5.11.1 CONTACT FORCE AND NORMAL REACTION when two bodies are kept in contact, Electro magnetic forces act between charged particles at the surfaces of the bodies. As a result each body exerts a force on the other body. The resultant molecular force exerted by the surface of one body on the contacting surface of other body is called contact force. Contact forces make an action -reaction pair. Normal force Contact force '——~$ friction ee 7 Contact forceHowever the direction of a contact force may not be perpendicular to contact surface. It can be resolved into two components. One, parallel to contact surface is called friction and the other, perpendicular to contact surface is called normal seaction force. From the above discussion we can conclude that normal reaction and friction are Electro- magnetic in nature. Normal reaction is normal to the contact surface and friction is parallel to contact surface. (Normal reaction is the normal component of contactYforce acting on a body placed on a rigid surface perpendicular to the plane of contact. Normal reaction depends only on the nature of the contact. It does not depend on the area of contact. Even in the event of the motion of the body on the surface this force remains unaltered. Consider a body of mass ‘m’ lying on a horizontal surface. The weight of the body acts vertically down on the surface. The surface exerts a reaction “N” of same magnitude on the body in a direction vertically up, then N = mg When a body of mass “m” is on an inclined plane of inclination “@”, the weight of the body “mg” acts vertically downwards, and a component mg cos @ perpendicular to the inclined plane is exerting a contact pressure, then N = mg Cos® 5.11.2 FRICTION According to Newton’s first law of motion, a body continues in its own state of rest or state of uniform motion unless it is compelled by an external force to change that state. Let us see some examples, (a) If a body is placed on horizontal surface and a small force is applied on it, then it may not move, It is clear that some or other force must have come into play to keep the body in equilibrium (b) If the'force on the above body is suffi- ciently large then it is set into motion, If the appli- cation of force is ceased then the body moves through some distance on the surface and comes to hault. Here also some force must have come into play to stop the motion of body on the surface (©) If we stop peddling our bicycle, then it comes to rest after traveling certain distance In all these examples, the force that is impending motion or that is opposing relative motion and is called friction or frictional force. Friction may be defined as the opposing force which comes into play tangentially between two surfaces so as to destroy the relative motion between them, Friction is the contact force along common tangent to the contacting the surfaces to destroy the relative motion between them. Friction non ~ conservative force. $.11.3 CAUSES OF FRICTION 1. A surface though appears smooth by visual inspection or by touch, but when viewed under a powerful microscope, it consists of a large number of surface irregularities. 2. When two bodies are placed one above the other, these surface irregularities interlock together and oppose any attempt to bring a relative motion between the bodies. 3. At the actual points of contact a sort of “welding” takes place due to intermolecular forces. 4. When a sufficiently large force is applied, these irregularities will be sheared off and breaking of welded joints takes place and the body starts sliding on the surface. 5. Friction is due also to cohesive or adhesive force among molecules at close proximity. 6. Friction is regarded as electromagnetic force. 5.11.4 ADVANTAGES AND DISADVANTAGES OF FRICTION A) ADVANTA( OF FRICTION : ‘The advantage of friction can be understood if we imagine the situation when friction is absent, The surfaces in contact will slip on each other without opposition in the absence of friction. This mayUsing ball bearings: The free wheels g¢ vehicles like bicycles, two wheelers, moto, cars, shafts of motors, dynamos etc., are provided with ball bearings to reduce friction by replacing sliding with rolling. This is because, friction while in rolling motion is less than friction while in sliding motion. 