14-LOM-01-THEORY

LOM PARTICLE DYNAMICS In this chapter we shall observe the motion of bodies along with the cause of motion under assump- tion of mass being constant and all parts of body has same state of motion so that we may treat bodies as point masses called particle. NEWTON’S FIRST LAW OF MOTION: Every body in the universe continues it’s state of motion (rest or uniform velocity) until some external source acts up on it to change its state of motion. Action of the external source is a physical quantity known as force. FOR EXAMPLE: A book kept on the table remains there, until we lift it. MOMENTUM: It is the product of mass and velocity of the body usually denoted as . ( p) Since velocity is a vector quantity and mass is a scalar thus momentum is a vector quantity → → p  mV Illustration : A 2 Kg ball is moving on a floor (taken as X – Y plane) with velocity given as m/s at a moment. Find the magnitude of momentum at that moment. →  →  2(5ˆi  5ˆj) kg m/s → V  (5i  5 j) =10iˆ  10 ˆj → | p |  | 10iˆ  10 ˆj |  10 kg m/s NEWTON’S SECOND LAW OF MOTION The time rate of change of momentum of a body is directly proportional to the applied force. → dp F dt → (Rate of change of a vector quantity is also a vector thus force is a vector)  K dp dt K  1 [K is proportionality constant equals one in S.I. unit] → → →  F  dp  m dv dt dt [when mass is constant]  →  ma →  dvx ˆ dvy ˆ dvz ˆ   F  m i   dt dt j  k dt   F iˆ  F ˆj  F kˆ  m dvx iˆ  m dvy ˆj  m dvz kˆ x y z dt dt dt  Fx ax  m dvx , dt  Fx m F  m dvy , y dt a  Fy y m F  m dvz z dt a  Fz z m [Component Form] [For motion in a plane only two components Say. X-Y are needed] Illustration: A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m/s . If the mass of the ball is 0.15 kg and duration of interaction between ball and bat is 0.5 s. Determine the average force imparted to the ball. Solution: Let us consider initial motion direction as X-axis → → ˆ ˆ Ns pi  mu  0.15  (12i)  1.8i → → ˆ ˆ p j  mv  0.15  (12i )  1.8i Ns → → → ˆ  p  p j  pi  3.6 i Ns → Average force Fav   t →  Fav  3.6iˆ 0.5  (7.2iˆ) N . Illustration: Momentum of a block as function of time is given as p  (10t i  5t j) kgm/s Find the force acting on block at t  2 sec. → → Solution: F  dp  dt d (10t 2 iˆ  5t ˆj ) dt  →  d (10t 2 iˆ)  d (5t ˆj) F dt dt  →  10 ˆi d (t 2 )  5 ˆj d (t) F dt dt  →  20 t ˆi  5ˆj  →  (40iˆ  5 ˆj) N. At t = 2 s. Illustration: A block of mass 2 kg is initially moving on a floor (taken as X – Y plane) with velocity 10 iˆ . At t  0 a constant force of magnidue 10 N parallel to y-axis begins acting on block for duration of 3 second and then ceases. (a) Find the velocity of block at t = 2 s. (b) Find the velocity of block at t = 4 s. Solution: (a) 0  t  3 sec → F  10 j Velocity at t  0 , → ˆ u  10i → → F ˆj 2 For 0  t  3 a   5 m m/s As acceleration is constant v  u  at v  10iˆ  (5 ˆj)t v  10iˆ  10 ˆj m/s (b) At  t  3 , let velocity of block be → v1  10iˆ  15 ˆj for t  3 → F  0 a  0 This means velocity of block remains constant that is v1  (10i  15 j)ms TRANSLATIONAL EQUILIBRIUM When force on body is zero then body is said to be under transnational equilibrium. →  →  ˆ  ˆ  ˆ → th  Fx  Fx  0, Fy  Fy  0, Fz  Fz  0  a x  0 , a y  0 , a z  0 * If body in equilibrium is at rest then it is called static equilibrium * If body is under equilibrium moving with uniform velocity then it is called dynamic equilibrium. NEWTON’S THIRD LAW OF MOTION: To every action (Force) there is an equal and opposite reaction (Force). Force never occurs singly in nature. Force is the mutual interaction between two bodies. Mutual forces between two bodies are always equal and opposite mA mB FBA A B → → Let body B exert force FAB on body A, body A exert a force FBA on body B. These forces are → → related by FAB  FBA The negative sign represents opposite direction. Some important points about IIIrd law. * Choice of term action and reaction is arbitrary and there is no cause effect relationship * Action reaction forces act on different bodies. * While studying motion of a body say A, only force acting on body A should be considered not the force applied by body A on other body or bodies. Illustration: An object far away from the surface of the earth has acceleration of 6 m/s2 towards center of earth. Find