Topic: Friction, Exams of Mechanics

Download Topic: Friction and more Exams Mechanics in PDF only on Docsity! Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Course Material Engineering Mechanics Topic: Friction by Dr.M.Madhavi, Professor, Department of Mechanical Engineering, M.V.S.R.Engineering College, Hyderabad. Contents Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 PART – I : Introduction to Friction 1. Definition 2. Types of Friction 3. Mechanism of Friction 4. Laws of Friction 5. Values of Coefficient of Friction 6. Friction Angles 7. Types of Friction Problems 8. Problems Involving Dry Friction. Session-II : Applications of Friction 9. Wedges 10. Connected Bodies 11. Ladder 12. Belts Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Referring the graph we may now recognize three distinct types of problems. Here, we have static friction, limiting friction and kinetic friction. P( Applied Force) 1. Static Friction: If in the problem there is neither the condition of impending motion nor that of motion then to determine the actual force, we first assume static equilibrium and take F as frictional force required to maintain the equilibrium condition. Here, we have three possibilities. (i) F < Fmax = Body is in the static equilibrium condition which means body is purely at rest. (ii) F = Fmax = Body is in limiting equilibrium condition which means impending motion and hence F = Fmax = µsN is valid equation. (iii) F > Fmax= Body is in motion which means F = Fk = µkN is valid equation the condition is impossible, since the surfaces cannot support more force Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 than the maximum frictional force. Therefore, the assumption of equilibrium is invalid, the motion occurs. 2. Limiting Friction: The condition of impending motion is known to exist. Here a body which is in equilibrium is on a verge of slipping which means the body is in limiting equilibrium condition. “It is the maximum value of friction force that the surface can exert on the block and is designated as Fmax.”.This mainly depends on roughness of the materials of the surfaces and of the normal contact force which these surfaces exert on each other. Fmax = µsN is valid equation. 3. Kinetic Friction: The condition of relative motion is known to exist between the contacting surfaces. So, the body is in motion. Kinetic friction takes place Fk = µkN is valid equation. 4.0 LAWS OF FRICTION 1. The frictional force is always tangential to the contact surface and acts in the direction opposite to that in which the body tends to move. 2. The magnitude of frictional force is self-adjusting to the applied force till the limiting frictional force is reached and at the limiting frictional force the body will have the impending motion. 3. Limiting frictional force Fmax is directly proportional to normal reaction (Fmax = µsN). 4. For a body in motion, kinetic frictional force Fk developed is less than that of limiting frictional force Fmax and the relation Fk = µkN is applicable. 5. Frictional force depends upon the roughness of the surface and the material in contact. 6. Frictional force is independent of the area of contact between the two surfaces. 7. Frictional force is independent of speed of the body. Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 8. Coefficient of static friction µs.is always greater than coefficient of kinetic friction µk . (µk may be 25% smaller than µs.in general). Note: µs. & µk are dimensionless. 5.0 . Values of Coefficient of Friction Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Tan Θ = µs In previous discussion, we had Tanø = µs which shows Angle of friction ø = Angle of Repose Θ The above relation also shows that the angle of repose is independent of weight of the body it depends on µ. Cone of Friction: When the applied force P is just sufficient to produce the impending motion of given body, angle of friction ø is obtained which is the angle made by resultant of limiting frictional force with normal reaction as shown in Fig. If the direction of applied force P is gradually changed through 3600, the resultant R generates a right circular cone with semi vertex angle equal to ø. This is called Cone of Friction. Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 FOUR DIFFERENT SITUATIONS MAY OCCUR WHEN A RIGID BODY IS IN CONTACT WITH A HORIZONTAL SURFACE. First case: No friction, Px = 0 ; F=0 Second case : No motion, Px < Fm There is no evidence that the maximum frictional force has been reached. F ≠ Fm = μs N Third case: Motion impending, Px = Fm ;if the body just about to slide F= Fm= μs N may be used. Fourth case: Motion, Px > Fm ; Fm = μK N Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 First case: No friction Second case: No motion ; tan𝛷s = Fm/N = μs N / N = μs Third case: Motion impending Fourth case: tan𝛷k = Fk/N = μk N / N = μk Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 7.0 Types of Friction Problems The first step in solving a friction problem is to identify its type 1) In the first type of problem, the condition of impending motion is known to exist. Here a body which is in equilibrium is on the verge of slipping, and the friction force equals the limiting static friction . The equations of equilibrium will holds good. 2) In the second type of problem, neither the condition of impending motion nor the condition of motion is known to exist. To determine the actual friction conditions, we first assume static equilibrium and then solve for the friction force F necessary for equilibrium. Three outcomes are possible: a) F < () Here the friction force necessary for equilibrium can be supported and therefore the body is in static equilibrium as assumed. We emphasize that the actual friction force F is less than the limiting value (i.e, ) and that F is determined by the equations of equilibrium. b) F = () Since the friction force F is at its maximum value , motion impends. The assumption of static equilibrium is valid. c) F > () This condition is impossible, because the surfaces cannot support more force than the maximum . The assumption of equilibrium is therefore invalid, and motion occurs. The friction force F is equal to . 