4. Streamlining: Automobiles and aeroplanes are specially designed with curved surfaces, so that the air layers may get streamlined during 4. Vehicles move on the roads without slipping the motion and hence reduce friction Smooth surface rough surface ‘The frictional force provides necessary torque for the front wheel, and rotates it. There fore on front wheel fric- tion is backwards. However if the bicycle is not pedaled friction on both wheels will be opposite to the direction its motion. ie, in backward direction 5.11.7 VARIATION OF FRICTION WITH APPLIED FORCE, Consider a block “BY which is at rest on a horizontal table as shown in figure. A small pan is attached to the block by means of a thread passing over a frictionless pulley. When the weight in the pan is increased the applied force also increases, the static friction also increase in ‘equal magnitude. Let the static friction reach its maximum value for an applied force F,. This value ofnumerically equal to limiting friction f, . Therefore the net force on the body is zero and the body is in equilibrium. Ifthe applied force is increased slightly more than f, the body starts sliding on the surface. However, the magnitude of force required to keep the body sliding under friction with constant velocity is slightly less than the force required to start the motion. The friction in this condition is called dynamic friction, Hence dynamic friction is less than the limiting friction for the same body on the same surface. Even if the applied force is increased the dynamic friction remains constant. ‘Smooth sliding F, (at rest) Here we notice that until the static friction reaches its maximum value applied force is equal proportional to the frictional force. Thus the angle made by the straight line with X-axis is equal to 45°. Slope of the line “m= tan@ = tan45 = 1 The static friction is always equal to the force or component of force parallel to the two surfaces in contact in magnitude, The static friction is always opposite to the applied force in direction, The minimum tangential force needed to move the body is equal to the limiting friction The kinetic frictional force is always opposite 40 the direction of the velocity of sliding. The dynamic friction is equal to the applied force in magnitude only when it is sliding with constant velocity. 5.11.8 LAWS OF FRICTION 5.11.8(A) LAWS OF FRICTION- STATIC FRICTION 1. Static friction always acts in a direction, opposite to that in which the body tends to move. 2. Static friction is a self adjusting force, 3. The magnitude of static friction is always equal to the force which tends to move the body. 4, The force of static friction is independent of area of contact. : 5. The value of limiting frictional force is directly proportional to the normal reactions faN =f, =sN . Where “Ll,” is a proportionality constant called “coefficient of static friction”. ‘The ratio of maximum static friction to normal reaction is called “coefficient of static friction”, 5.11.8(B) LAWS OF FRICTION: KINETIC FRICTION 1. Kinetic friction always acts in a direction opposite to the direction in which the body is mo 2. Kinetic friction is independent of area of contact. 3. For small velocities kinetic friction is indepen- dent of velocity, where as at high velocities owing to heat energy produced a sort of “welding” takes place which increases the friction. 4. Kinetic friction is directly proportional to normal reaction, fy ON => fx =I, N Where “iy.” is a proportionality constant called “coefficient of kinetic friction”. ji n=] ‘The ratio of kinetic friction to normal reaction is called “coefficient of kinetic friction”. 5.11,8(C) LAWS OF FRICTION - ROLLING FRICTION 1. Rolling friction increases with increase in area of contact, 2. Rolling friction is inversely proportional to radius, 3. Rolling friction is directly proportional to normal reaction. f,aN=f, =yN Where “Ht, ” is a proportionality constant called “coefficient of rolling friction”.