Chapter-6 Newton’s Laws of Motion

NEWTON'S LAWS OF MOTION CHAPTER-6 NEWTON’S LAWS OF MOTION 1. FORCE A pull or push which changes or tends to change the state of rest or of uniform motion or direction of motion of any object is called force. Force is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Effect of resultant force : (1) (2) (3) (4) may change only speed may change only direction of motion. may change both the speed and direction of motion. may change size and shape of a body kg  m Unit of force : Newton and (MKS System) s2 g  cm s2 1 Newton = 105 dyne dyne and (CGS System) Kilogram force (kgf) : The force with which earth attracts a 1kg body towards its centre is called kilogram force, thus Forece in newton kgf = g Dimensional Formula of force : [MLT–2 ] 1.1 Fundamental Forces All the forces observed in nature such as muscular force, tension, reaction, friction, elastic, weight, electric, magnetic, nuclear, etc., can be explained in terms of only following four basic interactions: (A) Gravitational Force : The force of interaction which exists between two particles of masses m 1 and m2, due to their masses is called gravitational force. m1m2 r r3 = position vector of test particle ‘T’ with respect to source particle ‘S’. and G = universal gravitational constant = 6.67 × 10–11 Nm2/kg2. F = −G (i) It is the weakest force and is always attractive. (ii) It is a long range force as it acts between any two particles situated at any distance in the universe. (iii) It is independent of the nature of medium between the particles. An apple is freely falling as shown in figure, When it is at a height h, force between earth and apple is given by F= GMem (Re + h)2 where Me – mass of earth, Re – radius of earth. It acts towards earth’s centre. Now rearranging above result, NEWTON'S LAWS OF MOTION F=m GMe Re 2 2  Re  .   .  Re + h   Re  F = mg    Re + h  2 Here h << Re, so  GMe  g = 2  Re   Re Re + h 1  F = mg This is the force exerted by earth on any particle of mass m near the earth surface. The value of g = 9.81 m/s2 ~ − 10 m/s2  2 m/s2 ~ − 32 ft/s2. It is also called acceleration due to gravity near the surface of earth. (B) Electromagnetic Force : Force exerted by one particle on the other because of the electric charge on the particles is called electromagnetic force. Following are the main characteristics of electromagnetic force (a) These can be attractive or repulsive. (b) These are long range forces (c) These depend on the nature of medium between the charged particles. (d) All macroscopic forces (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the force of friction, normal reaction, muscular force, and force experienced by a deformed spring are electromagnetic forces. These are manifestations of the electromagnetic attractions and repulsions between atoms/molecules. (C) Nuclear Force : It is the strongest force. It keeps nucleons (neutrons and protons) together inside the nucleus inspite of large electric repulsion between protons. Radioactivity, fission, and fusion, etc. result because of unbalancing of nuclear forces. It acts within the nucleus that too upto a very small distance. (D) Weak Force : It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range of weak force is very small, in fact much smaller than the size of a proton or a neutron. It has been found that for two protons at a distance of 1 Fermi : FN : FEM : FW : FG : : 1 : 10–2 : 10–7 : 10–38 1.2 Classification of forces on the basis of contact : (A) Field Force : Force which acts on an object at a distance by the interaction of the object with the field produced by other object is called field force. Examples (a) Gravitation force (b) Electromagnetic force (B) Contact Force : Forces which are transmitted between bodies by short range atomic molecular interactions are called contact forces. When two objects come in contact they exert contact forces on each other. Examples : (a) Normal force (N) : It is the component of contact force perpendicular to the surface. It measures how strongly the surfaces in contact are pressed against each other. It is the electromagnetic force. A table is placed on Earth as shown in figure Here table presses the earth so normal force exerted by four legs of table on earth are as shown in figure. Now a boy pushes a block kept on a frictionless surface. NEWTON'S LAWS OF MOTION Here, force exerted by boy on block is electromagnetic interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal force. A block is kept on inclined surface. Component of its weight presses the surface perpendicularly due to which contact force acts between surface and block. Normal force exerted by block on the surface of inclined plane is shown in