3) In the third type of problem, relative motion is known to exist between the contacting surfaces, and thus the kinetic coefficient of friction clearly applies. 8.0 Problems Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 1) Will the 200N block be held in equilibrium by the horizontal force of 300N, if µ=0.3. Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Sol: Assume that 300N force is sufficient to hold the block from sliding down the plan. Let F acts down the plane. F + 200 sin= 300 cos F = 300 cos-200sin=160N For balance to exist, a frictional resistance of F=160kg would be required acting down plane. -N + 300sin + 200cos = 0 N=323N However, the maximum value obtained (limiting friction) Fi = µN = 0.3 x 323 = 97N The value of F necessary to hold the block from moving up the plane is 160N. Therefore it means that the block will move up the plane. Practice Problems Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 FBD of A Downward force = f > downward force ‘A’ will not move. 2. A straight circular cylinder of weight W rests on a V-block having an angle 2α as shown in figure. If the coefficient of friction is µ between contacting surfaces, find the horizontal force P necessary to cause slipping to impend. Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Solution to Problem2 : FBD: PART-II: APPLICATIONS OF FRICTION 1. WEDGES 2. CONNECTED BODIES 3. LADDER Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 4. BELTS 1. Wedges A wedge is one of the simplest and most useful machines. A wedge is used to produce small adjustments in the position of a body or to apply large forces. Wedges largely depend on friction to function.  In the figure, the block A supports a load W and is to be raised by forcing the wedge under it. The contact reactions between the blocks at this common surface are not only equal and oppositely directed on the free body diagram of each block; they also act so that their tangential or frictional components along the common contact surface oppose the impending motion of each block relative to the other.  In order for the wedge to slide out of its space, slippage must occur at both surfaces simultaneously, otherwise the wedge is self-locking. Method I: Equations of equilibrium using resolution method: Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 Sine rule : Note: 1. When only three forces act on a free body,it is usually best to apply the Sinelaw to the triangle formed by the force polygon. 2. When more than three forces are involved,of which only two are unknown,it is suggested that force summations to be taken with respect to perpendicular axes,one of which coincides with one of the unknown forces. Square Threaded Screws: Course Material on Friction by Dr.M.Madhavi,Professor,MED | 2°17 A square thread power screw Development of a Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 If Q is the force applied on the arm to produce torque for lifting the body, the force P,acting in a plane perpendicular to the axis of the thread and at the mean radius of the thread. Qa = Pr ( a= arm length ; r = mean radius; a>r) P = Qa / r In the first case, with motion impending up the incline, Q = F tan(ϕ + α) In the second case, with motion impending down the plane, Q = F tan(ϕ - α) Note: 1. It is evident that if the screw is to be self-locking, the angle of friction, ϕ must be larger than the pitch angle, α. 2. In case of over hauling , ϕ < α Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 So taken along the strut gives Bx = Ax FBD of block B: …………………….. (1) Substituting in equation (1) kg kg FBD of block A: Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 But kg kg kg Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 3. A uniform bar AB, 10m long and weighs 300N is hinged at B and rests upon a 500N block at A as shown in figure. If the coefficient of friction is 0.4 at all contact surfaces, find the horizontal force P required to start moving the 500N block. Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017  Many a times, we come across the uses of ladder for attending the higher heights. Ladders are used by painters and carpenters who want peg a nail in the wall for mounting a frame.  We observe that care is taken to place the ladder at appropriate angle with respect to ground and wall. We try to adjust the friction offered by the ground and wall in contact with ladder. Also sometimes we prefer to hold the ladder by a person for safety purposes.  The forces acting on ladder are normal reactions, frictional forces between the ground, the wall and the ladder, weight of the ladder and the weight of the man climbing the ladder.  Considering the free body diagram of ladder, we get general force system. The simplification of the system by considering equilibrium condition can be worked out by following equations: , and Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 1. A uniform ladder weighing 100N and 5 meters long has lower end B resting on the ground and the upper end A resting against a vertical wall. The inclination of the ladder with horizontal is 60˚. If the coefficient of friction at all surfaces of contact is 0.25, determine how much distance up along the ladder a man weighing 600N can ascent without causing it to slip. Sol. Consider the FBD of the ladder 0.25 Course Material on Friction by Dr.M.Madhavi,Professor,MED 2017 100 × 2.5 cos60˚+ 600 × d cos60˚- × 5 sin60˚ - × 5 cos60˚ = 0 100 × 2.5 cos60˚+ 600 × d cos60˚- 164.71 × 5 sin60˚ - 0.25 × 164.71× 5 cos60˚ = 0 d = 2.304m. 5. Belt Friction Belt or rope is wrapped around the pulleys to transmit power or effectively used for braking systems. In order to evaluate the effectiveness of the system, the tensions in the belt or rope are of importance.