the force exerted by the object on the earth. Solution: Force on object by the earth → FOE → mob .a ob  5  6(N)  30 N Force exerted by earth on object is 30 N. Force between earth and object makes an action reaction pair thus force exerted by object on earth is also 30 N directed away for the center of earth. STEPS FOR DYNAMICAL ANALYSIS BY LAWS OF MOTION. (a) Select a suitable reference frame and then assign the positive and negative direction for the coordinate axes. Reslolve the forces into their components taking any direction as +ve and those in opposite direction as – ve. (b) Considering the body as point mass, represent the forces with arrow sign showing direction. [Most suitable for motion in a plane] This representation is known as free body diagram abbreviated as F.B.D. Let us consider a case to understand step (a) and (b) A rectangular block of mass M lying on a smooth horizontal surface is being pulled by a force F. at angle  with horizontal. Forces acting on the block are 1. Applied force represented as F 2. Weight W = Mg in vertically downward direction Reaction of surface on block N in vertically upward direction F.B.D. is shown Break up of forces along axes N F  (Mg) W (Vertical) y x (horizontal) Fcos Illustration: A bucket full of water has mass 10 kg is being pulled through a rope from a well. Pulling force of rope on bucket at certain moment is 125 N in the direction vertical. Find the accel- eration of bucket at that moment showing all steps of dynamical analysis. (Take g = 10 m/s2) Solution: FP Fx  0 Fy  Fp  W  125 100 a  Fy  2.5 m / s2 . y m W(mg) Most important aspect in dynamical analysis is identification of forces. To identify the forces we divide them in two categories. FIELD FORCES: Those forces which do not require contact between the bodies to act example weight electromag- netic force etc CONTACT FORCES: Those forces which require contact between the bodies to act; example are normal reaction, tension spring force, friction, push or pull etc. Let us discuss these forces one by one WEIGHT : It is the force acting on a body by virtue of gravitational interaction. Near the surface of the earth it is the product of mass of the body M and gravitational acceleration g, thus weight W = Mg and acts vertically downward. NORMAL REACTION: It is the adjustable force that acts at the contact points in a direction normal to the surface. It exists in action reaction pair at contact points. In those cases when body is treated as particle normal reaction between body and the surface means resultant of the normal reaction at all points. Illustration: Find the normal reaction on a block of mass M lying over a horizontal surface Solution: Here the number of contact points between block and sur- faces are infinite so that normal reaction means the net normal reaction. Let it be N F.B.D. of block N W(Mg) As body is under static equilibrium net force on the body in vertical direction is zero N  Mg  0  N  Mg . Illustration: A round table with four symmetrical pointed legs is lying over a horizontal surface. Find the normal reaction of the surface on each leg, mass of the table is M. Solution: Since all four legs are identical normal reactions at all contact points are same and are in vertical direction. Let it be N 4N As table is at rest net force acting is zero (static equilibrium) 4N – Mg = 0 (Considering vertical direction) B C  N  Mg 4 Mg A D If in the same problem we are asked about net normal reaction, it is N  4N   Mg   Mg net 4  .  4  Illustration: A block of mass 4 kg lies over a horizontal surface (g = 10 m/s2) (a) Find the normal reaction between the block and the surface. (b) Now another block of mass 2 kg is placed over the first block. Find the normal reaction Solution: (a)  between the first block and the ground surface in new case. N N  Mg  0 N  Mg  40N (b) Let M offer a normal reaction N’ to m upwards equally, m will exert N’ downwards on M. F.B.D. of m F.B.D. of M Mg N' N mg  N  mg N' Mg N  N  Mg  N  (M  m)g  N = 60 N In case (a) and (b) between the same surfaces normal reaction take different value means it is adjustable. Exercises: A sphere of mass M lies between two smooth incline as shown. Find the normal reaction at contact points A and B. Solution: Let normal reaction at contact points A and B be NA and NB respectively F.B.D. of sphere is Mg Considering 2 forces in vertical direction NA cos 1  NB cos 2  Mg  0 ...