412123, 10:18 PM NLM - Friction, Chapter Notes, Class 11, Physics (ITJEE & AIPMT) Sign in Class 11 (https://edurewin/ex 28/Class-11) > NLM - Friction, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) NLM - Friction, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) 1 Crore+ students have signed up on EduRev. Have you? GY continue with Google Friction 1. Friction Friction is a contact force that opposes the relative motion or tendency of relative motion of two bodies. Consider a block on a horizontal table as shown in the figure. If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block in rest from moving is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of hitpssfedurev inl71591/NLM-Fricton-Chapter-Notes~Class-11~Physics~I 437412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) about to move. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort than to start its motion. So from this observation, we see that we have three states of block, first, block does not move, second, block is about to move and third, block starts moving. The friction force acting in three states are called static frictional force, limiting frictional force and kinetic frictional force respectively. If we draw the graph between applied force and frictional force for this observation its nature is as shown in figure. 1.1 static frictional force al" Dattoregion Kinet region When there is no relative motion between the contact surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph ab will have a slope of 45° The Direction of Static Friction The direction of static friction on a body is such that the total force acting on it keeps it at rest with respect to the body in contact. The direction of static friction is as follows. For a moment consider the surfaces to be frictionless. In absence of friction the bodies will start slipping against each other. One should then find the direction of friction as opposite to the velocity with respect to the body applying the friction. hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 2187412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) 1.2 Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. fig 2 N= fy = HAN Here mg is a constant the value of which depends on nature of surfaces in contact and is called as ‘coefficient of static firction’. 1.3 Kinetic Frictional Force Once relative motion starts between the surface in contact, the frictional force is called as kinetic frictional force. The magnitude of kinetic frictional force is also proportional to normal force. f= HN From the previous observation we can say that 1, < 1, Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., when relative motion starts a constant frictional force starts opposing its motion. Direction of Kinetic Friction The kinetic friction on a body A slipping against another body B is opposite to the velocity of Awith respect to B. It should be carefully noted that the velocity coming into picture is with respect to the body applying the force of friction. hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 397412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) AR, oR Suppose we have a long truck moving on a horizontal road. A small block is placed on the truck which slips on the truck to fall from the rear end. As seen from the road, both the truck and the block are moving towards right, of course the velocity of the block is smaller than that of the truck. What is the direction of the kinetic friction acting on the block due to the truck ? The velocity of the block as seen from the truck is towards left. Thus, the friction on the block is towards right. The friction acting on the truck due to the block is towards left. Ex.1 Find the direction of kinetic friction force ira] [tka] Varis (a) on the block, exerted by the ground. (b) on the ground, exerted by the block. FEIN 5m/s Sol. (a) 1 ka] Sm” (by Hit tobleck watto ground where f and fa are the friction forces on the block and ground respectively. Ex.2 The correct relation between magnitude of f4 and fz in above problem is : (A) f4 > f2 (B) f2>f4 (C) f4 =f2 (D) not possible to decide due to insufficient data. hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 4137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Sol. By Newton's third law the above friction forces are action-reaction pair and equal but opposite to each other in direction. Hence (C) Also note that the direction of kinetic friction has nothing to do with applied force F. [A } +10 ms Ex3 3 }->20mis EMIT All surfaces are rough. Draw the friction force on A & B Kinetic friction acts to reduce relative motion. Summary We can summarise the laws of friction between two bodies in contact as follows: (i) If the bodies slip over each other, the force of friction is given by f= HN where N is the normal contact force and yy is the coefficient of kinetic friction between the surfaces. (ii) The direction