Newton's Laws of Motion & Friction PDF

TG: @Chalnaayaaar PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Newton's Laws of motion & Friction ENGLISH MEDIUM  TG: @Chalnaayaaar All rights including trademark and copyrights a...

TG: @Chalnaayaaar PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Newton's Laws of motion & Friction ENGLISH MEDIUM  TG: @Chalnaayaaar All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reproduced, adapted, abridged or translated, transcribed, transmitted, stored or distributed in any form retrieval system, computer system, photographic or other system or transmitted in any form or by any means whether electronic, magnetic, chemical or manual, mechanical, digital, optical, photocopying, recording or otherwise, or stood in any retrieval system of any nature without the written permission of the Allen Career Institute Private Limited. Any breach will entail legal action and prosecution without further notice. 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Any violation or infringement of the propriety rights of Allen shall be punishable under Section- 29 & 52 of the Trademark Act, 1999 and under Section- 51, 58 & 63 of the Copyright Act, 1957 and any other Act applicable in India. All disputes are subjected to the exclusive jurisdiction of courts, tribunals and forums at Kota, Rajasthan only. Note:- This publication is meant for educational and learning purposes. All reasonable care and diligence have been taken while editing and printing this publication. ALLEN Career Institute Private Limited shall not hold any responsibility for any error that may have inadvertently crept in. ALLEN Career Institute Private Limited is not responsible for the consequences of any action taken on the basis of this publication. ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Isaac Newton (1642 - 1727) Isaac Newton was born in Woolsthorpe, England in 1642, the year Galileo died. His extraordinary mathematical ability and mechanical aptitude remained hidden from others in his school life. In 1662, he went to Cambridge for undergraduate studies. A plague epidemic in 1665 forced the university town to close and Newton had to return to his mother's farm. There in two years of solitude, his dormant creativity blossomed in a deluge of fundamental discoveries in mathematics and physics : binomial theorem for negative and fractional exponents, the beginning of calculus, the inverse square law of gravitation, the spectrum of white light, and so on. Returning to Cambridge, he pursued his investigations in optics and devised a reflecting telescope. In 1684, encouraged by his friend Edmund Halley, Newton embarked on writing what was to be one of the greatest scientific works ever published : The Principia Mathematica. In it, he enunciated the three laws of motion ® and the universal law of gravitation, which explained all the three Kepler's laws of planetary motion. The book was packed with a host of path-breaking achievements : basic principles of fluid mechanics, mathematics of wave motion, calculation of masses of the earth, the sun and other planets, explanation of the precession of equinoxes, theory of tides, etc. In 1704, Newton brought out another masterpiece Optics that summarized his work on light and colour. The scientific revolution triggered by Copernicus and steered vigorously ahead by Kepler and Galileo was brought to a grand completion by Newton. Newtonian mechanics unified terrestrial and celestial phenomena. The same mathematical equation governed the fall of an apple to the ground and the motion of the moon around the earth. Galileo Galilei (1564 - 1642) Galileo Galilei, born in Pisa, Italy in 1564 was a key figure in the scientific revolution in Europe about four centuries ago. Galileo proposed the concept of acceleration. From experiments on motion of bodies on inclined planes or falling freely, he contradicted the Aristotelian notion that a force was required to keep a body in motion, and that heavier bodies fall faster than lighter bodies under gravity. He thus arrived at the law of inertia that was the starting point of the subsequent epochal work of Isaac Newton. Galileo's discoveries in astronomy were equally revolutionary. In 1609, he designed his own telescope (invented earlier in Holland) and used it to make a number of