‘The ratio of rolling friction to normal reaction is called “coefficient of rolling friction”. Hs > > He The value of coefficient of friction depends op the nature of the materials in contact and surface finish and is independent of area of contact and mass of the body. . vis pure ratio and hence has no units or dimensions. The value of jt has a characteristic value for the given pair of surfaces. In general coefficient of friction is less than cone but in some cases like in the case of highly polished copper plates it is greater than one . Dust and impurities on the surface drastically effect the value of frictional force and also the coefficient of friction. * Note 5,57 ; The laws of friction given above do not have the status of fundamental laws like those for gravitational, electrical and magnetic forces. They are empherical relations that are only approximately true in limited domains. Yet they are very useful in practical calculations in mechanics *Note : 5.58: To find direction of frictional force. First we assume that no friction is present, then in absence of fric- tional force, we find relative rubbing of each body on each surface and finally, we apply frictional force in such aman- ner to oppose the relative rubbing. ie., frictional force acts, opposite to the direction of relative motion. Work done by kinetic friction is converted to , $0 in this case there is always a loss in mechanical energy. $11.9 ANGLE OF FRICTION When a body is in contact with a surface, the angle made by the resultant of normal reaction and the limiting friction with the normal reaction is called “Angle of Friction (6)”. mg Let us consider a rectangular block of mass ‘m’ on a rough horizontal surface in limiting friction under the applied force ‘F’ The forces acting in the system are weight of the body “mg” acting verti- cally downwards, Normal reaction “N” vertically up and limiting friction “f,” in a direction opposite to the direction in which the body tends to move. If a parallelogram is constructed taking limiting friction and normal reaction as the adjacent sides, its diagonal represents their resultant. The angle made by this resultant with the normal reaction is called “angle of friction (6)”. In fig. OA represent N, OB represents f,. Completing the parallelogram OACB, OC gives the resultant of N and fe In> Hs = tand, ‘The greater angle of friction the greater is the value of coefficient of friction. Note: 5.60 : Cone of friction : When a body is subjected to horizontal force P and when the motion is impending, the frictional force is equal to the limiting friction and the angle made by the resultant with the normal is equal to the. angle of friction. Ifthe force P is changed gradually through 360? in same horizontal planeas shown in Fig. the resultant RY generates a right circular cone with semi-vertex angle equal to angle of friction. This cone is known as cone of friction re ve eee je Be 58 () © + Note S.61 : The forces exerted by the surface on the body are friction “,” and normal reaction “N”. As these two are mutually perpendicular, the net contact force acting on the body is given by R=Mgy ite +1 mgyTan?gel (+: Fg = mgseeg = "| k= mg: aca5.11.10 MOTION OF A BODY ON ROUGH HORIZONTAL SURFACE Applying a Horizontal Force : Consider a body of mass “m” lying on @ horizontal surface of coefficient of static friction 41°; Let a horizontal force “F” is applied on the body as shown in the figure is increased gradually, is too small, the body does not slide, = Frictional force is static friction and itis equal to force that tends to move the body . (F,,) but F,,=F => f= F to bring it into motion, For the body to come in to motion the applied force “F” must be at least equal to the limiting friction “f,”. Thus, when F=f, F= U.N; F= mg) (.7N=mg} Consider a body of mass “rh” placed on a rough horizontal surface of coefficient of kinetic friction “}1,”. A horizontal force “F” greater than limiting friction is applied on it so that the body comes in to motion. Now kinetic friction acts on it in a direction opposite to its motion. If “N” is the normal reaction, then the kinetic friction is given by wm N=Heme If “a” is the acceleration produced in the body, the resultant force acting on the body is given by Tp =F-f, > ma=F-h, FoR _Foume] mg Note 5.62 : The distance travelled and velocity acquired jn 2 given interval of time "1" can be obtained from the - ic kinematic equations s=ut45a0° and v=u+at. Note 5.63 : A force “F just enough to set the block into motion is aplied on the block. This by definition is equal to maximum static friction(f, = #mg)- If this force is continued, even after the block starts moving, the body now has to overcome kinetic friction f, = mg) which is less than the applied force. Thus, there is a net resultant force which produces acceleration in the body. Then, N f, =Pamg, renee eG) i Fy =(tismg—Hyme) mma=(s Hx) mg mg : (Us-be)e / Problem - 5.70: A body of mass 60 kg is pushed up with just enough force to start it moving on a rough | | surface with p, = 0S and jt, =04 and the force | | continues to act afterwards. What is