UNIT 3 FRICTION. Structure. Objectives. 3.3 Problems Involving Dry Friction. 3.1 Introduction Objectives. 3.2 Laws of Friction

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1 UNIT 3 FRICTION Structure 3.1 Introduction Objectives 3.2 Laws of Friction 3.3 Problems Involving Dry Friction 3.4 Inclined Plane 3.5 Wedge Friction and Screw Friction Wedge I<nct~on Screw Fr~ctlon 3.6 Belt and Rope Friction 3.7 Summary 3.8 Key Words 3.9 Answers to SAQs INTRODUCTION In the preceding units. it was assumed that the surfaces in contact Were smooth and the forces between the bodies act normally to the surface of contact. However, in practice, it is impossible to have perfectly smooth surfaces. There always exists microscopic roughness which tends to prevent any possible sliding motion between the two bodies. In this unit, you will learn the laws governing the dry friction formulated by Coulomb and their applications in different situations. You will study the relation between the normal and frictional fofe6 at a point of contact between two non-lubricated rigid bodies. You will also study (he frictional forces in various devices such as inclined plane, wedges, screws, belts and rope drives. The study of frictional forces is essential to solve the practical problems in Engineering Mechanics. Objectives After studying this unit, you should be able to * understand the laws of dry friction, * determine the magnitudes of frictional forces in different situations, * work out friction-related quantities such as coefficient of friction, angle of friction, angle of repose to solve problems involving dry friction, and apply laws of friction to various devices like inclined plane, wedges, screws, belts and rope drives. As a resull of studies carried out by Coulomb in 1781, he reported that : I. The total amount of friction that can be developed is independent of the magnitude of the area of contact. 2. The total frictional force that can be developed is proportional to the normal force transmitted across the surface of contact. 3. Fur low velocities, the total amount of friction that can be developed is practically independent of velocity. However, it is less than the frictional force corresponding to impending motion. Consider a body of weight W resting on a floor. Let P be the force applied as shown in I $A j p F Figure 3.1. As forces P and Ware concurrent, there must be a third force equal in \ / magnitude and opposite in nature to the resultant of P and W to keep the body in N 'R equilibrium. Let it be denoted by R. The normal and frictional components of R are represented by Nand F respectively. Figure 3.1

2 As P increases F will also increase correspondingly till the limiting condition of impending motion. Tile maximum value sf F that can be developed is called limiting static friction and is proportional to the normal reaction N. Mathematically, F = N :. F = p N. where p is constant and is called the coefficient of static friction. The angle between the normal reaction N and the resultant reaction R is called the angle of friction. If it is denoted by 9, then we get, E Therefwe, the tangent of angle of friction is equal to the coeflcient offiiction. It is our common experience that when the body begins to move, there is a decrease in i- frictional effect from the limiting static friction. An idealised plot of this action as a function of time is shown in Figure 3.2. a t - This shows that there is a drop from the limiting frictional effect to a frictional effect that is 32 constant with time. It is independent of the velocity of the object. Generally, the coefficients of friction for dynamic conditions are about 25 percent less. Table 3.1 gives the values of static coefficients of friction : Surface of contact Table 3.1 Steel on cast iron 0.40 Copper on steel Hard steel on hard steel Mild steel on mild steel 0.57 Rope on wood 0.70 Wood on wood (Source : F.P. Bowden and D.Tabor, 1950, The Friction and Lubrication of Solids, Oxford University Press, New York) 3.3 PROBLEMS INVOLVING DRY FRICTION Just before the conditions of impending motion, the bodies are in equilibrium. Using the equations of equilibrium, you will be able to work out the unlcnown frictional forces and hence determine the various friction-related quantities. The most important thing is to ascertain the direction of frictional forces which always oppose possible, impending or actual relative motion at the contact surfaces. Example 3.1 Find the force P needed to start block B shown in Figure 3.3 moving to the right if the coefficient of friction is 0.3 for all surfaces of contact. Block A weighs 80 N and block B weighs 160 N. P

