull the body of weight 50 N on a rough horizontal surface is 20 N when it is applied at an angle of 25 o with the horizontal figure. Determine the coefficient of friction and magnitude of reaction ‘R’ between the body and horizontal surface. Solution: Consider the free body diagram of the block. It is subjected to 1. Weight 50 N vertically downwards 2. Normal reaction ‘R’ vertically upwards 3. Horizontal component of applied pull 4. Force of friction in opposite direction to the motion 5. Vertical component of applied pull

24. For equilibrium Σ F x = 0 Σ F y = 0 ⇒ 20 cos 25 o - µ.R = 0 …(1) 20 sin 25 o + R – 50 = 0 …(2) where µ is coefficient of friction. From equation (2), we get R = 50 – 20 sin 25 o = 41.55 N Putting the value of R in equation (1), we get µ = 20 cos 25 o / 41.55 = 0.43

25. The ladder is a device for climbing or scaling on the roofs or walls. There are two long pieces in the ladder to which many small cross pieces called rugs are connected at equal distances to work as steps.

26.  A ladder AB is shown in figure. The coefficient of friction between ladder and wall, and ladder and floor is 0.30. A man weighing 600N is standing at the mid length of ladder. Calculate reactions at floor and wall. Neglect the weight of ladder. Fw 600N Ff A 2m 5m Rw Rf C D B

27.  Here horizontal and vertical component are zero. ∴Rf+Fw=600 Rw=Ff ∴600=5Rw+0.6Rw ∴Rw = 107.14N ∴Fw=0.3*107.1 =32.14N ∴Rf=600-Ff=567.85N Ff=0.3*567085=170.35N But, Ff=Rw =107.14N

28.  A wedge is usually of a triangular or trapezoidal in cross section.  Wedges are used to adjust the elevation or provide stability for heavy objects such as this large steel vessel

29.  Problems of equilibrium of forces acting on the wedge are solved by any one of the following methods. 1. Lami’s thereom 2. Resolution of forces 3. Graphical method Body 1 2 w 3 α p 4 wedge

30.  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. Pull W T 2 (Tight side)  T 1 (Slack side)

31.  Belts are used for transmitting power or applying brakes. Friction forces play an important role in determining the various tensions in the belt. The belt tension values are then used for analyzing or designing a belt drive or a brake system.

32.  Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to  radians.  If the belt slips or is just about to slip, then T 2 must be larger than T 1 and the friction forces. Hence, T 2 must be greater than T 1.  T 2 = T 1 e   where  is the coefficient of static friction between the belt and the surface

33. Belt Analysis

34.  The load weighs 100 lb and the  S between surfaces AC and BD is 0.3. Smooth rollers are placed between wedges A and B. Assume the rollers and the wedges have negligible weights Find: The force P needed to lift the load. Plan:  1. Draw a FBD of wedge A. Why do A first?  2. Draw a FBD of wedge B.  3. Apply the EofE to wedge B. Why do B first?  4. Apply the EofE to wedge A.

35. The FBDs of wedges A and B are shown in the figures. Applying the EofE to wedge B, we get  +  F X = N 2 sin 10  – N 3 = 0  +  F Y = N 2 cos 10  – 100 – 0.3 N 3 = 0 Solving the above two equations, we get N 2 = 107.2 lb and N 3 = 18.6 lb Applying the EofE to the wedge A, we get  +  F Y = N 1 – 107.2 cos 10  = 0; N 1 = 105.6 lb  +  F X = P – 107.2 sin 10  – 0.3 N 1 = 0; P = 50.3 lb 10º N2N2 A N1N1 F 1 = 0.3N 1 P N2N2 10º B F 3 = 0.3N 3 N3N3 100 lb

36.  The screw, bolt, nut etc. Are used in various machines and structures for fastenings.  Due to movement of screw up or down in the nut, friction experienced on the contact of two surfaces. Such friction is screw friction.  The principal on which screw jack works is similar to that of an inclined plane. Therefore force applied on lever is consider in horizontal direction. P=W tan(α+Φ)

37. A double square-thread power screw (Fig. 6.11) has the major diameter d = 40 mm and the pitch p = 6 mm. The coefficient of friction of the thread is m = 0:08 and the coefficient of collar friction is mc = 0:1. The mean collar diameter is dc = 45 mm. The external load on the screw is F = 8 kN. Find: a) the lead, the pitch (mean) diameter and the minor diameter.

38.  Solution minor diameter=d-p=40-6=34mm pitch diameter =d-p/2=40-3=37mm lead diameter=2p=12mm