(i) Considering force in horizontal direction NB sin 2  NA sin 1  0 Solving equation (i) and (ii) we get ...(ii) N  Mgsin 2 N  Mgsin 1 TENSION: sin(1  2 ) sin(1  2 ) . When a body is connected through a string or rope a force may act on the body by the string or rope due to the tendency of inextension. This force is called tension.. While pulling a bucket of water from well, you exert upward force on bucket through tension only. * Tension acts at, all contact points between body and rope/string along the length of the string to the either side of contact point. * If some force is applied on string/rope Tension in the string/rope become equal to applied force at that point. Illustration: A boy is holding a bucket full of water through a light string as shown. Mass of the bucket is M. Find the (a) Tension in the string (b) Force applied by boy on string Solution: Mg F.B.D. OF BUCKET (a) Let tension at point of contact between string and bucket be T0 As bucket is under static equilibrium. So, T0  Mg  0  T0  Mg . (b) Since string is light and no other force is applied in between string. T  T0 throughout the Force applied by boy say F = – (force applied by string on boy)  T0  F  T0 (As string applies force as tension) -ve sign indicates opposite direction, if we consider only magnitude F  T0 * On any cross-section of string/rope tension acts at either side. * On the left part, due to right part it acts towards right * On the right part, due to left part it acts towards left. Consider cross-section AB of a string. About AB let string be divided in two parts as I and II. If tension at AB is T, means on part I it acts to the right as force T on part II it acts to the left as force T. A T T B * In case of rope (string with mass) tension may be different at different points. Illustration: A vehicle of mass M is lying over smooth horizontal surface. A light string starting from point A and passing the points B, C and D of vehicle is acted upon by a horizontal force as shown. A D F 30º B 30º C (a) Represent the tension forces acting at different point in a free body diagram. (b) Find the acceleration of the vehicle. Solution: See point (a) Tension act at all contact points at either side along string length means at point A it act only in one direction as shown.   60º T At point B it acts in two directions as shown so on for point C and D T T  T B T T (a) Now free body diagram for the whole vehicle T T (b) Fx  T Mg , T  F (applied) Fy  0 (As vehicle does not move vertically) x  Fx  F  Ma x a x  M . Illustration: Block A and B are joined through a rope R and suspended from the ceiling as shown under gravity ( g  10 m / s 2 ) through a light string. Find: (a) Tension in the light string. (b) Tension at the mid point of rope. mA=2kg mR=2kg mB=4kg Solution:(a) To get tension T in the light string Block A, rope R and block B can be treated as single body of mass M  m A  m R  m B  8 kg T Free body Diagram of this body  T  (mA  mR  mB )g  0  T  (m A  m R  m B )g  80 N MAg MRg MBg (b) Let tension at the midpoint of rope be T ’. About mid point system is divided in two parts. Free body diagram of lower part (on lower part tension acts upward) T  m R g  m 2 Bg  0 T (Net vertical force being zero) T'  T  g m R  m  MR g    2   T ' = 50 N. MBg Alternative Method: Free body diagram of upper part, as body is under equilibrium On upper part tension acts downward m  m R g  A   2  downward T  T  m  m R g T  A   2  T  T  m  m R g   A  T'  2  T  80  30N (M + MR )g SPRING FORCE: A 2 = 50 N. F Consider a light spring tied to a vertical wall, which is being pulled to right and the final elongation of spring is x and at that moment force applied is say F. What is the force applied by spring on stretching agent? Applying IIIrd law of Newton you can easily say that spring also applies force F on stretching agent. Ideal spring follows Hooke’s law which says that force applied by spring on bodies connected to it is proportional to extension or compression (Change over natural length) and is always opposite to extension or compression. So F  x  F  kx Where k is a constant that is characteristic