3 Solution - Figure 3.3 (a) shows the free body diagrams for block A and block B. As the possible motion of the block B is tow-ards right, the direction of frictional forces FA and FB are shown acting towards left, for block B. When block B moves towards right the relative motion of block A will be towards left and hence the frictional force FA for block A is shown acting towards right. Now the conditions of equlibrium can be applied to determine the force P. For Block A : CF,=O :. FA = T cos 30" = T XF, = 0 Also For Block B : XF,=O Example 3.2 = = N (3.6) Therefore, a force of N is needed to move the block B to the right. A force of 200 N inclined at 600 to the horizontal is applied to the block A weighing 400 N. Determine whether block A moves if the coefficient of friction is 0.5. If not, then find out the maximum value of the coefficient of friction when it is just on the point of moving... Free bod diagram of bkka

4 Solution There is a possibility of movement of block a towards right. Therefore the direction of frictional force FA will be towards left. = sin 60" = N This is the limiting static friction that will be developed between the surfaces of contact. As the horizontal component of P (i.e. 200 cos 60" = 100 N) is less than N, the block A will not movc if the coefficient of friction is 0.5. But, if the limiting static friction is less than the horizontal component of P then the block A will move.... FA < 100 N = Therefore, if the coefficient of friction is or less, then the block A will move under the given conditions. Figure 3.6 SAQ 1 (d) Mark the directions of frictional force and n o d reaction at the surface of contact for block A shown in Figure 3.5. Determine the frictional force developed. What is the value of limiting frictional force if coefficient of friction is 0.3? A ladder weighing 80 N rests at a comer as shown in Figure 3.6. What is the minimum angle possible before slipping occurs? The coefficient of static friction at A is 0.2 and at B is 0.3. Two blocks; A weighing 30 N and R weighing 50 N are on rough horizontal surface as shown in Figure 3.7. Rr.d the minimum value of P just to move the system. If the coefficient of hiction between block A and the ground is 0.28 and that between block B and the ground is Also find the tension in the string. For the system shown in Figure 3.8. find the value of load W so that the blocks A and B are just on the point of sliding. The coefficient of fnction between the blocks and the ground is The weights of the blocks are 800 Pu' each.

5 (e) A force P is applied at an angle a to the packing crate measuring 0.5 m x 0.8 m as shown in Figure 3.9. If the coefficient of friction is 0.3, determine the largest allowable value of the angle a and the corresponding value of P if it moves to the left without tipping. One of the simplest engineering device of lifting loads to higher altitudes is the inclined plane. The component of the weight of body along the inclined plane opposes if the body has to move up or helps if it has to move down. Consider a body resting on an inclined plane which is just on the point of moving down. The maximum angle of inclination at which this happens is called the angle of repose. As the body is just on the point of moving down, the direction of frictional force will be acting up the inclined plane. Figure 3.10 Resolving the forces along and the normal to the inclined plane and applying the conditions of static equilibrium, we get, ZF, = 0 :. I;, = Wsina But p=-= FA NA Wsina = tan a W cos a This shows that the tangent of the angle of repose is equal to the coefficient of friction. Also, we know, p = tan 4 = angle of friction Therefore, angle of repose is equal to the angle of friction. If the angle of inclination a of the plane is less than the angle of repose I$, the body will be at rest and an external force will be required to move it. If a > I$ then the body will run down the plane and an external force will be required to prevent the body from running down. Example 3.3 A body weighing 50(9 N is resting on an inclined plane making an angle of 30" with the horizontal. The coefficient of friction is 0.3. A force P is applied parallel to and up the inclined plane. Determine the least value of P when the body is just on the point of (i) (ii) Casc I : moving down. and Case 11 : moving up.

6 Solution Here, the angle of '(0.3) Case I : When the body is just on the point of moving down, the frictional force will be acting upwards. Figure Resolving all forces parallel to the inclined plane we get Pl + F,- Wsin30 =0 = Fl Resolving all forces normal to the inclined plane, we get, Putting this in Eq. (3.7), we get, Nl - W C ~S 30" = 0 ;. Nl = 500 cos 30" = 433 N (3.8) :. Fl = p Nl = 0.3 x 433 = N Pl = = N Case I1 : When the body is just on the point of moving up, the frictional force will be acting downwards. Figure 3.12 Resolving all forces parallel to the inclined plane, we get P,- F,- Wsin 30 =0 :. P, = 500 sin 30" + F, = F2