startling observations : mountains and depressions on the surface of the moon; dark spots on the sun; the moons of Jupiter and the phases of Venus. He concluded that the Milky Way derived its luminosity because of a large number of stars not visible to the naked eye. In his masterpiece of scientific reasoning : Dialogue on the Two Chief World Systems, Galileo advocated the heliocentric theory of the solar system proposed by Copernicus, which eventually got universal acceptance. With Galileo came a turning point in the very method of scientific inquiry. Science was no longer merely observations of nature and inferences from them. Science meant devising and doing experiments to verify or refute theories. Science meant measurement of quantities and a search for mathematical relations between them. Not undeservedly, many regard Galileo as the father of modern science. 164  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical NEWTON'S LAWS OF MOTION ARISTOTLE'S FALLACY According to Aristotle, a constant continuous force is required to keep a body in uniform motion. This is called Aristotle's Fallacy. 1. NEWTON'S FIRST LAW OF MOTION (OR GALILEO'S LAW OF INERTIA) Every body continues its state of rest or uniform motion in a straight line unless compelled by an external force to change its state.  This law defines the force and states that "force is a factor which can change the state of object." Definition of force from Newton's first law of motion "Force is the push or pull which changes or tends to change the state of rest or of uniform motion". 2. FORCE ® Any push or pull which either changes or tends to change the state of rest or of uniform motion (constant velocity) of a body is known as force. Effects of Resultant force :- A non zero resultant force may produce the following effects on a body : (i) It may change the speed of the body.   (ii) It may change the direction of motion. (iii) It may change both the speed and direction of motion. (iv) It may change the size or/and the shape of the body.   Units for measurement of force :- Absolute units Other units (i) N (M.K.S) kg-wt or kg-f (kg-force) (ii) dyne (C.G.S) g-wt or g-f Relation between above units :- 1 kg-wt = 9.8 N 1 g-wt = 980 dyne 5 1 N = 10 dyne 3. INERTIA Inertia is the property of a body due to which it opposes any change in its state. Mass of a body is the measure of its inertia of translational motion. It is difficult to change the state of rest or uniform motion of a body of heavier mass and vice-versa.  Mass of a body is quantitative or numerical measure of a body's inertia.  Larger the inertia of a body, more will be its mass. Inertia of rest : It is the inability of a body to change its state of rest by itself. Examples :  When we shake a branch of a mango tree, the mangoes fall down.  When a bus or train starts suddenly, the passengers sitting inside tends to fall backwards.  When a horse starts off suddenly, its rider falls backwards.  A coin is placed on cardboard and this cardboard is placed over a tumbler such that coin is above the mouth of tumbler. Now if the cardboard is removed with a sudden jerk, then the coin falls into the tumbler.  The dust particles in a blanket fall off when it is beaten with a stick. 165 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Inertia of motion : It is the inability of a body to change its state of uniform motion by itself. Examples :  When a bus or train stops suddenly, the passengers sitting inside lean forward.  A person who jumps out of a moving train may fall in the forward.  A bowler runs with the ball before throwing it, so that his speed of running gets added to the speed of the ball at the time of throw.  An athlete runs through a certain distance before taking a long jump because the velocity acquired during the running gets added to the velocity of athlete at the time of jump and hence he can jump over a longer distance.  A ball is thrown in the upward direction by a passenger sitting inside a moving train. The ball will fall :-  back to the hands of the passenger, if the train is moving with constant velocity.  ahead of the passenger, if the train is retarding (slowing down) behind of the passenger, if the train is accelerating (speeding up) ® Inertia of direction : It is the inability of a body to change its direction of motion by itself Examples :  When a straight running car turns sharply, the person sitting inside feels a force radially outwards.  