the acceleration of the body ? | | Sol. m=60kg; Ht, =05 5 4 =04 Force f= ma=mi(}i,- fl.) 8 jae He | | (05-04) 98 = 0.98 mvs? / Problem - 5.71: A block of mass 10kg pushed by a Jorce F on a horizontal rough plane moves with an | | acceleration of Sms". When force is doubled, its | acceleration becomes 18ms*. Find the coefficient of | Jriction between the block and rough horizontal plane. (g=10ms*). Sol. On a rough horizontal plane, acceleration of a block | of mass ‘m’ is given by a= Ht, Initially, P= Fy a=5 ms* He (10), ii) (: =10ke) | 0 When force is doubled i.c., P= 2F; a= 18 ms, i) 2F We= Fy ~He 10) Multiplying Eq Gi) with 2 and subtracting from Eq, (i) 8244 (10)=94 = rok5.11.11 PUSHING & PULLING OF A LAWN ROLLER i) A Roller on Horizontal Surface Pushed by an Inclined Force : i ~F cosé F sin mg When a lawn roller is pushed by a force ‘F’, which makes an angle @ with the horizontal, the component of force acting vertically downwards is F sin @. The horizontal component F cos@ pushes the roller to the right The weight ‘mg’ of the lawn roller acts vertically downwards. Therefore the normal reaction N of the surface on the roller is given by N=mg+Fsin@. Then the frictional force acting towards left is given by f, =HN =|, (mg +Fsin6) Where |, is the coefficient of rolling friction between the roller and horizontal surface. P, and concluded that pushing is difficult than pulling. (6x) pulling is easier than pushing. Application-5.3 Applying an Inclined Pulling Force : Consider a body of mass “m” lying on a horizontal surface of coefficient of static friction “,”. Let an inclined force “F” be applied on the body so as to pull it on the horizontal surface as shown in the figure, The forees acting on the body are; 1) Weight ofthe body “mg” acting vertically down. 2) Normal reaction “N” exerted by the surface oon the body vertically up. EISEN TK 3) Frictional force “f,” opposing the motion, 4) The applied pulling force “F” acting at ay angle “Q” with the horizontal. The applied force “F” can be resolved in to two components “F Cos®” and “Fsin®”. The ~ body is in contact with the surface, there by N+FSing =mg= N =mg-F Sing Force that tends move the body, F,,, = Fcos6 () F sino If F< f,then body doesn't side = frictional force=F,= F cos For the body to be pulled F,= f =F Cos =f, FCos@ = IL, N= FCos® = pt, (mg-F Sing) FCosQ= |, mg ~ jt, F Sing FCos@ +, F Sing = n, mg F (CosQ + pt, Sin8) = Hy mg -__limg “:Tan d=, sing ‘cosh fos Sind, sia) cos mgSind F=—_mesing (coscos + SindSing) For F to be minimum cos(8—4) should be maximum =3cos(0-9)=1=30-$=0,0=9 T++Enin =mgsin® (or)mgsing From the figure, This is the least force required to move the body on a horizontal force. Application-5.4 : Applying an Inclined Pushing Force : Consider a body of mass “m” lying on a horizontal surface of coefficient of static friction “m,”. Let an inclined force “F” is applied on the body so as to push it on the horizontal surface as shown in the figure. Fsin® ‘The forces acting in the system are’; 1) Weight of the body “mg” acting vertically down, 2) Normal reaction “N” exerted by the surface on the body vertically up. 3) Frictional force “f,” opposing the motion. 4) The applied pushing force “F” acting at angle “Q” with the horizontal. The applied force “F” can be resolved in to two components “F Cos 9” and “F Sin@ ”. The body is in contact with the surface, there by N= mg +F Sing— (1) Force that tends to move the body, F.,= F cos IFF.,< fthen body doesnt side = frictional force = F.,= F cos For the body to be pulled F,= f, = Fos =f, F.CosQ = HN = F Cos = Ht, (mg + F Sing) FCos@ = It, me +p, F Sing ME bn F Cos ~ 1, F Sing =p, mg F (Cos ~ H, Sin8) = Ht, mg __mgSing “cos(O# 9) (cos 0cos@ —sin sind =cos(8+6)) For F to be minimum @=(0 = MBSING agtang > From the figure , Tand= |, => [Fria = Hors Note 5.64: When pushing force F makes an angle @! with vertical then the minimum value of F for which the body bmg, sind —p, cos Here F must be positive for which moves is given by F= sin! —p.cos6! > 0 => tan! >, (or) @ >tan"(q) = 6' > where 6 =tan"'(\) :. for angle @' <6 no motion will take place however large the force Fis. Application-5.5 : Sliding of a Chain On A Horizontal Table: Consider a uniform chain of mass “m” and length “L” lying on a horizontal table of coefficient of friction “}1,". When I/nof its length is hanging from the edge of the table, the chain is found to be about to slide from the table, Weight of the hanging part of