7 Resolving all forces normal to the inclined plane, we get, Putting the value of F2 in Eq. (3.9), we get, SAQ 2 Thus, it is seen that when the force applied is or mote fhe hody will move up and when it is N or less it will move down. When the force applied is between N and N, the body will neither move upw@ nor downwards, experiencing variable friction varying from N adtittg upwards decreasing to zero and then increasing upto N acting downwards. The mistake most frequently made in the solution of problem involving friction is to write the frictional force in the form F.= p N. It is to be remembered that d y h cue of impending or actual sliding motion of bodies with respect to one another he frichonal force will be maximum i.e. F,, = p N. In all other cases, the frictional force acting on a body is found by solving the equations of static equilibrium of the body. (a) A crate weighing 5 kn is kept on an inclined plane making an angle of 30" with the horizontal. 1. Determine the value of P (a horimntal force) required to move the crate up the plane. 2. What is the minimum value of P required to keep the crate from sliding down the plane? 3. For what range of values of P will the crate remain in equilibrium position shown in Figure \ (b) (c) 4. If P = 5.2 kn, fmd the magnitude and direction ~igum 3.13 of the frictional force acting on the crate. If the weight of block Q is 1.2 kn, find the minimum value of weight of block P to maintain the equilibrium as shown in Figure Blocks M and N rest on an inclined plane as shown in Figure The coefficient of friction between block M and block N is 0.4 ahd hiit between block Nand the plane is 0.5. If the weight of block M is 600 N and that of N is 800 N, fur what value of 0, the motion of one or both of the blocks is impending? If 0 = 15", find frictional forces between M and N and between Nand the plane.

8 3.5 WEDGE FRICTION AND SCREW FRICTION Wedge Friction Wedges are generally used to move heavy loads by applying a force which is considerably smaller than the weight of the load. Owing to the friction existing between the surfaces in contact, a wedge remains in place after being forced under the load. Therefore, for small adjustments in the position of heavy pieces of machinery wedges are extensively used. The problems involving wedges can be solved by applying the friction laws to the various parts of the device. The example given below will illustrate the procedure to solve problems involving wedges. Example 3.4 A block weighing 800 N is raised up with the help of two 6" wedges B and C of negligible weights as shown in Figure If the coefficient of static friction is 0.25 for all surfaces of contact, determine the smallest force P to be applied to raise the block A. Solution qq F l g 3.16 ~ Let us first draw the free body diagrams for block A and wedges B and C. Block A is in contact with the vertical wall and the horizontal surface of block B. Therefore, the nonnal reactions Nl and N2 will be perpendicular to the respective surfaces of contact as shown in Figure The direction of movement of block A with respect to the wall being vertically upwards, the direction of frictional force will be vertically downwards as friction opposes the motion. Similarly, the motion of block A with respect to wedge B being towards right, the direction of frictional force at the surface of contact of block A and the wedge will be towards left as shown in the free body N1 diagram of block A. - In the limiting condition, we know F1 = pnl and F2 = pn2. Therefore there are only two unknowns : Nl and N2 and two equations of equilibrium being available, we can find the values of Nl and N2. N2 D,=o :. N,-F2=0 efp dbloela :. N, = F2 = 0.25 N2 (3.11) DY=0 ;. N2-W-Fl=O ;. N Nl = 0 :. N (0.25 N2) = 0 :. N2 ( ) = 800 i ~3.17:FrccBodyhgram N3 BIgplpf18:FrccBodyDlPgRln ot Wedge B. N (3.12) ' N Now let us draw the free body diagram of wedge B. Keeping in mind that the frictional forces oppose the motion and the normal reactions, as the name suggests, are perpendicular to the surfaces of contact, the various forces acting on wedge B will be as indicated in the free body diagram of wedge B shown in Figure Applying the equations of equilibrium. we get, DY=0 :. N3co~60-Nz-F3sin60=0 :. N3 cos 6" - N N3 sin 6" = 0 ( as F3 = 0.25 N3) N N3 = 0. N N (3.13) '