Rotating wheels of vehicle throw out mud, mudguard fitted over the wheels prevent this mud from spreading.  When a knife is pressed against a grinding stone, the sparks produced move in the tangential direction. 4. MOMENTUM The total quantity of motion possessed by a moving body is known as the momentum of the body. It is the product of the mass and velocity of a body. It is a vector quantity whose direction is along the instantaneous velocity.   momentum p = m v SI Unit : kg–m/s –1 Dimension : [M L T ] 5. NEWTON'S SECOND LAW OF MOTION According to Newton, the rate of change of momentum of any system is directly proportional to the applied external force and this change in momentum takes place in the direction of the applied force.  dp d    dv  dm =F = (mv) or = F m +v (general form) dt dt dt dt  dm  dv If m = constant then =0 If v = constant then =0 dt dt     dm dv  ⇒= = ma F m ⇒ F=v (e.g. conveyor belt, rocket propulsion) dt dt Illustrations Illustration 1.  ˆ N produces an acceleration of A force F = (6iˆ − 8ˆj + 10k) 2 2 m/s in a body. Calculate the mass of the body. Solution   From Newton's IInd law |F||ma| = = ma   |F| |F| 62 + 82 + 102 ⇒ Acceleration a = ⇒ = m = = 10 kg m a 2 166  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical Illustration 2. A 5kg block is resting on a frictionless plane. It is struck by a jet, releasing water at the rate of 3kg/s emerging with a speed of 4m/s. Calculate the initial acceleration of the block. Solution dm Force exerted on block F = v = 4 × 3 = 12 N 5kg dt F 12 so acceleration of the block = = 2 a = 2.4m/s m 5 Illustration 3. A force of 50 N acts in the direction as shown in figure. The block is of mass 5kg, resting on a smooth horizontal surface. Find out the acceleration of the block. 50N 60° ® Solution a 50 3 5kg Horizontal component of the force = 50 sin 60° = 2 component of force in the direction of acceleration 50 3 1 acceleration of the block, a = = ×= 5 3 m/s2 mass 2 5 6. IMPULSE When a large force is applied on a body for a very short interval of time, then the product of force and time interval is known as impulse.      dp   = dI Fdt = dp =  dt F   Units of impulse = N–s or kg-m/s. 1 1 –2 1 1 1 –1 Dimensions of impulse = [F] [t] = [m] [a] [t] = [M L T T ] = [M L T ] Case-I : If this force is working from time t1 to t2, then integrating the above equation, we get -   t2  p2    ∫ ∫ I = F dt = dp = p2 − p1  t1 p1 ⇒ Impulse = Change in momentum Case-II : If a constant or average force acts on body, then :-    t2  p2    t2 p2  ∫ ∫ I = Favg dt = dp  ⇒= avg ∫ dt I F= t1 ∫ dp  p1 t1 p1         =I F= [ ]t 2 avg t t1 [p]pp 2 1 ⇒ I = Favg (t2 – t1 )= ( p2 − p1 )    I =Favg ∆t =∆p 167 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical 6.1 Impulse-Momentum Theorem The impulse of force is equal to the change in momentum. This relation is known as impulse-momentum theorem. 6.2 Law of Conservation of linear momentum If net external force on a system is zero then the linear momentum of the system remains constant. According to Newton's IInd Law.   dp Fext = dt   If Fext = 0 then  dp     =0 ⇒ p = constant or pinitial = p final dt      ⇒ p1 + p2 + p3 + p4 +...... + pn = constant (Conservation of linear momentum )    ∆ ( p1 + p2 +..... + pn ) = ® ⇒ 0    ⇒ ∆p1 + ∆p2 +..... + ∆pn =0   for two-particles system p1 + p2 = constant   ∆p1 + ∆p2 = 0   ∆p1 = −∆p2 (change in momentum of I particle = –change in momentum of II particle) st nd GOLDEN KEY POINTS Important points about Newton's Second Law of Motion  Newton's first law of motion defines force and second law of motion measures force. It gives the units, dimensions and magnitude of the force. Unit of force = (unit of mass) × (unit of acceleration) = 1kg × 1 m/s2 = 1N 2 1N = 1 kg-m/s 1 dyne = 1 g-cm/s2 1 1 –2 1 1 –2 Dimensions of force :- = [m] [a] = [M ] [L T ] = [M L T ]  If a particle moves uniformly, means velocity = constant      dv d(constant)   =a = = 0 So, F = ma = m × 0 = 0 dt dt It means that, in the absence of external force, a particle moves uniformly. This is Newton's first law of motion. It means that, we can derive mathematically Newton's first law of motion with the help of Newton's second law of motion,  Accelerated motion is always due to an external force.  