figure. Force acts perpendicular to the surface Example 1. Two blocks are kept in contact on a smooth surface as shown in figure. Draw normal force exerted by A on B. Solution : In above problem, block A does not push block B, so there is no molecular interaction between A and B. Hence normal force exerted by A on B is zero. Note : Normal is a dependent force, it comes in role when one surface presses the other. (b) Tension : Tension in a string is a electromagnetic force. It arises when a string is pulled. If a massless string is not pulled, tension in it is zero. A string suspended by rigid support is pulled by a force ‘F’ as shown in figure, for calculating the tension at point ‘A’ we draw F.B.D. of marked portion of the string; Here string is massless. F.B.D. of marked portion T A F  T=F String is considered to be made of a number of small segments which attracts each other due to electromagnetic nature as shown in figure. The attraction force between two segments is equal and opposite due to Newton’s third law. NEWTON'S LAWS OF MOTION For calculating tension at any segment, we consider two or more than two parts as a system. Here interaction between segments are considered as internal forces, so they are not shown in F.B.D. (C) Frictional force : It is the component of contact force tangential to the surface. It opposes the relative motion (or attempted relative motion) of the two surfaces in contact. 2. THIRD LAW OF MOTION : To every action, there is always an equal and opposite reaction. Newton’s law from an 1803 translation from Latin as Newton wrote “To every action there is always opposed an equal and opposite reaction : to the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.” 2.1 Important points about the Third Law (a) The terms ‘action’ and ‘reaction’ in the Third Law mean nothing else but ‘force’. A simple and clear way of stating the Third Law is as follows : Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A. (b) The terms ‘action’ and ‘reaction’ in the Third Law may give a wrong impression that action comes before reaction i.e. action is the cause and reaction the effect. There is no such cause-effect relation implied in the Third Law. The force on A by B and the force on B by A act at the same instant. Any one of them may be called action and the other reaction. (c) Action and reaction forces act on different bodies, not on the same body. Thus if we are considering the motion of any one body (A or B), only one of the two forces is relevant. It is an error to add up the two forces and claim that the net force is zero. However, if you are considering the system of two bodies as a whole, FAB (force on A due to B) and FBA (force on B due to A) are internal forces of the system (A + B). They add up to give a null force. Internal forces in a body or a system of particles thus cancel away in pairs. This is an important fact that enables the Second Law to be applicable to a body or a system of particles. 3. SYSTEM : 3.1 Two or more than two objects which interact with each other form a system. Classification of forces on the basis of boundary of system : (A) Internal Forces : Forces acting each with in a system among its constituents. (B) External Forces : Forces exerted on the constituents of a system by the outside surroundings are called as external forces. (C) Real Force : Force which acts on an object due to other object is called as real force. An isolated object (far away from all objects) does not experience any real force. NEWTON'S LAWS OF MOTION 4. 4.1 FREE BODY DIAGRAM A free body diagram consists of a diagrammatic representations of a single body or a subsystem of bodies isolated from its surroundings showing all the forces acting on it. Steps for F.B.D. Step 1 : Identify the object or system and isolate it from other objects clearly specifying its boundary. Step 2 : First draw non-contact external force in the diagram. Generally it is weight. Step 3 : Draw contact forces which acts at the boundary of the object or system. Contact forces are normal, friction, tension and applied force. In F.B.D, internal forces are not drawn, only external are drawn. Example 2. Solution : A block of mass ‘m’ is kept on the ground as shown in figure. (i) Draw F.B.D. of block. (ii) Are forces acting on block action–reaction pair. (iii) If answer is no, draw action reaction pair. (i) F.B.D. of block (ii) ‘N’ and mg are not action-reaction pair. Since pair act on different bodies, and they are of same nature. (iii) Pair of ‘mg’ of block acts on earth in opposite direction. and pair of ‘N’ acts on surface as shown in figure. Example 3. Two sphere A and B are placed between two vertical walls as shown in figure. Draw the free body diagrams