9 Dx=0 :. F2+F3co~60+N3~in60-P=0 :. P = F2 + F3 cos 6" + N3 sin 6" = 0.25 N N3 cos 6' + N3 sin 6" Putting the values of N2 and N3, we get, P = 0.25 x x CO$ 6" sin 6' = N (3.14) Therefore, a force of N will be required to raise the block A. You can solve the problem by constructing the triangle of forces R1, R2 and W for block A where R1 is the resultant reaction of Nl & F, and R2 is the resultant reaction of N2 & F2. As the three forces keep the block in equilibrium, the forces must the concurrent. This can be solved graphically or by using Lami's theorem fm three COflCluTent forces. fjy applying Lami's Theorem, we get, where W - R1 - R2 sin (90" + 2$) sin (180"-@) - sin (90"- Q).'. R, = angle of friction = tan ' p = tan- (0.25) = " W cos $ W cos " - cos 241 cos " Similarly, for wedge B, there are three forces acting: R,, R3 and P where, R2 = resultant reaction of N2 and F2 R3 = resultant reaction of N3 and F3 Tlle three forces acting will be as shown in Figure By applying Lami's theorem, we get, R2 - P - R3 sin (90" + $ + 6") - sin (180" ") - sin (90" + 0) :. P = R2 sin (2 $ + 6") cos ( $ + 6") sin " cos " Screw Friction A screwjack is a device used for lifting or lowering heavy loads by applying comparatively smaller efforts at the end of the lever. The thread of a screw jack may be considered as an inclined plane wound round a cylinder and the principles used in solving problems on inclined plane can be applied to solve problems involving screw friction. If a is the angle of

10 the inclined plane and $ is the angle of friction, we know that the horizontal force required to pull the load up is given by P= Wtan(a+$) (3.15) This force P which drags the load along the inclined plane is related to the force P, applied at the end of the lever of the screw jack. This can be found out by taking moments of the forces about the centre line of the cylinder. If I is the length of the lever and r is the mean radius of the screw, we get, Plxl = P xr r :. P1 = 7 W tan (a + $) When there is one complete revolution the load is raised through one pitch i.e. centre to centre distance between two consecutive threads. and tan$=cl Using these relations we can work out the horizontal effort P, required to rais'e the load up. If the load remains in position even after removal of the effort P1 the screw jack is said to be self-locking. It does not work in reverse direction because the angle of inclination, in such cases, will be less than the angle of friction :. tan a c p i.e. p > tan a P 'Therefore, if the coefficient of friction is greater than -, the screw jack will be 2nr self-locking. To lower the load, the effort P at the thread required will be W tan (I$ - a) hence the effort at the end of the lever will be given by SAQ 3 (a) Two 8" wedges are used to push a block horizontally as shown in Figure If the coefficient of friction is 0.25 for all surfaces of contact, determine the minimum load P required to push the block weighing 6 kn. (b) Two wedges lift a heavy block of 8 kn as shown in Figure If the angle of wedges is 10' and the coefficient of friction is 0.3 for all surfaees of contact, find the value of P required to drive the wedges under the load. (c) The pitch of the thread of a screw jack is 5 rnrn and its mean diameter is 60 mm. The coefficient of friction is Find the force that should be

11 (d) applied at the end of the lever 200 mrn long measured from the axis of the screw (i) to raise a load of 20 kn and (ii) to lower the same load. 'Ihe pitch of a square threaded screw jack is 8 mm and the mean diameter is 50 mm The length of the lever is 400 mm. If a load of 2 kn is to be lifted, what force at the end of the lever will be required? Take p = 0.2. State with reasons whether the screw is self-locking or not. 3.6 BELT AND ROPE FRICTION A belt or a rope with a pulley is a device in which friction is utilised in raising a load, transmitting power or applying brakes to stop motion. Consider a flexible belt wrapped around a portion of a drum. Let the angle of contact be a. Let TI and T, be the tensions in the belt, as shown in the Figure 3.23 (a), when the motion is just developed between the belt and the drum. If T, > T2, then the impending motion-of the Eigure 3.23 (a) Figun 3.23 (b) belt will be clockwise relative to the drum Consider a small element of the belt, of length ds subtending an angle d0 at the centre. The free body diagram of this element is shown in Figure 3.23 (b). T and T + dt are the tensions at P and Q respectively and dn is the normal reaction. The friction force is (p dn) acting as shown. Resolving all forces tangentially and applying equation of equilibrium, we get, d0 :. dt = p dnr (:. d0 is small, cos 1 -t 1) Similarly, resolving all forces in nonnal direction, we get, For small angles, sin de :. Tde = dnr (neglecting very small quantity dt. -) (3.18) 2