of kinetic friction on a body is opposite to the velocity of this body with respect to the body applying the force of friction. (iii) If the bodies do not slip over each other, the force of friction is given by hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 597412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) where Ue is the coefficient of static friction between the bodies and N is the normal force between them. The direction and magnitude of static friction are such that the condition of no slipping between the bodies is ensured. (iv) The frictional force f, or f; does not depend on the area of contact as long as the normal force N is same. Ex.4 A block of mass 5 kg is resting on a rough surface as shown in the figure. It is acted upon by a force of F towards right. Find frictional force acting on block when (a) F=5N (b)25N ()50N(ms = .6, mj = 0.5) lg = 10 ms F Sol. Maximum value of frictional force that the surface can offer is fmax = flim = HsN [Mo hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 6137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Therefore, it F< f,,,, body will be at rest and f = F or F > f,,, body will more and f = f, (a) F = 5N < Frag So body will not move hence static frictional force will act and , f=f=5N (b) F=25NF,,,, So body will move and kinetic frictional force will act, its value will be f,=1,N = 0.5 x 5 x 10 = 25 newton Ex.5 A block having a mass 3 kg is initially at rest on a horizontal surface. The coefficient of static friction 1g = 0.3 between the block and the surface and x is 0.25. A constant force F of 5ON, acts on the body at the angle @ = 37° What is the acceleration of the block Sol. We have two possibilities here, the block may remain at rest, or it may accelerate towards the right. The decision hinges on whether or not the x-component of the force F has magnitude, less than or greaer than the maximum static friction force. The x-component of F is hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 737412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) The x-component of F is F, = Fcos é = (50 N) (0.8) = 40 N To find f, ...., We first calculate the normal force N, whether or not the block accelerates horizontally, the sum of the y-component of all the forces on the block is zero. N-Fsin@-mg orN = Fsin 6 + mg=(50N) (0.6) + (3kg)(9.8ms7) = 59.4.N The maximum static frictional force f,max = y,.N = (0.3) (59.4N) = 17.8 N This value is smaller than the x-component of F; hence the block moves. We now interpret the force f In the figure as a Kinetic frictional force. This value is obtained as fi = ty N = (0.25) (59.4 N) = 14.8 N Therefore resultant force in the x-direction is SF .=Feos6-f =40N-14.8N=25.2N Then the acceleration ‘a’ of the block is 252N 3kg 84ms2 Think : What would happen if the magnitude of F happened to be less than f,.max but larger than fy? hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 8137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Best Study Material for Class 11 Exam (/courses/626_Chemistry-Class-11-Notes--Questions--Videos-MCQs-JEE--NEET) raj ‘Endich” engishciass 11 > aca” (/courses/3160_English-for-Class-11--XI---NCERT--Commerce--Humanities- Docs--Videos--Tests) A crash csst . (/courses/3453_English-Snapshot-for-Class-11-XI--CBSE-and-NCERT-Docs-- Videos--Tests) Explore courses —> (/explore/28/Class-11) Ex.6 In the previous example, suppose we move the block by pulling it with the help of a massless string tied to the block as shown here. What is the force F required to produce the same acceleration in the block as obtained in the last example ? * oa m [se Sol. We are given that, hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 9137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) NN 4F sino m = 3kg, 1, = 0.3, 1, and a = 8.4 ms* In order to determine the force F, we first draw the FBD as shown below The equations of motion therefore, are N+ Fsiné=mg mg N= mg -Fsin 6 Fos 8-f=ma and where f = 11, N before the start of the motion, once motion is set, f = 1,N. Hence, force F which produces a = 8.4 m/s? is given by Fcosé - u, (mg - F sin 8) =ma 25, 8 = 370, F cos masnkmg _ 40-4+0.25%9.8) 44 ogy ofp eer cosésyksing §— 08+ 0.25x0.6 7B: Fsin® works out to be less than mg. Otherwise we would lift the block up in the above analysis Comment It is easier to pull then to push. Only about 34 N force is required to pull than 50 N required during pushing why ? Because, when we pull at an angle, the effective normal force N by which block is pressing down on surface is reduced and consequently friction is reduced. Just the contrary happens when you are pushing. 2.Minimum force required to move the pa A body of mass m rests on a horizontal floor with which it has a coefficient of static friction m. Itis desired to make the body slide by applying the minimum possible force F. F pas Fig. A Let the applied force F be at angle f with the horizontal R F R= Normal force hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 10137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) For vertical equilibrium, R+Fsing=mg o, R=(mg-Fsini) —...