of both the spheres. Solution : F.B.D. of sphere ‘A’ : F.B.D. of sphere ‘B’ : (exerted by A) Note : Here NAB and NBA are the action–reaction pair (Newton’s third law). NEWTON'S LAWS OF MOTION 5. NEWTON’S LAWS OF MOTION : 5.1 First Law of Motion Each body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise. Newton’s first law is really a statement about reference frames in that it defines the types of reference frames in which the laws of Newtonian mechanics hold. From this point of view the first law is expressed as: If the net force acting on a body is zero, it is possible to find a set of reference frames in which that body has no acceleration. Newton’s first law is sometimes called the law of inertia and the reference frames that it defines are called inertial reference frames. Newton’s law from an 1803 translation from Latin as Newton wrote “Every body preserves in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.” Examples of this law : (a) A bullet fired on a glass window makes a clean hole through it while a stone breaks the whole of it. The speed of bullet is very high. Due to its large inertia of motion, it cuts a clean hole through the glass. When a stone is thrown, it inertia is much lower so it cannot cut through the glass. (b) A passenger sitting in a bus gets a jerk when the bus starts or stops suddenly. 5.2 Second Law of Motion : The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts. Newton’s law from an 1803 translation from Latin as Newton wrote “The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.” Mathematically dp F= dt F = ma Or where p = mv , p = Linear momentum. In case of two particles having linear momentum P1 and P2 and moving towards each other under mutual forces, from Newton’s second law ; dp2 d 1 + =0 (p1 + p2 ) = F = 0  dp dt dt dt F1 + F2 = 0 5.3  F2 = −F1 which is Newton’s third law. Important points about second law (a) The Second Law is obviously consistent with the First Law as F = 0 Implies a = 0. (b) The Second Law of motion is a vector law. It is actually a combination of three equations, one for each component of the vectors : dp y dp dp Fy = = ma y Fx = x = max Fz = z = maz dt dt dt This means that if a force is not parallel to the velocity of the body, but makes some angle with it, it changes only the component of velocity along the direction of force. The component of velocity normal to the force remains unchanged. (c) The Second Law of motion given above is strictly applicable to a single point mass. The force F in the law stand for the net external force on the particle and a stands for the acceleration of the particle. Any internal forces in the system are not to be included in F. NEWTON'S LAWS OF MOTION (d) The Second Law of motion is a local relation. What this means is that the force F at a point in space (location of the particle) at a certain instant of time is related to a at the same point at the same instant. That is acceleration here and now is determined by the force here and now not by any history of the motion of the particle. 5.4 Applications of Newton’s Laws (a) When objects are in equilibrium To solve problems involving objects in equilibrium: Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the free-body diagram for the object. Label all external forces acting on it. Step 3 : Choose a convenient coordinate system and resolve all forces into rectangular components along x and y direction. Step 4 : Apply the equations  Fx = 0 and  Fy = 0 . Step 5 : Step 4 will give you two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities. Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities. Example 4. A ‘block’ of mass 10 kg is suspended with string as shown in figure. Find tension in the string. (g = 10 m/s2) Solution : F.B.D. of block Fy = 0 T – 10 g = 0  T = 100 N Example 5. The system shown in figure is in equilibrium. Find the magnitude of tension in each string ; T1, T2, T3 and T4. (g = 10 m/s–2) Solution : F.B.D. of block 10 kg T 0 = 10 g T0 = 100 N NEWTON'S LAWS OF MOTION F.B.D. of point ‘A’ Fy = O T2 cos 30º = T0 = 100 N T2 = 200 3 N Fx = O T1 = T2 sin 30º = 200 1 100 = N. 2 3 . 3 F.B.D. of point ‘B’ Fy = O  T4 cos600 = T2cos300 and Fx = O T3 = Example 6. 200 3 N,  T3 + T2 sin300 = T4sin600 T4 = 200 N Two blocks are kept in contact as shown in figure. Find 50N 0 100 N 10kg 20kg /////////////// 30 (a) forces exerted by surfaces (floor and wall) on blocks. (b) contact force between two blocks. Solution : F.B.D. of 10 kg block N1 = 10 g = 100 N .....(1) N2 = 100 N .....(2) N4 F.B.D. of 20 kg block  N3 = 100 – 25 = 75 N 50 N 0 N2 = 50 sin 30º + N3 30 .....