Law of Conservation of Linear Momentum (COLM) is applicable in the direction in which external force is zero, i.e. dp x If Fx = 0 ⇒ = 0, then p x = constant dt dp y If Fy = 0 ⇒ 0, = then p y = constant dt dp z If Fz = 0 ⇒ = 0, then p z = constant dt 168  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical  The slope of momentum-time graph is equal to the force on the particle p dp e.g. Here, = F = slope = tan θ dt θ t  Area under the force-time graph represents impulse or change in momentum. F Area = I or ∆p = +ve e.g. t Area = I or ∆p = –ve Here, I or ∆= p ∫ Fdt = Area under F-t graph ∆p ® Note :- Since Favg = ∆t therefore, for a certain momentum change if the time interval is increased, then the average force exerted on body will decrease. Examples (i) A cricketer lowers his hands while catching a ball so as to increase the time interval of momentum change consequently the average reaction force on his hands decreases. So he can save himself from getting hurt. (ii) Shockers are provided in vehicles to avoid jerks. (iii) Buffers are provided in boggies of train to avoid jerks. (iv) A person jumping on a hard cement floor receives more injury than a person jumping on muddy or sandy road. Illustrations Illustration 4. A hammer of mass 1 kg moving with a speed of 6 m/s strikes a wall and comes to rest in 0.1 s. Calculate. (a) impulse of the force (b) average retarding force that stops the hammer. (c) average retardation of the hammer Solution (a) Impulse = F × ∆t = m (v – u) = 1 (0 – 6) = – 6 N-s Impulse 6 (b) Average retarding force that stops the hammer F = = = 60 N time 0.1 F 60 (c) Average retardation = a = = 60 m/s2 m 1 Illustration 5. mv cosθ A ball of 0.20 kg hits a wall with a velocity of 25 m/s at an angle of 45°. If the ball rebounds at 90° to the direction of incidence, calculate the magnitude of change in v θ momentum of the ball. θ −mv cosθ Solution Change in momentum = (–mv cos 45°) – (mv cos 45°) = – 2mv cos 45° v  θ=45° 1 ∆p = 2mvcos45° = 2 × 0.2 × 25 × = 5 2 N-s 2 169 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Illustration 6. A cricket ball of mass 150 g is moving with a velocity of 12 m/s and is hit by a bat so that the ball gets turned back with a velocity of 20 m/s. If the duration of contact between the ball and bat is 0.01 s, find the impulse and the average force exerted on the ball by the bat. Solution According to given problem change in momentum of the ball ∆p = pf – pi = m(v – u) = 150 × 10–3 [20 – (–12)] So by impulse-momentum theorem, Impulse Ι = ∆p = 4.8 N–s Ι ∆p 4.80 And by time average definition of force in case of impulse F= av = = = 480 N ∆t ∆t 0.01 Illustration 7. Figure shows an estimated force–time graph for a base ball struck by a bat. From this curve, determine (a) Impulse delivered to the ball (b) Average force exerted on the ball. B ® 18000 Solution (a) Impulse = Area under F-t curve Force 12000 1 (in N) = Area of ∆ABC = 18000 × (2.5 – 1) 2 6000 A C = 1.35 × 104 kg–m/s 0 1 1.5 2 2.5 3 Ιmpulse 1.35 × 104 Time (in sec.) (b) Average force = = = 9000 N Time (2.5 − 1) BEGINNER'S BOX-1 1. A force of 72 dynes is inclined at an angle of 60° to the horizontal, find the acceleration in a mass of 9 g which moves under the effect of this force in the horizontal direction. 2. A constant retarding force of 50N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. What time does the body take to stop ? 3. A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 25 s in the direction of the motion of the body. What is the magnitude and direction of the force? 4. A body of mass 5kg is acted upon by two perpendicular forces of 8N and 6N, find the magnitude and direction of the acceleration. 5. A ball of mass 1 kg dropped from 9.8 m height, strikes the ground and rebounds to a height of 4.9 m. If the time of contact between ball and ground is 0.1 s, then find impulse and average force acting on ball. 6. A machine gun fires a bullet of mass 50 gm with velocity 1000 m/s. If average force acting on gun is 200 N then find out the maximum number of bullets fired per minute. 7. A machine gun has a mass of 20 kg. It fires 35 g bullets at the rate of 400 bullets per minute with a speed of 400m/s. What average force must be applied to the gun to keep it in position ? 