() For horizontal equilibrium i.e. when the block is just about to slide, Fcos 6 =uR wlll) Substituting for R, F cos = u' (mg -F sing) or F=umg/ (cos 6+ wsino) for minimum F (cos¢ + sing) is maximum, => Letx=cosd+usind a Hsing + nos di for maximum of x, & 0 tan 4 =n and at this value of 6 -. - ima ot he 3. Friction as the component of contact force : When two bodies are kept in contact, electromagnetic forces act between the charged particles at the surfaces of the bodies. As a result, each body exerts a contact force on other The magnitudes of the contact forces acting on the two bodies are equal but their directions are opposite and hence the contact forces obey Newton's third law. {-nomal force N F =contact forceg~ The direction of the contact force acting on a particular body is not necessarily perpendicular to the contact surface. We can resolve this contact force into two components, one perpendicular to the contact surface and the other parallel to it. The perpendicular hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ wer412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) component is called the normal contact force or normal force and parallel component is called friction. Contact force = = N {when f,,, = 0} Fema Fe mae = VIN? +N? {when fing. = HN} NSF. ¢ yu? +4)N O 400N = 20g cos 45° + 400 sin 45° = 421.4N The frictional force experienced by the block Fo yN =p X 421.4 = 421.4 WN. ‘As the accelerated motion is taking placed up the plane. 400 cos 45° - 20g sin 45° - f = 20a 400 | 20x98 BR (400196 4.) 1 _ 2828-1386-12 =| 2 =e --12] x Ss = 0.3137 MRR "a 4214 +4214 =20a= 20x0.6=12 The coefficient of sliding friction between the block and the incline = 0.3137 5. Angle of Repose: hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 14137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Consider a rough inclined plane whose angle of inclination q with ground can be changed. A block of mass m is resting on the plane. Coefficient of (static) friction between the block and planes p. For a given angle q, the FBD (Free body diagram) of the block is Where f is force of static friction on the block. For normal direction to the plane, we have N=mg cosé AS € increases, the force of gravity down the plane, mg sin 8, increases. Friction force resists the slide till it attains its maximum value. Fag = HN = mg Cos 6 Which decreases with 6 (because cos @ decreases as @ increases) Hence, beyond a critical value 9 = 6,, the blocks starts to slide down the plane. The critical angle is the ‘one when mg sin 8 is just equal of f.,., ie., when mq sin &, = \. mg cosé, or tan 8, where &. is called angle of repose If 6 > 8, block will slide down.For 6 < @, the block stays at rest on the incline. 6. Two blocks on an inclined plane: Consider two blocks having masses m4 & mg placed on a rough inclined plane. m4 & mg are the friction coefficient for m4 & mg respectively. If N is the normal force between the contact surface of my &m2 Now three condition arises. hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 18137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) G) If y, = y= wthen N=0 because, Both the blocks are in contact but does net press each other. = sin €-«mgcosé (a,,a, are acceleration of block u, & jt, respectively) Gi) IF, < p, then’ N =O because, there is no contact between the blocks. a, =gsin 8-4, g cos 8 sin 6~ 1,.g cos 8 Gi) IF w, > u, then Ne Ex.10 Mass m4 & mz are placed ona rough inclined plane as shown in figure. Find out the acceleration of the blocks and contact force in between these surface. Sol. As we know if y1 > 2 both will travel together so ay=ag=a F.B.D which is equivalent to 3asin374fi+f) ” Now F.B.D of 1 kg block is 8-72 = 3.6 m/sec hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 16137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) rs gsin37°+N-4=(1)a N=3.6+ 4-6=1.6 Newton 7. Range of force F for which acceleration of body is zero. \ Download the notes oa © NLM - Friction, Chapter Notes, Class 11, Physics (IIT-JEE & AIPMT) Ex.11 Find out the range of force in the above situation for which 2kg Gre block does not move on the incline. Sol. F.B.D of 2 kg block Now take different value of F hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 7137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Force (F) Friction Type Acceleration Direction & magnitude ‘Ss @ F=on m/s? 7 ww Kinetic > = (i) F=4n or ‘Om/s? J an s Static S = Git) om/s? Joan 2s eo we (iv) F m/s? o Static e (v) F=16N Om/s? fa e Static (in this condition friction change its direction to stop relative slipping) ie (vi) F=20Nn ¥ omis? Jf an Static = (vil) F=24N 2m/s? vf aN By Kinetic hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 18137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) From the above table block doesn't move from F = 4N (mgsin@ - umgcos6) to F = 20N (mgsind + umgcos8). So friction develope a range of force for which block doesn't move "@ : If Friction is not present then only for F = 12N the block will not move but friction develop a range of force 4N to 20N to prevent slipping. So we can write the range of force F for which acceleration of the body is zero. mg sin 6 — umg cos 8 < F 100N 77 fare=90 N bee Sol. (i) Let us assume that system moves towards left then as it is clear from FBD, net force in horizontal direction is towards right. Therefore the assumption is not valid. 