the chain‘Weight of the chain lying on the table = tol) When the chain is about to slide from edge of the table, the weight of the hanging part of the chain is equal to frictional force between the chain on The table and the table surface. Thus, 1=y,(n=1) => “The maximum fractional length of chain g from the edge of the table in equilibrium Application-5.6 : Connected Bodies : ‘Two objects of masses m, and m,are connected by a very light string passing over a clamped light smooth pulley. The object of mass m, is on rough horizontal table and the object of mass m, is hanging vertically. The coefficient of friction between m,and the table is mg Let the acceleration of the system of two objects be a Considering the forces on m, we can write the equation for its acceleration ‘a’ -as ma=mg-T (0 Considering the forces on m, = mya=T=pme ~ ‘Adding Eqs (i) and (ii), we get (am, +g) =(m, —HM)g Application-5.7 :Vebi Horizontal surface: ming To Rest On A A vehicle of mass “m” is moving on a horizontal rough road with a velocity “V". The coefficient of friction between the tyres and the road is “". On applying brakes it is brought to rest owing to friction. Here, the loss of kinetic energy of the vehicle is equal to work done wgainst friction, If “S” is the distance the vehicle travels before coming to rest, W=AKE s-(Lm? 13 ) = Humes =5mv? One can say that the safe maximum speed of the vehicle for which it comes to rest within a distance of *S” is given by V = (2na5} If “Vis the minimum time in which the body is to be brought (o rest, then impulse Yeas f,t=mv-0 Hymgt=mv =>Example 25. Find the acceleration 4, dy, 4 of the three blocks shown in figure 5.108, if a horizontal force of 10 N is applied on (a) 2kg block (b) 3 kg block (c) 7 kg blockTake g = 10 mvs? Hy=0.2 2ke a Figure. 5.108 Sol. (a) When force of 10 N is applied on 2 kg block. The limiting frictional force between 2 kg and 3 kg blocks f,= 0.2 x 2g = 0.2 x2 10=4N. 10 N= 2kg - a 4N 4N=<— 1 3ke | The Figure. 5.109 The limiting frictional force between 3 kg and 7 kg blocks that can be A=03%5g=03% 5x 10=15N As applied force 10 N is greater than f; but less than /;, so 2 kg block will slide over 3 kg but 3 kg will move together with 7 kg, Thus we have; 10-4 2 a= =3 mis? 4 7 and a= 43> 5 5 = OA mist, Ans, (b) When force of 10 N is applied on 3 kg block : As the applied force is less than the friction between 3 ky and 7 kg blocks (that can be 15 N) so the blocks will move together as one unit with an accelerationAns, 10N=* Figune. 5.110 Now find pseudo force on 2 ke block because of acceleration of kg block ay) Frege = 2 x ‘63 g aN js smaller than the friction between 2 kg and 3 ky fae cae ¢ N), so 2 kg will move together with other aay (©) When force of 10 N is applied on 7 kg block. Supposing 3 kg and 2 kg blocks move together with 7 kg block. The acoslerioaGl ——— whole system , @= 55357 Figure. 5.111 e 2x5 5. The pseudo force on 2 kg block = 0 2 ws 3 24347 6 sake wn5.11.12 BODIES IN CONTACT WITH VERTICAL SURFACE Consider a body of mass “m” in contact with a vertical wall. Let “1” be the coefficient of friction between the surfaces in contact. A horizontal force “F” is applied as shown in the figure to hold the body under equilibrium. The normal reaction “N” exerted by the wall on the body is equal to “F”. Weight “mg” of the body acts vertically down trying to pull the body down where as the frictional force “f? acts vertically up trying to prevent the body from sliding. The limiting friction f, =sN = UF If mg < i then the block is at rest and friction is : Static and is equal to weight .i.e , f= mg. If mg > f, then the block slides down . mg -f=ma For the body to be under equilibrium f=meg NS LL, N=me Nw Bare UF = me ( N=F) e pals ME I,Note 5.62: Ifa book is held between two hands and pressed using a force “F” by each hand, then weight is balanced by the total frictional force wef, mg=2,F ames 2s Application-§,13: A block of mass ‘m’ is pressed against vertical rough wall of coefficient of friction ‘}1* with a horizontal force ‘F’. a) The minimum force to be applied parallel to the wall so as o moved the body upwards is given by =(mg+uF)| ag = (nef) = (me b) Minimum work done to move up the body through a distance ‘S? is given by W=F,;,8=(mg+f)s| Application-5.14 : ‘A block of mass m kg is pushed up against a wall by a force P that makes an angle ‘Q” with the horizontal as shown in figure. The coefficient of static friction between the block and the