8. A force of 10N acts on a body for 3 µs. If mass of the body is 5 g, calculate the impulse and the change in velocity. 9. A body of mass 0.25 kg moving with velocity 12 m/s is stopped by applying a force of 0.6 N. Calculate the time taken to stop the body. Also calculate the impulse of this force. 170  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical 10. A batsman deflects a ball by an angle of 90° without changing C 800 its initial speed, which is equal to 54 km/hr. What is the 600 impulse imparted to the ball ? (Mass of the ball is 0.15 kg) Force F(N) 400 11. The magnitude of the force (in newton) acting on a body A B varies with time t (in microsecond) as shown in fig. AB, BC 200 and CD are straight line segments. Find the magnitude of the D 0 2 4 6 8 10 12 14 16 total impulse of the force (in N–s) on the body from t = 4 µs Time (µs) to t = 16 µs. 7. ROCKET PROPULSION R Case-I : If rocket is accelerating upwards, then - O Rocket C Net upwards force on rocket = ma vrel dm (wt. = mg) K dt ® E dm v rel – mg = ma T dt Where vrel is the relative velocity of the ejected mass w.r.t rocket gases Case-II : If rocket is moving with constant velocity, then a = 0 vrel dm dm dt vrel = mg dt Illustrations Illustration 8. A 600 kg rocket is set for a vertical firing. If the exhaust speed of gases with respect to rocket is 1000 m/s, then calculate the mass of gas ejected per second to supply the thrust needed to overcome the weight of rocket. Solution dm Force required to overcome the weight of rocket F = mg and thrust needed = vrel dt dm dm mg 600 × 9.8 so vrel = mg ⇒ = = = 5.88 kg/s dt dt v rel 1000 8. NEWTON'S THIRD LAW OF MOTION According to Newton's third law, to every action, F F there is always an equal (in magnitude) and A B opposite (in direction) reaction. This law is also known as action-reaction law. F F A B  Here, F12 (force on first body due to second body)  The forces between two is equal in magnitude and opposite in direction to F21 objects A and B are equal and opposite, whether they are (force on second body due to first body). attractive or repulsive.   F12 = −F21 171 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Important points about Newton's III law (i) We cannot produce a single isolated force in nature. Forces are always produced in action-reaction pair. (ii) There is no time gap in between action and reaction. So we cannot say that action is the cause and reaction is its effect. Any one force can be action and the other reaction. (iii) Action - reaction law is applicable on both the states either at rest or in motion. (iv) Action and reaction is also applicable between bodies which are not in physical contact. (v) Action - reaction law is applicable to all the interaction forces eg. gravitational force, electrostatic force, electromagnetic force, tension, friction, viscous force, etc. (vi) Action and reaction never cancel each other because they act on two different bodies. Examples : Walking, Swimming, Recoiling of gun when a bullet is fired from it, Rocket propulsion. ® GOLDEN KEY POINTS First law : If no net force acts on a particle, then it is possible to select a set of reference frames, called inertial reference frames, observed from which the particle moves without any change in velocity. Second law : Observed from an inertial reference frame, the net force on a particle is equal to the d(mv) rate of change of its linear momentum w.r.t. time : dt Third law : Whenever a particle A exerts a force on another particle B, simultaneously B exerts a force on A with the same magnitude in the opposite direction. Newton's III law can be derived from principle of conservation of linear momentum. If two particles of masses m1 and m2 are moving under the action of their mutually interacting forces with each other, such that no external force acts on the system, then ∴ momentum of system remains constant. → → → → → → ∆p1 ∆p i.e. ∆p1 + ∆p 2 =0 ⇒ ∆p1 = − ∆p 2 ⇒ = − 2 ∆t ∆t → → F 12 = − F 21 st nd nd st Force on 1 due to 2 = –Force on 2 due to 1 Inertial and Gravitational mass F (1) The ratio of force applied on a particle to its acceleration is known as inertial mass m i = a (2) The ratio of gravitational force to gravitational acceleration is known as gravitational mass F mg = g It is experimentally proved that both masses are equal i.e. mi = mg 172  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical 9. FREE BODY DIAGRAM A diagram showing all external forces acting on an object is called "Free Body Diagram" (F.B.D.) In a specific problem, we are required to choose a body first and then we access the different forces acting on it, and all the forces are drawn on the body, The resulting diagram is known as free body diagram (FBD). For example, if two bodies of masses m and M resting on a smooth floor are in contact and a force F is applied on M from the left as shown in figure (a), the free body diagrams of M and m will be as shown in figures (b) and (c). N1 N2 F f1 M f1 m F M m Mg mg (a) (b) (c) Important Point : ® Two forces in Newton's third law never occur in the same free–body diagram. This is because a free–body diagram shows forces acting on a single object, and the action–reaction pair in Newton's third law always act on different objects. 10. NORMAL REACTION When a stationary body is placed on a surface then, that surface exerts a contact force on that body which is perpendicular to the surface and towards the body. This force is known as Normal Reaction e.g. A block of mass "m" is placed on a horizontal table. Then the forces exerted on it are : (1) Downward gravitational force of attraction on body due to earth, means weight of the block (=mg). (2) Upward normal reaction exerted by the surface of table on the block. Since the body is in equilibrium, therefore net force on it is zero. N    ∴N+W= 0   N = −W m   N=W  N = − W  ∴ N= mg [ W = mg] W = mg 10.1 Effective or Apparent weight of a man in lift N Case-I : If the lift is at rest or moving uniformly (a = 0), then Lift Wapp = N N = mg Wactual = mg Man So, Wapp = Wactual Case-II : If the lift is accelerating upwards, then - mg Net upward force on man = ma N N – mg = ma N = mg + ma ∴N = m (g +a) a motion Wapp or N = m (g +a) So, Wapp > Wactual mg 173 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Case-III : If the lift is retarding upwards, then N mg – N = ma mg – ma = N a motion ∴ Wapp or N = m (g – a) So, Wapp < Wactual mg Case-IV : If the lift is accelerated downwards, then - N mg – N = ma N = mg – ma a motion Wapp or N = m (g - a) ® So, Wapp < Wactual mg Case-V : If the lift is retarding downwards, then - N N – mg = ma N = ma + mg a motion Wapp or N = m (g + a) So, Wapp > Wactual mg Two Special Cases of downward acceleration (IV Case) I Special Case :- If the lift is falling freely, it implies that its acceleration is equal to the acceleration due to gravity. i.e. a = g, then - Wapp. = m (g–a) = m (g–g) = m × 0 = 0 ∴ Wapp. = 0 It means the man will feel weightless. This condition is known as Condition of Weightlessness. The apparent weight of any freely falling body is zero. II Special Case :- If the lift is accelerating downwards with an acceleration which is greater than 'g', then the man will move up with respect to the lift and stick to the ceiling. 10.2 Bird-Cage Problem : I. A bird is sitting on the base in an air tight cage. Now, if the bird starts flying, then – (i) Weight of system will not change, if the bird flies with constant velocity. (ii) Weight of system will increase, if the bird flies with upward acceleration. (iii) Weight of system will decrease, if the bird flies with downward acceleration. II. A bird is sitting on the base in a wire cage. Now if the birds flies upward its weight will decrease in all the cases. 174  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical 11. MOTION OF BODIES IN CONTACT (CONTACT FORCE) Two bodies in contact : a F N N m1 m2 Free body diagrams : a a F - N = m1 a.....(1) F m1 S m2 N = m2 a.....(2) N On adding the above equations F = m1a + m2a F = a (m1+m2) ® F Fnet ⇒ a=....(3) OR a= m1 + m2 m total putting the value of 'a' from equation (3) in (2), we get m2 F N= (here N = contact force) m1 + m2 Three bodies in contact : a N2 N2 N1 N1 F m1 m2 m3 F – N1 = m1 a....(1) a a N1– N2 = m2 a....(2) a m3 N2 = m3 a....(3) F m1 N1 N1 m2 N2 N2 On adding the above equations F = m1a + m2a + m3a F = a (m1+m2+m3) F FNet ⇒ a=....