(3) N2 N3 and N4 = 50 cos 30° + 20 g N4 = 243.30 N Example 7. Find magnitude of force exerted by string on pulley. Solution : F.B.D. of 10 kg block : T = 10 g = 100 N F.B.D. of pulley : Since string is massless, so tension in both sides of string is same. Force exerted by string = (100)2 + (100)2 = 2 100 N Note : Since pulley is in equilibrium position, so net 20 g NEWTON'S LAWS OF MOTION forces on it is zero. Hence force exerted by hinge on it is 100 2 N. (b) Accelerating Objects To solve problems involving objects that are in accelerated motion : Step 1 : Make a sketch of the problem. Step 2 : Isolate a single object and then draw the free-body diagram for that object. Label all external forces acting on it. Be sure to include all the forces acting on the chosen body, but be equally carefully not to include any force exerted by the body on some other body. Some of the forces may be unknown; label them with algebraic symbols. Step 3 : Choose a convenient coordinate system, show location of coordinate axes explicitly in the free-body diagram, and then determine components of forces with reference to these axes and resolve all forces into x and y components. Step 4 : Apply the equations Fx = max and Fy = may Step 5 : Step 4 will give two equations with several unknown quantities. If you have only two unknown quantities at this point, you can solve the two equations for those unknown quantities. Step 6 : If step 5 produces two equations with more than two unknowns, go back to step 2 and select another object and repeat these steps. Eventually at step 5 you will have enough equations to solve for all unknown quantities. Example 8. A force F is applied horizontally on mass m1 as shown in figure. Find the contact force between m1 and m2. Solution : Considering both blocks as a system to find the common acceleration. Common acceleration F a= .....(1) (m1 + m2 ) To find the contact force between ‘A’ and ‘B’ we draw F.B.D. of mass m2. F.B.D. of mass m2 Fx = max N = m2 .a m2F N= (m1 + m2 ) Example 9. Solution : The velocity of a particle of mass 2 kg is given by. Find the force acting on the particle. From second law of motion : F= dP d d = (mv) = 2. ( atiˆ + bt 2 ˆj )  F = 2aiˆ + 4btjˆ dt dt dt Example 10. A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s2 by an external force F0. (a) What is F0 ? (b) What is the net force on rope ? (c) What is the tension at middle point of the rope ? (g = 10 m/s2) Solution : For calculating the value of F0, consider two blocks with the rope as a system. F.B.D. of whole system NEWTON'S LAWS OF MOTION (a) F0 – 100 = 10 × 2 F = 120 N .....(1) (b) According to Newton’s second law, net force on rope. F = ma = (2) (2) = 4 N .....(2) (c) For calculating tension at the middle point we draw F.B.D. of 3 kg block with half of the rope (mass 1 kg) as shown. T – 4g = 4.(2) ; T = 48 N Example 11. Solution : A block of mass 50 kg is kept on another block of mass 1 kg as shown in figure. A horizontal force of 10 N is applied on the 1kg block. (All surface are smooth). Find (g = 10 m/s2) (a) Acceleration of block A and B. (b) Force exerted by B on A. (a) F.B.D. of 50 kg N2 = 50 g = 500 N along horizontal direction, there is no force aB = 0 (b) F.B.D. of 1 kg block : along horizontal direction 10 = 1 aA. aA = 10 m/s2 along vertical direction  N1 = N2 + 1g = 500 + 10 = 510 N N1 N2 10N ////////////////////////////////////////// 1g Example 12. A horizontal force is applied on a uniform rod of length L kept on a frictionless surface. Find the tension in rod at a distance ‘x’ from the end where force is applied. Solution : Considering rod as a system, we find acceleration of rod a = F M now draw F.B.D. of rod having length ‘x’ as shown in figure. Using Newton’s second law M F – T =   x.a  L Example 13. T = F− M F x.  L M x T = F(1 − ) . L One end of string which passes through pulley and connected to 10 kg mass at other end is pulled by 100 N force. Find out the acceleration of 10 kg mass. (g = 9.8 m/s2) NEWTON'S LAWS OF MOTION Solution : Since string is pulled by 100 N force. So tension in the string is 100 N. F.B.D. of 10 kg block 100 – 10 g = 10 a 100 – 10 × 9.8 = 10 a a = 0.2 m/s2. Example 14. Two blocks m1 and m2 are placed on a smooth inclined plane as shown in figure. If they are released from rest. Find : (i) Acceleration of mass m1 and m2 (ii) Tension in the string (iii) Net force on pulley exerted by string Solution : F.B.D. of m1 : m1gsin – T = m1a 3 g–T = 2 3a .....(1) F.B.D. of m2 : T – m2gsin = m2a 3 g = 1.a 2 T – 1. .....(2) Adding eq.(1) and (2) we get a = 0 Putting this value in eq.(i) we get T= 3g , 2 F.B.D. of pulley FR = FR = Example 15. 2T 3 g 2 A 60 kg painter stands on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. NEWTON'S LAWS OF MOTION Solution : 6. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter. (ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1s? (iii) What force should he apply now to maintain the constant speed of 1 m/s? The free body diagram of the painter and the platform as a system can be drawn as shown in the figure. Note that the tension in the string is equal to the force by which he pulls the rope. (i) Applying Newton’s Second Law 2T – (M + m)g = (M + m)a 2T − (M + m)g or a = M+m Here M = 60 kg; m = 15 kg; T = 400 N g = 10 m/s2 2(400) − (60 + 15)(10) a= = 0.67 m/s2 60 + 15 (ii) To attain a speed of 1 m/s in one second, the acceleration a must be 1 m/s 2. Thus, the applied force is 1 1 F= (M + m) (g + a) = (60 + 15) (10 + 1) = 412.5 N 2 2 (iii) When the painter and the platform move (upward) together with a constant speed, it is in a state of dynamic equilibrium. Thus, 2F – (M + m) g = 0 (M + m)g (60 + 15)(10) or F = = = 375 N 2 2 WEIGHING MACHINE : A weighing machine does not measure the weight but measures the force exerted by object on its upper surface. Example 16. A man of mass 60 Kg is standing on a weighing machine placed on ground. Calculate the reading of machine (g = 10 m/s2). Solution : For calculating the reading of weighing machine, we draw F.B.D. of man and machine separately. F.B.D. of man F.B.D. of weighing machine N = Mg Here force exerted by object on upper surface is N Reading of weighing machine N = Mg = 60 × 10 N = 600 N. NEWTON'S LAWS OF MOTION 7. SPRING FORCE : Every spring resists any attempt to change its length; when it is compressed or extended, it exerts force at its ends. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm-1). When spring is in its natural length, spring force is zero. F=0 Example 17. Solution : F = – kx Two blocks are connected by a spring of natural length 2 m. The force constant of spring is 200 N/m. Find spring force in following situations : (a) If block ‘A’ and ‘B’ both are displaced by 0.5 m in same direction. (b) If block ‘A’ and ‘B’ both are displaced by 0.5 m in opposite direction. (a) Since both blocks are displaced by 0.5 m in same direction, so change in length of spring is zero. Hence, spring force is zero. (b) In this case, change in length of spring is 1 m. In case of extension or compression of spring, spring force is F = Kx = (200).(1) F = 200 N Example 18. Force constant of a spring is 100 N/m. If a 10 kg block attached with the spring is at rest, then find extension in the spring. (g = 10 m/s2) Solution : In this situation, spring is in extended state so spring force acts in upward direction. Let x be the extension in the spring. F.B.D. of 10 kg block : Fs = 10g  Kx = 100  (100)x = (100)  x = 1m Two blocks ‘A’ and ‘B’ of same mass ‘m’ attached with a light spring are suspended by a string as shown in figure. Find the acceleration of block ‘A’ and ‘B’ just after the string is cut. Example 19. NEWTON'S LAWS OF MOTION Solution : When block A and B are in equilibrium position F.B.D of ‘B’ .....(1) F.B.D of ‘A’ .....(2) When string is cut, tension T becomes zero. But spring does not change its shape just after cutting. So spring force acts on mass B, again draw F.B.D. of blocks A and B as shown in figure F.B.D. of ‘B’ F.B.D. of ‘A’ 7.1 Spring Balance : It does not measure the weight. t measures the force exerted by the object at the hook. Symbolically, it is represented as shown in figure. A block of mass ‘m’ is suspended at hook. When spring balance is in equilibrium, we draw the F.B.D. of mass m for calculating the reading of balance. F.B.D. of ‘m’. mg – T = 0 T = mg Magnitude of T gives the reading of spring balance. Example 20. Solution : A block of mass 20 kg is suspended through two light spring balances as shown in figure. Calculate the (1) reading of spring balance (1). (2) reading of spring balance (2). For calculating the reading, first we draw F.B.D. of 20 kg block. NEWTON'S LAWS OF MOTION F.B.D of 20 kg. mg – T = 0 T = 20 g = 200 N Since both balances are light so, both the scales will read 20 kg 8. CONSTRAINED MOTION: 8.1 String Constraint : When two objects are connected through a string and if the string have the following properties: (a) The length of the string remains constant i.e. inextensible string. (b) Always remains tight, does not slacks. Then the parameters of the motion of the objects along the length of the string and in the direction of extension have a definite relation between them. Steps for String Constraint Step 1. Identify all the objects and number of strings in the problem. Step 2. Assume variable to represent the parameters of motion such as displacement, velocity acceleration etc. (i) Object which moves along a line can be specified by one variable. (ii) Object moving in a plane are specified by two variables. (iii) Objects moving in 3-D requires three variables to represent the motion. Step 3. Identify a single string and divide it into different linear sections and write in the equation format. 