wall is Jt The minimum value of P that allows the block to remain stationary is Psind PsinO+f=mg {=(mg-Psind) yN=(mg- Psin®) uPcosd=mg~Psind P[sind-+p.cos0]=mg Application-5.15 + Body placed in contact with inner W Hollow Cylinder = Consider a hollow cylinder of radius “R” rotating in a horizontal plane about a vertical axis passing through its centre and parallel to its lengt with a constant angular velocity “w”. A block of mass “m” is in contact with the inner wall of the all of Rotating cylinder. The forces involved in the system are as shown in the figure. Here, normal reaction N = mrw?, For the body to be under equilibrium, friction (f) = weight (mg) HAN = mg h,mro? 2 mg So 2ghige Thus the minimum angular velocity a= |e BR ‘The maximum time period of revolution of the eylinder T=2n, | 8 (é ‘Problem - 5.88: A 70kg man stands in contact against | the inner wall of a hollow cylindrical drum of radius | 3.m rotating about its vertical axis 200 rev/min, The coefficient of friction between the walland his clothing is 0.15, What is the minimum rotational speed of the cylinder to enable the man to remain stuck 0 the wall | (without falling when the floor is suddenly removed ? Sol, The titional force F (vertically upwards) opposes the weight mg. The man remains stuck tothe wall after the floor is removed if mg Sf, ie, mg ot ; then mgsin® > f, =(j!,mgcos®) and the body slides down on a rough inclined plane under kinetic friction, ‘The kinetic friction, =[1,N = [,mgcos® ‘The resultant force acting on the body down the plane is Fy =(mg sind f,) +. Fa = mgsin6 —j1,N = mgsin@ ~y,mgeos! +. Fy =mg(sin® - py, cos) The acceleration of the body F Fe meg(sinO—{1, cos®) as m m fa=e(sind—\1,c0s0) = If ‘v’ is the velocity of the body at point ‘B’ andthe body sas fom rst a A then by the equation v? = we get v? Dglsind- 000) Where ‘s’ is the distance travelled along the inclined plane Then {v= /2es(sinO 1, cos®) => The time taken by the body to slide down is given by s=ut+1/2a? “Here, w= 0 =0(t)+ petsind—H, cos8)t? 21 e(sin8 =p, c0s0) Note 5.68 : Let F” is the force to be applied parallel to the inclined plane to prevent the body from sliding down, Fy =0 F—mg(sin@-p1, cosd)=0 ng (sind—11, cos) Force "Fis required only ifangle of incinaion (0) is greater than angle of repose (0). ~ A(Probie 5.90: A body of mass ‘m’ slides down a ‘Smooth inclined plane having an inclination of 45° with ‘he horizontal, I takes 28 to reach the bottom. Ht the body is placed on a similar plane having coefficient Sriction 0.5 What is the time taken for it to reach the bottom ? Sol. Mass=m, @=45° Time taken to reach the bottom TT, = 2 sec jt = 0.5 Time taken by the body to touch the bottom without friction is 1, = [24 gsin® * | Time taken with friction is a ‘g(sin 0-41 cos 8) finO- cos sind sind ind Heos @ =2:828 see ——~ + SLA block of woudl of mass 0.5 ky is placed on a plane making 30° with:thejiarizontal. If | the coefficient of friction between the suzfaces of contact | | of the body and the plane is 0.2. What, tforce is required | 410 keep the body sliding down with uniform velocit Sol, Force required to keep the body siding down with constant velocity Given, m=0.5kg g=9.8 ms?; 0 =305 w= 02 F=mg(sin 8 - [t cos 8) = plane have coefficient of friction 0.5. If the normal | reaction is twice that of resultant downward force along | the incline. Find the angle between the inclined plane | and the horizontal. Sol, W=05. N=2mg(sin0-pcos8) mgcos, = 2mg(sin,~j.cos,) | | cos, =2c0s,(tan,— 1) i 1 fetan, 3 = tand=1=3 = 45° 5.11.14(B) BODY MOVING UP ON A ROUGH INCLINED PLANE Let us consider a rough inclined plane of coefficient of kinetic friction ‘y,". Let @ be the angle made by inclined plane with horizontal. A body of mass ‘m’ is placed at the bottom of the inclined plane, ‘A force ‘F’ is applied on the body en to the inclined plane to move the body uP te ol ie with uniform velocity. The forces involve ee system are shown in the figure. For the body ° = up the inclined plane with uniform velocity, tl he ore or sum of components of forces acting Up the incine plane must be equal to the sum of forces or sum of components of forces acting down the plane. F,=0 = Fin ~(mgsinO + f,)=0 gsinO-+f, =mgsinO+p,N =mgsin6+y4mgcos0_( N-=mgcos0} F=me(sin0 +11, c0s6) In order to move the body up the inclined plane with uniform acceleration ‘a’ the force to be applied must be greater than the sum of opposing forces by an amount equal to ‘ma’, Fy =F=mg(sind +p