12 From Eq. (3.17) and Eq. (3.18), we get, Integrating both sides around the portion of belt in contact with the drum, we get Tl :. log, - = pa T2 Thus, it is seen that the ratio of tensions depends only on the angle of wrap a and the coefficient of friction p. The torque developed by the belt as a result of friction is given as under: Torque = TI r - T2 r = (TI - Tz) r If the belt or rope is actually slipping, the coefficient of static friction should be replaced by coefficient of kinetic friction. The belts used are sometimes V-shaped. The contact between belt and pulley takes place along the sides of the grove. If /3 is the angle between these two surfaces of contact then the magnitude of the total friction force acting on the element will be 2 df and the sum of the components of the normal forces is 2 dn sin t, Proceeding on similar lines, we get, therefore, 3.7 SUMMARY In this unit, you have studied laws of friction and problems involving dry friction. The relative sliding motion of one body on another body is resisted by forces called friction forces. The sense of these friction forces is such as to oppose the impending or actual sliding motion. When there is no impending motion, the friction forces must be found by using the equations of static equilibrium. The limiting static friction is reached when relative sliding motion of the surfaces is impending and is given by: where p is the coefficient of static friction and N is the normal reaction. When sliding motion actually occurs, the retarding friction force has the magnitude pk N where p, is the coefficient of kinetic friction. The angle between the normal reaction Nand the resultant reaction R is called the angle of friction when sliding motion of the surfaces is impending. This is related to the_ coefficient of friction by: The maximum angle of inclination of the inclined plane, when the body kept on it is just on the point of moving down the plane, is called the angle of repose. Tbe angle of repose is equal to the angle of friction. You have also studied in this unit, the engineering applications where dry friction plays an irnprtant role e.g. wedges used to lift heavy loadsand screw jacks frequently used in presses and other mechanisms. By drawing free body diagrams indicating correct sense of friction forces and applying equations of equilibrium, you can analyse the engineering

13 applications where dry friction is involved. In case of belt and rope drives, over a curved 'surface, when sliding motion is impending the ratio of tensions is given by: -=p Tl T2 where TI = tension of the tight side T2 = tension on the slack side p = coefficient of friction a = angle of lap in radians In case of V belt the above formula is modified by multiplying a by cmec E, where the angle between two surfaces of contact forming V. 3.8 KEY WORDS Dry Friction Normal Reaction Limiting Frictional Force Coefficient of Static Friction Coefficient of Kinetic Friction Angle of Friction Angle of Repose Wedge : Friction depends on the condition of lubrication which exists between the surfaces of contact of two bodies. The frictional effects are reasonably predictable under dry conditions. In this unit, the problems related to friction when the surfaces of contact are dry, are considered. : When two bodies are in contacts and one body tends to slide over the other, the reaction force exerted by the force can be split into two components :(i) perpendicular to the surface of contact and (ii) parallel to the surface of contact. The perpendicular component is called the normal reaction and the parallel component is called the frictional force. : The maximum possible frictional force when the motion is impending is called the limiting frictional force. : The ratio of limiting frictional force to the normal \ reaction when motion is impending is called the coefficient of static friction. : The ratio of frictional force to the nonnal reaction when actual sliding motion takes place is called the coefficient of Kinetic friction. The coefficient of Kinetic friction is less than coefficient or static friction. : The angle between the n o d reaction and the resultant reaction is called angle of friction. The vector addition of normal reaction and the frictional force gives the resultant reaction. The tangent of the maximum value of angle of friction is equal to the coefficient of friction. :?be angle of inclination of an inclined surface when there is impending motion of the body kept on it is called the angle of repose. The tangent of the angle of repose is equal to the coefficient of static friction. : The wedge is a simple machine which is intended to transform an applied force into a force at approximately right angles to the direction of the applied force. The wedge angle is usually quite small. The'typical values for this angle are 5" to 10". is Friction 63