of the spring known as spring constant or force constant. In same case if you are asked about force applied by spring on wall? It is also equal to F. Let us see it. Let force applied by spring on wall be F ’ then wall will also apply force F’ on spring. Let us see forces acting on spring. Let mass of spring be m and acceleration a, to the right F' F F  F  ma = 0. [Here ma = 0 even if a  0 since m  0  F  F ma  0 ] Hence force applied by a light spring on connecting bodies at both ends is always equal and opposite and either of this force is known as spring force. * Spring constant of a spring is inversely proporational to its relaxed length K  1 . l Illustration: If a spring of spring constant K is cut in n equal parts what is spring constant for each part. Solution: Let initial length of spring be l the length of each part after cut is l  l n K  1 l  K  l K l / n  K  nK . Illustration: An ideal spring of spring constant k = 100 N/m is pulled at both ends along length with force 20 N what is the extension of spring? Solution: kx  20N (force at either end) F' K = 100 N/m F  x   20 m  0.2 m . 100 (20 N) (20 N)   FRICTION: Take a plane sheet of paper over a smooth table and place a book over it. Now pull the sheet slowly. What do you find? Book also moves along with the sheet, let us see the motion of book, initial velocity of book is zero but it moves due to some velocity, means its velocity has changed in horizontal direction means book has horizontal acceleration. Which indicates some horizontal force must have acted on the book. What is that force. That force is friction. If I ask you whether friction opposes motion of book your answer in this case is no, as friction is the force causing motion of book. But when we carefully see, we find that when sheet wanted to move, some horizontal force acted on sheet to oppose the motion of sheet but at the same time reaction of that horizontal force in opposite direction acted on book to move it. Now one thing you can observe that friction is the force that opposes the relative motion between the bodies. friction F Friction is the force that acts at the contact points to oppose the relative motion between the surfaces in a direction parallel to surface. It always acts as action reaction pair between surfaces. STATIC FRICTION : In cases where there is no relative motion between the surfaces in contact, friction is known as static friction. (abbreviated as fs ) Static friction is an adjustable force which can take any value from zero to certain maximum values. Under a given case the maximum possible value of this static friction is known as limiting static friction.  0  f s  Limiting static friction. (Liming static friction)  (Normal reaction between the surfaces in contacts) L.S.F.  s N where s is a constant, characteristic of the surfaces in contact thus 0  f s  s N Let us consider a case to understand it. A block of mass 2 kg lies over a rough horizontal surface. The coefficient of static friction between the block and surface is s  0.5 N N  Mg 0  f s  s N  10N NMogw a horizontal force say F is applied on block. F is gradually increased from zero Friction for F = 0 f = 0 for F = 5 N f = 5 N for F = 9.99 N f = 9.99 N See friction is adjustable for F = 10 N f = 10 N when F > 10 N relative motion starts and friction becomes kinetic. KINETIC FICTION : Whenever there is relative motion between the surfaces in contact, friction is known as kinetic friction (fk ) which is a constant force such that f k  k N where k is known as the coefficient of kinetic function between the surface. In most of the cases k  s but if in a problem only  is given s  k   Let us consider the same case as discussed earlier for for F  10N f k  k N  9.6N k  0.48 Friction suddenly decreases from 10 N to 9.6 N just as motion starts for F  fLSF f L.S.F. F fs reduces from fs to f k suddenly, plot of friction against applied force is as shown above Illustration : A block of mass 2kg is lying over a rough horizontal surface at rest with s  0.5 and k  0.45 . A horizontal force F acts on block such that F increases from 5 N to 12 N continuously in certain time and then dioceses from 12 N to 5 N continuously in certain time. [g = 10 m/s2] (a) Find the acceleration of block when F = 9.5 N (increasing) (b) Find