120N 10 20 4100N oon 60N Above assumption is not possible as net force on system comes towards right. Hence system is not moving towards left. (ii) Similarly let us assume that system moves towards right. hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 21137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) 120N 10 20 4100N con ‘60N Above assumption is also not possible as net force on the system is towards left in this situation. Hence assumption is again not valid. Therefore it can be concluded that the system is stationary r 120 N<—] 10 >> 20> 100N Fon=90 N 60 N Assuming that the 10 kg block reaches limiting friction first then using FBD's 1 tT 20/5 aN eT — 100N 120=T+90 > T=30N AlsoT + f= 100 30+f=100>f=70N which is not possible as the limiting value is 60 N for this surface of block. Our assumption is wrong and now taking the 20 kg surface to be limiting we have WPT T+ 20 120N f 60N 100N T+60=100N> T=40N Alsof+T=120N =>f=80N This is acceptable as static friction at this surface should be less than 90 N. Hence the tension in the string is T= 40 N 8. Pulley block System involving friction : * If friction force is acting and value of acceleration of a particle is negative, then it means direction of friction force is opposite to that what we assumed and acceleration would be having a different numerical value. hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 22197412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Ex.15 Two blocks of masses 5 kg and 10 kg are attached with the help of light string and placed on a rough incline as shown in the figure. Coefficients of friction are as marked in the figure. The system is released from rest. Determine the acceleration of the two blocks. Sol. Let 10 kg block is sliding down, then acceleration of both the blocks are given by, 10gsin37*-p, x 10gcos37*-5g sin53° It means our assumed direction of motion is wrong and 5 kg block is going to slide down, if this would be the case, the direction of friction force will reverse and acceleration of blocks would be given by: uy x 10gc0s37° =10gsing7?_ = -ve 8 It means in this direction also there is no motion. So we can conclude that the system remains at rest and friction force is static in nature. 9. Two Block System: Ex.16 ey Take a Practice Test = Test yourself on topics from Class 11 exam hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 23137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) wos B| 2kg Al 4x9" +F frictionless Find out the maximum value of F for which both the blocks will move together Sol. In the given situation 2kg block will move only due to friction force exerted by the 4 kg block F.B.D. a[2ig os f+ 4kg +F f The maximum friction force exerted on the block Bis fax = HN fmax = (0.5) (20) = 10N So the maximum acceleration of 2 kg block is 2kg |}-—f,.. = 10N =sm/s? amax is the maximum acceleration for which both the block will move together. ie, for a £5 ms" acceleration of both blocks will be same and we can take both the blocks asa system. FB.D a ‘ + Smis* 6kg [Yor hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 24137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Fmax= 6% 5=30N for 0 30 both the blocks will move separately so we treat each block independently F.B.D of 2 kg block 2kg | —*f = 10N (Friction force) ag=5 m/s? F.B.D of 4 kg block =10N + . val 4kg | F = 36 36-10 262 aa =Fls Ex.18 hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 25137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) al ag] puros Al 4g F h=0.4 Find out the range of force in which both the blocks move together Sol. If f4 is friction force between block A & lower surface and fg is friction force between both the block's surface. FB.D £=10N én 4 max = m4N1 = (0.1) (60) = 6N f2 max = M2N2 = (0.5) (20) = 10N Upper 2kg block is move only due to friction force so maximum acceleration of that block is 2kg }-+f, =10N aon emis 2 This is the maximum acceleration for which both the blocks will move together. 