14 Belt Friction Angle of Wrap (lap) Ratio of Tension Vee Belt and Rope ANSWERS TO SAOs SAQ 1 (a) : When aflexible element such as a belt, rope or cable passes over a pulley or a drum, the distributed force which opposes motion of the bell or rope or cable is called belt friction. : The angle subtended by the belt which is in contact with the pulley or drum at the centre is called angle of wrap. : The tension TI in the belt on the side leaving the pulley is greater than the tension T2 on the side approaching, the difference being gradually taken up in friction between the belt and the rim of the pulley. The ratio of tension in the belt depends on the angle of wrap (lap) and the coefficient of friction. : When a rope passes through a V notch it comes in contact with two surfaces of Vee. The ratio of tensions in the rope depends also on the angle subtended by the two surfaces of Vee. Tbe directions of frictional force F and normal reaction N are marked in Figure for Answer to SAQ I (a). N is acting vertically upwards and F is horizontally acting towards left i.e. opposite to the direction of tending motion. Figure for Answer to SAQ 1 (a) The horizontal force which tends to move the blocka is 60 cos 45" = N towards right..= 'Ihe frictional force developed will also be N but acting towards left, when the motior~ is not impending. If the coefficient of friction is 0.3, the limiting frictional force will be 0.3 x nod reaction = 0.3 ( sin 45O ) = N. As the actual frictional force is less than limiting frictional force, the motion is not impending. (b) If slipping is impending, the various forces and reactions will be acting in the directions as shown in Figure for Answer to S AQ 1 (b). Taking moments of all forces about C, we get, I. I 1 - I NA -. sma + FA - cos a + FE- sina - NB- cosa = 0 * (1) and FB = limiting frictional force at B = 0.3 NB

15 (c) NB for Annrcr b SAQ 1 (b) FA = limiting frictional force at A = 0.2 NA Putting these values in equation (I), we get, NA NA sina+ 0.2NAcosa+ NAeina - - ma = sin a = ( ) cos a.-. tan a = :. a = 57' 26' 59" Applying equations of equilibrium to blocka we get NA = WA = 30 N and T = FA = 0.28 NA = 0.28 x 30 = 8.4 N.-. Tension in the string = 8.4 N N A Figam lor Aaracr to SAQ 1 (c) N B Now applying equation of equilibrium to block B, we get, NB + P sin 30' - WE = 0 (d) = P and ~cos 30 - FB - T = 0 (FB = 0.22 NB) ; P ( P) = 0 : P P = P = = P = 19.4/0.976 = N.-. The minimum value of P just to move the system is N. The free body diagram f a the hinge;rigid rods and blocks A & B are shown in Figure for Answer to SAQ 1 (d).

16 Figmm lor Answer b SAQ 1 (d) Applying equations of equilibrium, we get. CA~~6O0 - CB~~600 = 0 :. C, = C, = C (say) Applying equations of equilibrium to block A, we get, NA - WA - CA sin 60 = 0 (e) (Alternately, considering block B also, we will get W = N).= The value of load W, when the blocks are just on the point of sliding is N. The overturning moment due to force P about the left comer is P cos a (0.5) + P sin a (0.8) and the stabiiising moment due to weight W about the left corner is 0.4 W P :. W 2 - (0.5 cos a sin a) 0.4 As the block does not overturn,.-. W 2 (1.25 cos a + 2 sin a) P Applying the equations of equilibrium when the block is just on the point of sliding, we get, N+Psina-Br=O :. N-. W-Psina and F-Pwa= 0 But F=0.3N, (N: Normalreaction) :. 0.3 (W - P sin a ) = P cos a :. 0.3W = P (ms a sin a) :. W = (3.333 cos a + sin a) P (111)

17 P 0( I" \T 0.5m -0-8m- 4 /// N Figure for Answer to SAQ 1 (e) Comparing Equations (II) & (111), we get, 1.25cosa+2sina = 3.333cosa+sina :. sin a = cos a :. tan a = :. a = 64O 21' 32' and correspondingly P = W SAQ 2 (a) 1. When the crate is about to move up, applying equations of equilibrium after resolving the forces along and n o d to the inclined plane, we get, Figure for Answer to SAQ 2 (a) Pcos30 - Wsin30" - F = 0 : P = 5 x F = N and N- P sin30 - Wcos30 = 0 :. N = 0.5 P Putting this value in Equation (IV), he get, P = (0.5 P ) : P = :. P = N?his is the minimum value of P to move the crate in upward direction. 2. When the mate is just on the point of moving down, the direction of friction force F = 0.25 N will be acting upwards. Working on similar lies, we get, P = (0.5 P ) : P = :. P = N 'Ibis is the minimum value of P to keep the crate sliding down the plane. 3. Considering the above two results, it can be concluded the crate will be in equilibrium for the range of values of P between N and N. 4. If P = 5.2 N (assuming F acting upwards)