cosd)=ma 19(sin8 +11, c0s0)+ma Problem -5.93 :A block is placed on a rough inclined Plane of inclination 0 = 30°. If the force to drag it along the plane is to be smaller than to lift it. The coefficient of friction \. should be less than Sol.mg (sind + cos8) mgsin®-f =ma mgsin® — jt, [mgcos+F,]=ma i g[sind 11 co: m, to keep the block just stationary, ie. a= 0.minimum force required is given by me (FE) = ™(sind-p1, cos0) aia =p, (00H cos) (pe = nsec. My oblem - 594: A 3h Bock is to be moved up an | | inclined plane atan anglue 30°to he horizontal witha | | vetoci ity of Sms". If the frictional force retarding the | | motion is ISON find the horizontal force required to | move the block up the plane. (g=1ms?.) Sol. The force required to a body up an inclined plane is F=mg sing + frictional force | = 30(10) sin 307+150 = 300N. | | J IfPis the horizontal force, Pos 2003 = 346N ‘ ~ cos cos (Problem - 5.95 : Am insect eras up a hemispherical | | surface as shown sete igireheeeficen of rcton | | between the insectand the surface is 13. Ye ine joining | | the centre ofthe hemispericalsirface tote insect makes | an angle Q with the vertical, the maximum possible value of @ is given by Sol, N= JW = mg sind From (1) and (2) we get : gcos0 (1) | L Problem -5.96: A rough inclined plane is inclined at 30° to the horizontal as shown in the figure. A uniform chain of length Lis partly on the inclined plane and partly hanging from the top of the incline. If the | |e ficient of friction between chain and inclined plane is {the maximum length of the hanging partto prevent | the chain from falling vertically is Sal. wey oxg= o (L-x)gsin30 +t 0(L~ (tevBu)t agen) }gcos30 | | | Ayalon 5.97 :A block is placed on a ramp of para. | Ios shape given bythe equation y =X720-If = 08, | then the maximum height above the ground ‘at which | the block can be placed without slipping is Sul Let the block can be placed on the ramp at a height | inclination of the ramp at | Inabove the ground and 9 i the position. In the position the component of weight | | along the slope of ramp is mg sing downwards. | frictional force N= [1 mg cos In equilibrium, mg sin =p1,mgcosO tanO=,=0.5 but, y=.720 | ty | Slope tana = oT p08 28e5 | | From the figure maximum height, Application-5.2 A body slides down a rough inclined plane and then over a rough plane surface. It starts at A and stops at D. The coetficient of friction is : Sol. B ic D : V=,2e(6inO—peosd)ac «.CD= ug ZX (sin0-wWcos0)AC Ang HCD=AB-,BC ~. Application-5.23 : A body takes “n” times as much time to slide down a rough inclined plane as it takes to slide down an identical but smooth inclined plane If the angle of inclination of the inclined plane is “9”. What is the coefficient of friction between the body and the rough planeSol. Let | be the length of both the inclined planes and @ be the inclination. Then the times taken by a body sliding the smooth and rough inclined planes ae 2 g(sin@ ft, c0s0) {gpooth = treogh = Lnaoth sind and trough TE Frough = "smooth = ‘gsin@—Hl, cos) On solving we get |H, = tn Application-5.24 : A body is projected along a rough inclined plane of coefficient of kinetic friction ply. with an initial velocity “u”. It travels a distance “I” up the plane and retraces its path and reaches the same point of projection. Let its time of ascent is “t.” and time of descent is “1,” and time of descent is “n” times time of ascent, Then 2 8 21 Vg(sind—11, cos) = Yin +1, cos®) ee = solving we get Application-5.25 = A body is projected along a rough inclined plane of coefficient of Kinetic friction ju with an initial velocity “u. It travels a distance “I” up the plane and retraces its path and reaches the same point of projection with velocity ‘¥’ . During its ascent it moves with a retardation of (sind +1, c0s0), while coming down it moves with an acceleration of g(sin8 jl, c080) 2 a d $= Dind+ 4, c0s8) —2g(6ind~ 1, 0088) 2 (sind +p.c0s8) _v__ ftand+ i \? (sind, cos6)¥ YtanOHy Baws of MotionApplication-5.26 : A body is released from rest from the top of an inclined plane of length ‘L’ and angle of inclination ‘Q’. The top of plane of length Fen > lis smooth and the remaining part is rough. If the body comes to rest on reaching the bottom of the plane then find the value of coefficient of friction of rough plane. Sol. v=0 ungCosd L For smooth part : V= 2a, — n hem “| For rough part : 04? =2a, eae n a 1 2a, — e =F aN n an gsinO= -g(sind- jicos0)(n—1) gsin@[1+n-1]=ftgcos8(n—1)