2 Therefore for a £ 5ms “ we can take both the blocks as one system. FB f-6Ne 849 iE For F < 6N, Blocks will not move at all. Now the value of Fax for which both the blocks will move together. Fmax-6=6%5 Fmax = 36N hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 26137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Conclusion if ON < F< 6N No blocks will move 6N < F < 36 N Both blocks will move together F > 36 N Both move separately. Ex.19 #05. g[ 2kg LF A_4kg ticieness The lower block A will move only due to friction force F.B.D. tf 2g bor Akg [?tetional rorce) 7 fmax = MN = (0.5) (20) = 10N F.B.D. of 4kg blocks ae 4g [-* t= 10N The maximum acceleration of 4 kg block is 25m/s? This is the maximum acceleration for which both the blocks move together 25 mis kg LF Fmax for which both the blocks will move together Fmax=2-5*6=15N hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 27137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Ex.20 u0.5~p[ 2kg LF a 4kg [wot If f2is the friction force between A & B and fy is the friction force between A & floor fmax=6N f2 max = 10N Lower block A will move only due to friction force So amax for 4 kg block 4kg 4, = 10N Is This is the maximum acceleration for which both the blocks will move together —+rims kg |-#F 4, =6Ne—1 F-6=6x1 F=12N If F is less than 6N both the blocks will be stationary Conclusion : 0 12N=Both move separately Ex.21 Find the accelerations of blocks A and B for the following cases. hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 28137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) (A)my=Oandmg=0.1 — (P)aq=ag=9.5 m/s? (B)mg=Oandmy=0.1 (Qhaq=9m/s2, Fe ap=10m/s? 91° [tho A B (C) mq =0.1 and (R) aq = ag = g = 10 m/s” m2=1.0 (D) m4 = 1.0 and (S) aq = 1, ap = 9 m/s? m2=0.1 Sol. (a) R, (b)Q, ()P, (ds (i) FBD in (case (i)) {m4 =0,m2=0.1} o uN While friction's work is to oppose the relative motion and here if relative motion will start then friction comes and without relative motion there is no friction so both the block move together with same acceleration and friction will not come. A La + 4 mg mg hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 29137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) Friction between wall and block A oppose relative motion since wall is stationary so friction wants to slop block A also and maximum friction will act between wall and block while there is no friction between block. "@ : Friction between wall and block will oppose relative motion between wall and block only it will not do anything for two block motion. 1 7 © 10 an=9 m/s? ; ag = 10 m/s* 4 f (ii) LA, f % io o Friction between wall and block will be applied maximum equal to 1N but maximum friction available between block A and B is 10 N but if this will be there then relative motion will increase while friction is to oppose relative motion. So friction will come less than 10 so friction will be f that will be static. i f t 4, 10 10 by system (20-1) = 2x a= a==9.5 m/s 11-10 (iv) ay = 18 tenis? hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 30137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) 10 ops 10. Friction involving pseudo concept : Ex.22 What is the minimum acceleration with which bar A should be shifted horizontally to keep the bodies 1 and 2 stationary relative to the bar ? The masses of the bodies are equal and the coefficient of friction between the bar and the bodies equal to m. The masses of the pulley and the threads are negligible while the friction in the pulley is absent. see in fig. 7 Sol. Let us place the observer on A. Since we have non-inertial frame we have pseudo forces. For body *1' we have, hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 31137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) For body °2' we have, N=ma mg-T-mma=0 mg=Tmmg...(2) From (1) and (2) ana z z Find out the range of force for which smaller block is at rest with respect to bigger block. Sol. Smaller block is at rest w.r.t. the bigger block. Let both the block travel together with acceleration a F.B.D of smaller block w.r. to the bigger block. frmax =m x N N=ma f=mma...(1) t ma (Pseudo) L 2) from (1) & (2) hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 32197412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) a=g/m=20m/s? So F = 20 (Mm) = 20(5) = 100N If F* 100 N Both will travel together Ex.24 The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end as shown. The coefficient of friction between the box & the surface below it is 0.15. Ona straight road, the truck starts from rest and accelerates with 2 ms", At what distance from the starting point does the box fall off the truck (i.e. distance travelled by the truck) ?