18 (b) The negative sign indicates downward direction of the frictional force. Consider forces acting on block Q. Applying equations or equilibrium after resolving the forces normal and parallel to the inclined plane, we get, Figwe tor Answer to SAQ 2 (bl and T- ~sin35o+~~ = 0 :. T sin 35O X = 0.-. T = kn Now, consider forces acting on block P, Here T-Fp= 0 (c).-. To maintain the equilibrium, the minimum value of weight of block P is kn. If only block M is on the point of moving down then applying the equations of equilibrium, we get, But FM = 0.4 NM sin8 = 0.4~600~0~8 :. tan 8 = 0.4 :. 8 = 21" 48' 5" If both blocks together are on the point of sliding down the inclined plane, then working on similar lines and noting p = 0.5, we get, 1400 sin 8 = 0.5 x 1400 cos 8.-. tan 8 = 0.5 As this angle is greater than 21 48' 5" the block M will slide fmt. If 0 = 15O, the motion is not impending. Frictional forces can be worked out from equilibrium equations FM = 600sin 15'

19 Friction Figure for Answer to SAQ 2 (c) This is the frictional force between block M and block N. (Note this is less than limiting frictional force which is 0.4 x 600 cos 15O = N) Similarly, FN = 1400sin 15O (W = WM + WN = = 1400 N) SAQ 3 :. The frictional force between the block N and the inclined plane is N. (a) Here, the coefficient of friction is :. The angle of friction 4 = tan-' (0.25) = O for all surfaces. :. RA, RB, RC & RD are making an angle of ' to NA, NB, NC, ND respectively when motion is impending. The suffixes A, B, C and D refer to wedge A, wedge B, block C and ground D respectively. As wedge B is moving down, the frictional forces FA and Fc will act upwards. N A Ground D W = 6kN Flgare for Answer to SAQ 3 (a) Three forces; RB, RD and Ware acting on block C. Refer to Figure for Answer to SAQ 3 (a) and apply the Lami's theorem. We get..'. RB = W sin " sin 1 18SI72" ' (:. W = 6 kn)

20 At the contact surface of wedge B and block C, the reaction offered by B is equal to reaction offered by C... RB = Rc = kn Now, consider wedge B. Three forces; RA, RC and P are acting as shown in figure. Note, contact surface between wedge A and B is inclined at 8' to the vertical. Applying Larni's theorem, we get, P - Rc sin (180" " ") sin (90" ') :. P = Rc sin " and Rc = kn sin " (b) :.?be minimum load P required to push the block horizontal is kn. The coefficient of friction of all surfaces of contact is 0.3. Therefore, the angle of friction will be tan-' (0.3) i. e. 16' 41' 57" for all surfaces. Three forces RA, RC and W act on the heavy block as shown in Figure for Answer to SAQ 3 (b). (Note: A and C denote wedges and D ground) Rgm for Answer to SAQ 3 (b) Applying Lami's theorem, we get, RA (or Rc) - W sin (180" - 16" 41' 57" - 10" ) sin 2 (16" 41' 57" + 10")... RA = RC = 8 sin 153.3' sin 53,398" Consider left hand side wedge. Three forces; RA, RD and P act on it as shown ih Figure for Answer to SAQ 3 (b), we get, P - RA sin (180" - 2i$ - 10") sin (90" + +) RA sin ( ') :. P = - cos sin " cos " :. The value of P to drive the wedges under the load is kn.

21 (c) (i) 'he force PI to be applied at the end of the lever to raise the load up is given by r PI = Wtan(a + $) where and Here $ = tan-lcc p = pitch of the thread = 5 rnm d = mean diameter and I$ = tan-' 0.08 = 4O 34' 26" r = mean radius = 30 mm 1 = length of lever = 200 mm W = 20kN :. PI = - x - x tan (lo 31' 10" + 4O 34' 26") = 3 tan (6.093") = 0.32 kn = 320 N (ii) To lower the load, P2 required is given by X 20 tan (4'34' 26"- 1 31' 10") = 3 tan(3.054') = 0.160kN =160N (d) To lift a load, the force required at the end of the lever is given by r PI=: Wtan(a+$) Putting the values of a and $ we get, P1 = tan O = kn = N P As - c p (0.051 < 0.2). the screw jack is self locking. It cannot work in reverse XD direction as the angle of inclination is less than the angle of friction.