1+2 +3 +4 +5 +6 =  Step 4. Differentiate with respect to time d 1 d 2 d 3 + + + .... = 0 dt dt dt d 1 = represents the rate of increment of the portion 1, end points are always in contact with some dt d 1 = V1 + V2 dt Take positive sign if it tends to increase the length and negative sign if it tends to decrease the length. Here +V1 represents that upper end is tending to increase the length at rate V1 and lower end is tending to increase the length at rate V2. Step 5. Repeat all above steps for different-different strings. Let us consider a problem given below object so take the velocity of the object along the length of the string Here 1 + 2 = constant d 2 d 1 + =0 dt dt (V1 – VP) + (VP – V2) = 0 a + a2 V + V2 Vp = 1 Similarly, aP = 1 Remember this result 2 2 VP VP 1 V1 2 V2 NEWTON'S LAWS OF MOTION Example 21. Solution : Two blocks of masses m1 and m2 are attached at the ends of an inextensible string which passes over a smooth massless pulley. If m1 > m2, find : (i) the acceleration of each block (ii) the tension in the string. m2 The block m1 is assumed to be moving downward and the block m2 m1 is assumed to be moving upward. It is merely an assumption and it does not imply the real direction. If the values of a1 and a2 come out to be positive then only the assumed directions are correct; otherwise the body moves in the opposite direction. Since the pulley is smooth and massless, therefore, the tension on each side of the pulley is same. The free body diagram of each block is shown in the figure. F.B.D. of m2 F.B.D. of m1 Applying Newton’s second Law on blocks m1 and m2 Block m1 m1g – T = m1a1 ....(1) Block m2 –m2g + T= m2a2 ....(2) Number of unknowns : T, a1 and a2 (three) Number of equations: only two Obviously, we require one more equation to solve the problem. Note that whenever one finds the number of equations less than the number of unknowns, one must think about the constraint relation. Now we are going to explain the mathematical procedure for this. How to determine Constraint Relation ? (1) Assume the direction of acceleration of each block, e.g. a1 (downward) and a2 (upward) in this case. (2) Locate the position of each block from a fixed point (depending on convenience), e.g. centre of the pulley in this case. (3) Identify the constraint and write down the equation of constraint in terms of the distance assumed. For example, in the chosen problem, the length of string remains constant is the constraint or restriction. Thus, x1 + x2 = constant dx1 dx 2 Differentiating both the sides w.r.t. time we get + =0 dt dt Each term on the left side represents the velocity of the blocks. Since we have to find a relation between accelerations, therefore we differentiate it once again w.r.t. time. d2 x 2 d2 x 1 + =0 2 dt dt 2 Since, the block m1 is assumed to be moving downward (x1 is increasing with time) Thus  d2 x 1 = + a1 dt 2 and block m2 is assumed to be moving upward (x2 is decreasing with time) NEWTON'S LAWS OF MOTION d2 x 2 = – a2 dt 2 Thus a1 – a2 = 0 or a1 = a2 = a (say) is the required constraint relation. Substituting a1 = a2 = a in equations (1) and (2) and solving them, we get  Example 22. Solution :  m − m2   2m1m2  (i) a =  1 (ii) T =   g  g  m1 + m2   m1 + m2  A system of three masses m1, m2 and m3 are shown in the figure. The pulleys are smooth and massless; the strings are massless and inextensible. (i) Find the tensions in the strings. (ii) Find the acceleration of each mass. All the blocks are assumed to be moving downward and the free body diagram of each block is shown in figure. F.B.D. m3 F.B.D. m2 F.B.D. m1 F.B.D. of pulley Applying Newton’s Second Law to Block m1 : m1g – T1 = m1a1 ....(1) Block m2 : m2g – T1 = m2a2 ....(2) Block m3 : m3g – T2 = m3a3 ....(3) Pulley : T2 = 2T1 ....(4) Number of unknowns a1, a2, a3, T1 and T2 (Five) Number of equations : Four The constraint relation among accelerations can be obtained as follows For upper string x3 + x0 = c1 For lower string x2 – x0) + (x1 – x0) = c2 x2 + x1 – 2x0 = c2 Eliminating x0 from the above two relations, we get x1 + x1 + 2x3 = 2c1 + c2 = constant. Differentiating twice with respect to time, we get d2 x 3 d2 x 1 d2 x 2 + + 2 =0 dt 2 dt 2 dt 2 NEWTON'S LAWS OF MOTION or a1 + a2 + 2a3 = 0 Solving equations (1) to (5), we get ......