(4) or a= m1 + m2 + m3 m total Putting the value of 'a' from equation (4) in (3), we get Fm3 N2 = m1 + m2 + m3 Putting the value of 'a' from (4) in (1); F – m1a = N1 Fm1 (m2 + m3 )F N1 = F – ⇒ N1 = m1 + m2 + m3 m1 + m2 + m3 175 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Illustrations Illustration 9. Consider a box of mass 10 kg resting on a horizontal table and acceleration due to gravity to be 10 m/s2. (a) Draw the free body diagram of the box. (b) Find value of the force exerted by the table on the box. (c) Find value of the force exerted by the box on the table. (d) Does force exerted by table on the box and weight of the box form third law action-reaction pair? Solution N1 (a) N1 : Force exerted by table on the box.   (b) The block is in equilibrium. ∑ F =0 ⇒ mg − N1 =0 ⇒ N1 =100 N (c) N1 = 100 N : Because force by table on the box and force by box on table make mg = 100 N Newton's third law pair. ® (d) No Illustration10. a Two blocks of masses m = 2 kg and M = 5 kg are in contact on a frictionless table. A horizontal force F (= 35 N) is applied to m. F m M Find the force of contact between the blocks. Will the force of contact remain same if F is applied to M ? Solution As the blocks are rigid both will move with same acceleration under a the action of a force F a F NM NM M F 35 m =a = = 5 m/s² m+M 2+5 Force of contact NM = Ma = 5 × 5 = 25 N a a If the force is applied to M then its action on m will be Nm m Nm M F Nm = ma = 2 × 5 = 10 N. Note :- From this problem it is clear that acceleration does not depend on the fact that whether the force is applied to m or M, but force of contact does. Illustration 11. A spring weighing machine inside a stationary lift reads 50 kg when a man stands on it. What would happen to the scale reading if the lift is moving upward with (i) constant velocity, and (ii) constant acceleration? Solution (i) In the case of constant velocity a = 0 Therefore, N = mg so Wapp = Wact hence reading = 50 kg (ii) In case with upward acceleration N – mg = ma So N = mg + ma = m(g + a) So Wapp. > Wact 50(g + a)  a Hence scale reading will be = = 50  1 +  kg. g  g 176  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical Illustration 12. Three blocks of masses m1 = 1 kg, m2 = 1.5 kg and m3 = 2 F m2 m3 kg are in contact with each other on a frictionless surface as m1 shown in fig. Find the (a) horizontal force F needed to push the blocks as a single unit with an acceleration of 4 m/s² (b) resultant force on each block and (c) magnitude of contact forces between the blocks. Solution a a a (a) F = (m1 + m2 + m3) a F N1 N1 N2 N2 m3 m2 = (1 + 1.5 + 2) × 4 m1 = 4.5 × 4 = 18 N F - N1 = m1a N1 - N2 = m2a N2 = m3a (b) For m1 F – N1 = m1a = 1 × 4 ® ⇒ F – N1 = 4 N......(i) for m2, N1 – N2 = m2 a = 1.5 × 4 = 6 ⇒ N1 – N2 = 6N......(ii) for m3, N2 = m3a = 2 × 4 ⇒ N2 = 8N......(iii) (c) Contact force between m2 and m3 is N2 = 8 N and contact force between m1 and m2 is N1 = N2 + 6 = 8 + 6 = 14N. BEGINNER'S BOX-2 1. A person of mass M kg is standing on a lift. If the lift moves vertically upwards according to given v-t graph then find out v(ms–1) the weight of man at the following instants : (g = 10 m/s2) 20 (i) t = 1 second (ii) t = 8 seconds (iii) t = 12 seconds 1 2 8 10 12 14 t(s) 2. Find the acceleration of the system and the contact force between (i) 2kg and 3 kg blocks (ii) 3kg and 5 kg blocks 5kg 10N 2kg 3kg 3. Calculate : (i) asystem (ii) FDE (iii) FCD (iv) FBC (v) FAB corrosponding to the following diagram. A B C D E P m m m m m 177 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical 12. SYSTEM OF MASSES TIED BY STRINGS F 12.1 Tension in a String T T It is the intermolecular forces between the molecules of a string,which become active when the string is streched. Important points about the tension in a string : (a) Force of tension act on a body in the direction away from the point of contact or tied end of the string. a a (b) String is assumed to be inextensible so that the magnitude of F M2 M1 accelerations of the blocks tied to the strings are always same. T T (c) (i) If the string is massless and frictionless, tension throughout the string remains same. Massless and T frictionless string T ® T1 T1 (ii) If the string is massless but not frictionless, at every contact Massless string but T2 tension changes. there is friction between string and pulley T2 (iii) If the string is not light, tension at each point of the string T1 T2 will be different depending on the acceleration. There is friction between T3 string and pulley and string T4 used is not light. (d) If a force is directly applied to a string, say a man is F pulling a string from the other end