[Ignore the size of the box] Sol. In the reference frame of the truck FBD of 40 kg block 704 ma une bla] (psueds force) Net force = ma~ iN = 40x 2- 22 x 40 x 30 et forve = ma ~ N= 40x 2- 7 = 80-60 =P edaye MA gy > 80 ~ 60 = Agee = Sr = IS This acceleration of the block in reference frame of truck so time taken by box to fall down from truck Age xZXP = Siu = Ugtt So distance moved by the truck atxayxt al pies ZX 2x (20) = 20 meter. Ex.25 Mass mz placed on a plank of mass mq lying on a smooth horizontal plane. A horizontal force F = agt (ag is a constant) is applied to a bar. If acceleration of the plank and bar are a and a2 respectively and the coefficient of friction between m4 and mz is hips sfedurev in\71591/NLM-ricton~Chapter-Notes~Class-11—Physics~I 33137412123, 1018 PM NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) m. Then find acceleration a with time t. Sol. If F < mmgg then both blocks move with common acceleration, ie., a4 = a2 When F > mmgg, then Equation for block of mass m F-mmgg=mz2a2 (1) andmmgg=m4ay_—.(2) From equation (1) at - mmyg = mzaz ie, acceleration ag varies with time linearly, its slope positive and intercept negative. From equation (2) aq is independent of time. So, the graph between a & tis as follow. Want to become a Class 11 topper? The right platform can make all the difference. Get access to high-quality study material including notes,videos, tests & sample papers along with expert guidance, and a community of like-minded individuals. Take the first step towards success by [sevice] signing up on EduRev today. hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11Physics~ 3419742123, 10:18 PM NLM - Friction, Chapter Notes, Class 11, Physics (ITJEE & AIPMT) About this Document Q 25K views oe 4.91/5 rating oy 48 505 0? _ Learners Habitat (/485050?publi publ icetr ue) ‘true) [__semietesror Download free EduRev App Track your progress, build streaks, highlight & save important lessons and more! See adiliW (/ancroid-download? t=https%3A%2F%2Fplay.google.com%2Fstore%2Fapps%2Fdetails%3Fid%3Dcom.edurev%26referrer%3Dhttp%253 A%252F%252Fedurev.in%253Futm_source%253DPage%2526utm_campaign%253DPage_Google Play_btn_rt_sd_br %2526utm_medium%253Dedurev_website) /apps.apple.com/tt/app/edurev/id1529835494) Related Searches ‘Sample Paper, Physics (IIT-JEE & AIPMT), Class 11, video lectures, Physics (IIT-JEE & AIPMT), Viva Questions, pdf, Exa m, mock tests for examination, Previous Year Questions with Solutions, Chapter Notes, Chapter Notes, Important que stions, practice quizzes, NLM - Friction, ppt, Physics (IIT-JEE & AIPMT), Class 11, Semester Notes, past year papers, Cla ss 11, NLM - Friction, Chapter Notes, Summary, study material, Objective type Questions, NLM - Friction, MCQs, Extr a Questions, Free, shortcuts and tricks; hitpsifedurev in\71591/NLM-Friction~Chapter-Notes~Class-11~Physies~I 35137412123, 1018 M, NLM - Friction, Chapter Notes, Class 11, Physic (TEE & AIPMT) SCHOOLS = Medical (/explore/33/NEET? utm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) = Non Medical (/explore/32/JEE? utm_source=footer_generic&utm_mediu durev_website&utm_campaign=footer#courses) = Commerce (/explore/1030/Commerce-Class-11-Class-12? utm_source-footer_generic&utm_medium=edurev_website&utm_campaign-footer#courses) = Humanities (/explore/1025/Humanities-Class-11-Class-12? utm_source-footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) = CBSE Class 10 (/explore/27/Class-10? utm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) "CBSE Class 9 (/explore/26/Class-9? utm_source=footer_generic&utm_mediur -durev_website&utm_campaign=footer#courses) * CBSE Class 8 (/explore/25/Class-8? utm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) "CBSE Class 7 V/explore/24/Class-7? uutm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) "CBSE Class 6 (/explore/23/Class-6? utm_source=footer_generic&utm_medium=edurev website&utm_campaign=footer#courses) = CBSE Class 5 (/explore/22/Class-5? utm _source=footer_generic&utm_medium=edurev website&utm_campaign=footeri#courses) "CBSE Class 4 (/explore/21/Class-4? utm_source=footer_generic&utm_medium=edurev.website&utm_campaign=footeri#courses) "CBSE Class 3 (/explore/20/Class-3? utm_source=footer_generic&utm_medium=edurev website&utm_campaign=footeritcourses) "CBSE Class 2 (/explore/19/Class-2? utm_source=footer_generic&utm_medium=edurev.website&utm_campaign=footerifcourses) = CBSE Class 1 /explore/18/Class-1? utm_source=footer_generic&utm_medium=edurev.website&utm_campaign=footeri#courses) COMPETITIVE EXAMINATIONS: = UPSC (/studytube/UPSC-Bible-15-Steps-to-Clear-UPSC-Exam/3a7c93ec-47cf-487¢-9dec- fb23486a81f_t? utm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footer#courses) = NEET (/explore/33/NEET? utm_source=footer_generic&utm_medium=edurev_website&utm_campaign=footeri#courses) "JE (/explore/32/JEE? utm_source-footer_generic&utm_medium=edurev_website&utm_campaign-footer#courses) = CLAT (/explore/34/CLAT? utm_source-footer_generic&utm_medium=edurev_website&utm_campaign-footer#courses) "CAT (/exolore/69/CAT? hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11~Physics~I412123, 1018 PM NLM -Frietion, Chapter Notes, Class 11, Physics (ITJEE & AIPMT) oe (http://www favebookccom/TheEducationRevolution) *°@eduURE Q rssurwitverconvedurevane) © (https:/Awww.instagram.com/edurevin/) hitpssfedurev inl71591/NLM-Fricton-Chapter Notes~Class-11~Physics~I 31137