(5)   4m1m2m3 (i) T1 =   g ; T2 = 2T1  4m1m2 + m3 (m1 + m2 )   4m m + m1m3 − 3m2m3   3m1m3 − m2m3 − 4m1m2  (ii) a1 =  1 2 g  g ; a2 =   4m1m2 + m3 (m1 + m2 )   4m1m2 + m3 (m1 + m2 )   4m m − m3 (m1 + m2 )  a3 =  1 2 g  4m1m2 + m3 (m1 + m2 )  Example 23. The figure shows one end of a string being pulled down at constant velocity v. Find the velocity of mass ‘m’ as a function of ‘x’. Solution : Using constraint equation 2 x2 + b2 + y = length of string = constant Differentiating w.r.t. time :  dy     =v  dt  8.2 2 2 x +b 2 2  dx   dy  .2x  +  =0  dt   dt  v  dx  x 2 + b2   = −  2x  dt  Wedge Constraint : Conditions : (i) There is a regular contact between two objects. (ii) Objects are rigid. The relative velocity perpendicular to the contact plane of the two rigid objects is always zero if there is a regular contact between the objects. Wedge constraint is applied for each contact. In other words, Components of velocity along perpendicular direction to the contact plane of the two objects is always equal if there is no deformations and they remain in contact. NEWTON'S LAWS OF MOTION Example 24. A rod of mass 2m moves vertically downward on the surface of wedge of mass as shown in figure. Find the relation between velocity of rod and that of the wedge at any instant. Solution : Using wedge constraint. Component of velocity of rod along perpendicular to inclined surface is equal to velocity of wedge along that direction. u cos  = v sin  u = tan  v u = v tan  9. NEWTON’S LAW FOR A SYSTEM Fext = m1a1 + m2a2 + m3a3 + ...... Fext = Net external force on the system. m1, m2, m3 are the masses of the objects of the system and a1,a2 ,a3 are the acceleration of the objects respectively. Example 25. The block of mass m slides on a wedge of mass ‘m’ which is free to move on the horizontal ground. Find the accelerations of wedge and block. (All surfaces are smooth). Solution : Let . a  acceleration of wedge b  acceleration of block with respect to wedge Taking block and wedge as a system and applying Newton’s law in the horizontal direction Fx = m1 a1x + m2 a2x = 0 0 = ma + m(a – b cos ) .....(1) here ‘a’ and ‘b’ are two unknowns , so for making second equation, we draw F.B.D. of block. F.B.D of block. using Newton’s second law along inclined plane mg sin  = m ( b – a cos ) .....(2) Now solving equations (1) and (2) we will get a= mgsin  cos  gsin  cos  2gsin  = and b = m(1 + sin2 ) (1 + sin2 ) (1 + sin2 ) So in vector form : NEWTON'S LAWS OF MOTION  gsin  cos   ˆ a wedge = aiˆ =   i 2  1 + sin   a block = −  a block = (a – b cos) î – b sin ĵ gsin  cos  ˆ 2gsin2  ˆ ij . (1 + sin2 ) (1 + sin2 ) Example 26. For the arrangement shown in figure when the system is released, find the acceleration of wedge. Pulley and string are ideal and friction is absent. Solution : Considering block and wedge as a system and using. Newton’s law for the system along x-direction T = Ma + m (a – b cos ) F.B.D of m along the inclined plane mg sin  – T = m (b – a cos ) using string constraint equation.    .....(ii) 1 + 2 = constant d2 2 d2 1 + =0 2 dt dt 2 b–a=0 Solving above equations (i),(ii) & (iii) , we get mgsin  a= M + 2m(1 − cos ) 10. .....(i) .....(iii) NEWTON’S LAW FOR NON INERTIAL FRAME : FReal + FPseudo = ma Net sum of real and pseudo force is taken in the resultant force. a = Acceleration of the particle in the non inertial frame FPseudo = − m aFrame Pseudo force is always directed opposite to the direction of the acceleration of the frame. Pseudo force is an imaginary force and there is no action-reaction for it. So it has nothing to do with Newton’s Third Law. 10.1 Reference Frame: A frame of reference is basically a coordinate system in which motion of object is analyzed. There are two types of reference frames. (a) Inertial reference frame: Frame of reference either stationary or moving with constant velocity. (b) Non-inertial reference frame: A frame of reference moving with non-zero acceleration. NEWTON'S LAWS OF MOTION Example 27. A lift having a simple pendulum attached with its ceiling is moving upward with constant acceleration ‘a’. What will be the tension in the string of pendulum with respect to a boy inside the lift and a boy standing on earth, mass of bob of simple pendulum is m. Solution : F.B.D . of bob (with respect to ground) T – mg = ma T = mg + ma .....(i) With respect to boy inside the lift, the acceleration of bob is zero. So he will write above equation in this manner. T – mg = m. (0)  T = mg He will tell the value of tension in string is mg. But this is ‘wrong’. To correct his result, he makes a free body diagram in this manner, and uses Newton’s second law. T = mg + ma .....