the acceleration of block when F  10.5 N (increasing) (c) Find the acceleration of block when F = 9.5 N (decreasing). Solution: Between block and the surface F.B.D. of block N f F Limiting static friction (L.S.F.) Mg  s N  0.5  N  0.5 Mg = 10 N f k (kinetic friction)  K N  0.45 Mg = 9 N (a) When F is increasing until F > L.S.F. = 10N block does not move at all thus for F  10N friction is static and adjustable and equals applied force, acceleration is zero. (b) When F = 10.5 N > L. S. F. (10N) block moves and friction becomes kinetic f = fK a  F  f K m  10.5  9 2  1.5 2 = 0.75 m/s2 (c) Once the motion starts, friction becomes kinetic block will only be accelerating for F  9N , thus when force F = 9.5 N (decreasing) f = fK a  F  f K m  9.5N  9N 2 = 0.25 m/s2. Illustration: A block of mass m rests on a rough incline of angle of inclination  (a) Find the force of friction between block and surface . Solution: (b) show that coefficient of statics friction between block and incline. s  tan  (c) Force F parallel to incline is applied down the incline. Find the minimum value of F needed to move the block. (d) A force F parallel to incline is applied up the incline find the minimum value of F needed to move the block. N f Mg sin  Mg Mg cos (a) Block doesn’t slide along the incline means net force along the incline is zero Mg sin   f  0  f  Mg sin  (b) There is no motion of the block normal to the incline means N  Mg cos  = 0  N = Mg cos  f  L.S.F.  S N  S Mg cos   Mgsin   SMg cos   S  tan  (c) When block is moved down the incline friction acts along up the incline. For motion net force down the incline > 0 Mgsin   F  SMg cos   F  Mg(S cos  sin ) N f Mg sin (d) When block is moved up the incline, friction acts down N F the incline net force up the incline > 0 F – SMg cos   Mg sin   0 fS  F  Mg(sin   S cos ) . Mg sin Mg cos Illustration: A block of mass M is at rest on a rough floor. The coefficient of static friction between block and the surface is  A pulling force F is applied at angle  with the horizontal as shown: (a) For given  , find the minimum value of F needed to move the block (b) If  is also adjustable, find the minimum value of F needed to move the block. Solution: (a) F.B.D. of block Breakup of forces along horizontal and vertical component N F f fs F cos Mg Mg fs  L.S.F. = N  (Mg  Fsin ) for motion of block net horizontal force > 0 Fcos   (Mg  Fsin ) F(cos    sin )   Mg F   Mg cos   sin  (b) (b) Fmin F   Mg cos   sin  Mg cos    sin  where   tan 1  1     As sin(  )  1 F  Mg Thus minimum force required to move the block is Mg . PSEUDO FORCES: Let us consider a straight road over which a block of mass M is lying at rest. Shyam is moving towards the block with acceleration a on his bike, while Ram is just standing as shown. What is the acceleration of block as seen by Ram obviously zero. But the acceleration of block as seen by Shyam is a towards Shyam. The acceleration of Ram with respect to Shyam is (-a) while acceleration of Shyam with respect to Ram is (a). Ram and Shyam are two observers representing frames of reference with relative acceleration. Such frames are known as non inertial frames. Let us consider the forces acting on block as seen by Ram. No force in horizontal direction explains no acceleration. N Mg But when we consider block with respect to Shyam, it has horizontal acceleration a towards Shyam so the same forces cannot explain this horizontal acceleration. Applying IInd law of motion Shyam assumes presence of additional force in the horizontal direction equal to Ma given as → → Fp  M (a) This additional force acting on body when observed by accelerated N frame is known as pseudo force given as → → Pseudo Force Fp  M (a) Where M is the mass of the body being observed (not mass of the frame) Mg and → is the acceleration of the frame (not the acceleration of body) Illustration: A block of mass 2 kg is being pushed with acceleration 4 m/s2 and block is observed by Zaheer running towards the block with acceleration 4 m/s2. Find the magnitude of pseudo force acting on the block. → → Solution: Fp  mb (af )  2  (4)  8 N → | Fp | 8 N . * To apply laws of motion in an accelerated frame pseudo force essentially has to be taken into account. But there is no change in real forces in