(ii) By using this extra force, equations (i) and (ii) give the same result. This extra force is called pseudo force. This pseudo force is used when a problem is solved with a accelerating frame (Non-inertial) Note : Magnitude of Pseudo force = mass of system × acceleration of frame of reference. Direction of force: Opposite to the direction of acceleration of frame of reference, (not in the direction of motion of frame of reference) Example 28. A box is moving upward with retardation ‘a’ F1) are applied at the free ends of uniform rod kept on a horizontal frictionless surface. Find tension in rod at a distance x from end ‘A’, NEWTON'S LAWS OF MOTION (F2 − F1 ) .x L Answer : T = F2 − Solution : F −F a= 2 1 m L F2 F1 T x T – F1 = m2a T F1 F2 m1  T – F1 = F −F m (L − x) 2 1 L m m2 (m2 = m (L − x) ) L A 10 kg block kept on an inclined plane is pulled by a string applying 200 N force. A 10 N force is also applied on 10 kg block as shown in figure. Find : (a) tension in the string. (b) acceleration of 10 kg block. (c) net force on pulley exerted by string Answer : (a) 200 N, (b) 14 m/s2, (c) 200 2 N Solution : (a) T = 200 N  T – 10 – 50 = 10a  200 – 60 = 10a (c) (FR) = m = gsin 50  N (b) T – 10 – mgsin = ma  a= T= 20 0N 90° T 10 Problem 8. kg x x x   T = F1 +  1 −  (F2 – F1) = F1 + F2 – F1 – (F2 – F1) = F2 – (F2 – F1) L L  L 10N mg = 100 N 140 = 14 m/s2 10 (200)2 + (200)2 = 200 2 N Ans. 30° 90 200 o 200 N (FR ) Problem 9. A man of mass 60 kg is standing on a weighing machine (2) of mass 5kg placed on ground. Another similar weighing machine is placed over man’s head. A block of mass 50kg is put on the weighing machine (1). Calculate the readings of weighing machines (1) and (2) (g = 10 m/s2) Answer : 500 N, 1150 N N1= 50g Solution : weighing machine 1 N2 m1 g R1 = N1 = 50 × g = 500 N NEWTON'S LAWS OF MOTION where R1 = reading in weighing machine 1 N3= (50+5+60)g weighing machine 2 N4 m2 g R2 = N3 = (50 + 5 +60) g = 115 × 10 = 1150 N where R2 = reading in weighing machine 2 Problem 10. Answer : Solution : Two blocks are connected by a spring of natural length 2 m. The force constant of spring is 200 N/m. Find spring force in following situations : (a) A is kept at rest and B is displaced by 1 m in right direction. (b) B is kept at rest and A is displaced by 1m in left direction. (c) A is displaced by 0.75 m in right direction, and B is 0.25 m in left direction. (a) F = 200 N, (b) 200 N, (c) 200 N (a) Extension in spring = 1 m.  Fspring = Kx = 200 × 1 = 200 N 2m A B (b) Same extension same spring force in both directions Fspring = 200 N. (c) Both displacements of A of B are compressing the spring total compressing = 0.75 + 0.25 = 1 m.  Fspring = kx = 200 × 1 = 200 N. Problem 11. If force constant of spring is 50 N/m. Find mass of the block, if it is at rest in the given situation. (g = 10 m/s2) Answer : m = 10 kg Solution : T = 100 N  mg = 100 N 100 m= = 10 kg. g Problem 12. T mg Two blocks ‘A’ and ‘B’ of same mass ‘m’ attached with a light string are suspended by a spring as shown in figure Find the acceleration of block ‘A’ and ‘B’ just after the string is cut.. NEWTON'S LAWS OF MOTION Answer : Solution : g (upwards), g (downwards) When string is not cut : kx A m FBD of ‘A’ block kx = mg + T T mg ......(i) T B m mg F. B.D of ‘B’ block T = mg When string is cut : ......(ii) Kx A m aA mg FBD of ‘A’ block kx – mg = maA 2mg – mg = maA = aA = g (upwards) FBD of ‘B’ block maB = mg aB = g (downwards) B m aB mg Problem 13. Find the reading of spring balance in the adjoining figure, pulley and strings are ideal. Answer : 2F Solution : FBD of spring balance R R=T FBD of pulley T .....(i) NEWTON'S LAWS OF MOTION T F F T=2F R = 2F .....(ii) Problem 14. The figure shows mass m moves with velocity u. Find the velocity of ring at that moment. Ring is restricted to move on smooth rod. Answer : VR = Solution : Problem 15. Answer : Solution : u , VR = 2u cos  Velocity along string remains same . VR cos = u u VR = cos   = 60o VR = 2u In the system shown in figure, the block A is released from rest. Find : (i) The acceleration of both blocks ‘A’ and ‘B’. (ii) Tension in the string. (iii) Contact force between ‘A’ and ‘B’. g mg g g 2mg (i) ˆi − ˆj , î (ii) (iii) 3 3 3 3 3 (i) Let acceleration of blocks in x & y directions are b m B Taking both blocks as a system m A a NEWTON'S LAWS OF MOTION T b T T = 2mb .....(i) Taking A block : mg – T = ma .....(ii) From equations (i) & (ii) ; ma + 2 mb = mg a+2b=g .....(iii) From string constraint ; a=b .....(iv) g From equations (iii) & (iv);a = b = 3 Hence, acceleration of block A g g aA = bî – a ĵ aA = î– ĵ 3 3 g Acceleration of block B aB = bî = î 3 2mg (ii) T = 2mb = 3 (iii) For contact force between ‘A’ and ‘B’ mg FBD of block ‘A’ : N = mb N= 3 T A a mg T b N A m a mg Problem 16. A block of mass 2 kg is kept at rest on a big box moving with velocity 2iˆ and having acceleration −3iˆ + 4jˆ m/s2. Find the value of ‘Pseudo force’ acting on block with respect to box Answer : ˆ Fpseudo = −maframe = −2(−3iˆ + 4j) F = 6iˆ − 8ˆj .