any frame. a Illustration: A lift is moving upward with the acceleration a block of mass M is lying at the horizontal floor of lift. Find the nor- mal reaction between block and surface of lift. Solution: GROUND FRAME: In ground frame block has same acceleration as lift that is a upward. F.B.D. of block in ground frame N N– Mg = Ma  N  M(g  a) LIFT FRAME : Mg Lift is an accelerated frame (non inertial) but in the lift frame block is at rest, a = 0 F.B.D. of block. N FP  ma (downwards). N  Fp  Mg  0  N  Fp  Mg  M(g  a) Fp Normal reaction is a real force, is same in both the frame. Mg Illustration: A pendulum of mass m is hanging from the ceiling of a car having an acceleration a0 with respect to the road in the direction shown. Find the angle made by the string with the vertical. Solution : Since bob of the pendulum is stationary relative to car hence T sin ma (pseudo force) ...(i) T cos 0 = mg ...(ii) FP(ma0) Dividing (i) by (ii), we get tan  = a /g  θ  tan1 a 0 mg g DYNAMICS OF CIRCULAR MOTION : When a body moves on a circle or along arc of a circle that part of motion is known as circular motion. Direction along radius at any point of path is known as radial direction given as direction along tangent is known as tangential directions say eˆT . (ˆr) and ial (rˆ) Direction of velocity is along tangent only → ˆ v  r(eT ) If   constant v  | v | r [speed remains constant] →  a rˆ  a eˆ {a represents component of acceleration along radius, a represents component of a r T T r T acceleration along tangent) → | a | a r  2 r   v r (ˆr) {if  is constant, a r  constant} -ve sign indicates that ar is inward a  R d  R  dv {Rate of change of speed} T dt dt uniform circular motion when   constant d  0 dt means a T  0 mv 2 There is no tangential acceleration 2 Fr  ma r  r  m r FT  ma T Fnet  . Radial force required to move a body on the circular path is known as centripetal force. In case of uniform circular motion net force acts only along radius, inward as centripetal force. Observe it carefully that centripetal force is requirement of force that has to be provided by some real agent. CENTRIFUGAL FORCE: When we observe a body from a frame moving on circle, as the frame is under acceleration an additional force acts on observed body. When we observe another body also moving with observer this pseudo force is known as centrifugal force =  mv 2 → a r ) r radially outward. Illustration: A small block of mass M is moving over a smooth horizontal surface on a circular path at some moment. Length of string is L and velocity of block at some moment be v0 find the tension in the string. Solution: Here Tension provides, the required centripetal force Illustration : mv 2 T  r mv 2  0 . L O Solution: Find the tangential acceleration of the bob of mass at a moment when string makes angle  with vertical as shown. Let us make F.B.D. of bob, tension is radial, only force along tangent is FT  mg sin  Radial T  Tangential mg a  FT T m  g sin  . CONSTRAINT : When we study the motion of connected bodies, motion of these bodies may be interrelated accord- ing to existing physical conditions. Such a relation/equation is known as constraint relation. Simplest case of constraint is two blocks are connected with a light string and one of the block is being moved say A as shown. If block A moves through distance x block B also moves through the same distance x A  x B + l0 Differentiating, we get xA dx A  dx B dt dt  vA  vB [ vA and vB represents velocities of blocks A and B respectively] Again differentiating dxA  dxB dt dt  a A  a B [aA and aB represents accelerations of blocks A and B respectively] Here the constraint is length of string which is constant. Whenever bodies are attached with ideal nonstretchable strings then string length remaining constant serves as constraint. In such cases we may proceed as follows (i) Take distances of all bodies from a convenient point on ground frame (or a fixed frame) (ii) Relate displacement of individual bodies for any arbitrary case. (iii) Differentiate displacement relation with time to get velocity relation. (iv) Differentiate velocity relation with time to get acceleration relation. Illustration: Relate the acceleration of block A and B as shown. Solution: Let height of block A & B respectively be ground is fixed say H. y A and y B from ground. Height of pulley from Length of string (L) = (H  y A )  (H  y B )  2H  (y A  y B ) …(i) Differentiating eqn. (1) with time we get 0  2  0   dy A  dy B   dy A dt  dt   dy B dt dt   VA  VB V  dyA A dt represents velocity of block A in upward direction. V  dyB respects velocity of block B in upward direction. B dt Again differentiating we get dVA dt   dVB dt a A  a B - ve sign represent opposite direction of acceleration. Illustration: As shown in the figure blocks A and B are connected through light inextensible string if block A is moved to the right with acceleration of 5 m/s2. Find the acceleration of block B. 5 m/s2 P x0 Solution: Let at any moment distance of block A and B from wall be respectively x A Length of string (x B  x 0 )  x B  x A  L . 2x B  x A  L  x 0 Differentiating with respect to time we get and x B . 2 dx B  dx A  0 xA dt dt 5 m/s2  VA  2VB Again differentiation a A  2a B P x0 xB a   a A   5 m / s2 . B 2 2 Means block B is accelerated to the left with acceleration of 2.5 m/s2. CONSTRAINT FOR BODIES OF FIXED LENGTH For bodies of fixed length or an inextensible string passing through fixed bodies, component of velocity along the length A are equal. While one end moves towards string/body other  v away from the string/body. A B  vB VA and VB are respectively the instantaneous velocities of blocks VA cos 1  VB cos 2 PROOF : Displacement of A along length + displacement of B along length = 0 [in case of small time interval dt] VA ( cos 1 )dt  VB cos 2 dt  0  VA cos 1  VB cos 2 . Illustration : A sphere of mass MS is lying between a friction less vertical wall and smooth wedge. Wedge lies on smooth horizontal surface as shown. When sphere is released, find the relation between the acceleration of sphere and wedge. Solution: Here constraint is geometry of the arrangement. Sphere can move only downward while wedge can move horizontally. Let at some moment, sphere has moved through distance y vertically downward, while wedge has moved distance x to the right, so that tan   y / x Here  is angle of inclination of wedge a constant differentiating with respect to time we get we get dy  dx     tan  dt  dt  VS  Vw tan  Again differentiating with respect to time, we get dVS  dVw tan  dt dt  a S  a w tan  . STEPS TO SOLVE THE MOTION OF CONNECTED BODIES: (a) Separate the system into an isolated body. (b) Make Free Body Diagram of individual bodies to asses the forces. (c) Find the relation among the motion of different bodies through constraint (d) Frame the equation of motion in suitable direction and solve. Illustration : Find the acceleration of blocks A, B and C in given figure all surfaces are smooth and string is ideal . Given light. m A  4 kg m B  4 kg mC  2 kg all pulleys are smooth and pulley P is Solution: (a) We have to see the motion of blocks A, B and C. (b) Tension in the light string connecting blocks A and C and pulley P remains same say T and tension in string connecting pulley P to block B is say T  . F.B.D. of block A F.B.D. of block C xA xC NA NC A T T C T T xB mAg P mCg T' F.B.D. of pulley P F.B.D. of block B B 2T T' T' mBg Since pulley P is light T  2T {why?} (c) x A  x C  2x B = L (approximately net length of string) Differentiating w.r.t. time dx A dt  dx C dt  2dx B  0 dt  VA  VC  2VB Again differentiating with respect to time dVA dt  dVC dt  2 dVB  0 dt Let us assume acceleration of block A to the right as aA block C to the lift as aC and acceleration block B downward as aB then dVA dt  a A dVC dt  a C and a  dVB B dt   a A  a C  2a B  0 ...(i) (d) Block A accelerates to right T  m A a A Block C accelerates to left T  m Ca C and for block B m Bg – T( 2T)  m Ba B ...(ii) ...(iii) ...(iv) (i) , (ii), (iii) and (iv) have four unknowns and 4 equations thus we can solve it. using values of a A , a B , a C respectively from equations (ii), (iii) and (iv) in (i) T T  2T    m A m C  2g   m   0 B   T  8 / 7 g Again using equations (ii), (iii) and (iv) respectively a  2 g A 7 a  4 g C 7 a  3 g . B 7 Brain Teaser: A monkey is climbing on a rope that goes over a smooth light pulley and supports a mirror of equal mass at the other end. Show that whatever force the monkey exerts on the rope, the monkey. can not escape from its image.