with some force, T then tension will be equal to the applied force T Box T=F irrespective of the motion of the pulling agent, irrespective of whether the box moves or not, man Man moves or not. Tmin Tmax (e) String is assumed to be massless unless stated, hence M F tension in it remains the same every where and equal to the applied force. However, if a string has a mass, tension at different points will be different being maximum (= applied force) at the end through which force is applied and minimum at the other end connected to a body. (f) In order to produce tension in a string two equal and opposite stretching T T F F forces must be applied. The tension thus produced is equal in magnitude to either applied force (i.e., T = F) and is directed inwards opposite to F. (g) Every string can bear a maximum tension, i.e. if the tension in a string is continuously increased it will break beyond a certain limit. The maximum tension which a string can bear without breaking is called its "breaking strength". It is finite for a string and depends on its material and dimensions. 178  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical 12.2 Motion of Connected Bodies Two Connected Bodies : a T T F m1 m2 Free body diagram :- a a F – T = m1 a....(1) F m1 T T m2 T = m2 a....(2) On adding the above equations F = m1a + m2a F = a (m1+m2) ® F Fnet a=...(3) Or a= m1 + m2 m total Putting the value of 'a' from equation (3) in (2), we get - m2 F T= m1 + m2 Three Connected Bodies : Free body diagrams :- T1 T1 m2 T2 T2 m3 F m1 a a a T1 T1 T2 T2 m3 F m1 m2 F – T1 = m1a.....(1) T1– T2 = m2a.....(2) T2 = m3a.....(3) On adding above equations F = (m1 + m2 + m3) a F Fnet ∴ a= or a= m1 + m2 + m3 m total Put the value of 'a' in equation (1) T1 = F – m1a m1F m F + m 2 F + m 3 F − m1F ⇒ T1 = F – = 1 m1 + m2 + m3 m1 + m 2 + m 3 T1 = ( m2 + m3 ) F m1 + m 2 + m 3 Similarly, putting the value of 'a' in equation (3) m3F T2 = m1 + m 2 + m 3 179 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical 12.3 Tension in a Rod Given L F a M L Mass of Rod = M T F T Length of Rod = L x a F – T = ma T F ⇒ T = F – ma x F  F ⇒T=F–m  a = M  M   ® M M  Mass of length 'L' = M ∴ Mass of unit length = ∴ Mass (m) of length 'x' = x L L Put this value of 'm' in equation (1) M  F  x T=F–  x ⇒ = T F 1 −   L  M  L  12.4 Bodies Hanging Vertically T3 T3 m3 Since all the bodies are in equilibrium, therefore net force m3 m3g + T2 on each body is zero T2 T2 T1 = m1g....(1) m2 m2 T2 = T1 + m2g m2g + T1 T1 ⇒ T2 = (m1 + m2)g....(2) T1 m1 T3 = T2 + m3g m1 ⇒ T3 = (m1 + m2 + m3)g m1g 12.5 Bodies Accelerating Vertically Upwards For m1 T1 – m1g = m1a F T3 ⇒ T1 = m1 (g+a)....(1) T3 a m3 m3 m3g + T2 For m2 T2 – m2g – T1 = m2a a T2 T2 ⇒ T2 = m2a + m2g + T1 m2 m2 a ⇒ T2 = m2 (g + a) + m1 (g + a) (from equation 1) m2g + T1 T1 T1 ⇒ T2 = (m1 + m2) (g + a)....(2) m1 a m1 For m3 F = T3 and T3 – m3g – T2 = m3a m1 g ⇒ T3 = T2 + m3g + m3a ⇒ T3 = (m1 + m2) (g + a) + m3 (g + a) (from equation 2) ⇒ T3 = (m1 + m2 + m3) (g + a)....(3) 180  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical 12.6 Bodies Accelerating Vertically Downwards For m1 m1g – T1 = m1a F T3 a m3 ⇒ T1 = m1 (g – a)....(1) T3 m3 m3g + T2 For m2 m2g + T1 – T2 = m2a a T2 T2 m2 a m2 ⇒ T2 = m2g – m2a + T1 m2g + T1 T1 ⇒ T2 = m2 (g – a) + m1 (g – a) (from equation 1) T1 m1 a m1 ⇒ T2 = (m1 + m2) (g – a)....(2) m1g For m3 F = T3 and m3g + T2 – T3 = m3a ® ⇒ T3 = m3g – m3a + T2 ⇒ T3 = m3 (g – a) + (m1 + m2) (g – a) (from equation 2) ⇒ T3 = (m1 + m2 + m3) (g – a)....(3) GOLDEN KEY POINTS  If several spring balances are connected in series, then the reading of each balance is the same and is equal to the applied load (Note : Spring balances have negligible mass so they are assumed massless) e.g. S1 = 10 kg–wt S2 = 10 kg–wt 10 kg  If several spring balances are connected in parallel and symmetrically to the load, then the applied load is applied Load equally divided in all the balances; so the reading of each balance will be = no.of balances e.g. S1 = 5 kg-wt S2 = 5 kg-wt 10 kg 181 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical Illustrations Illustration 13. Three blocks, are connected by strings as shown in the figure below, and are pulled by a force T3 = 120 N. If m1 = 5 kg, m2 = 10 kg and m3 = 15 kg. Calculate the acceleration of the system and tensions T1 and T2. a a a T1 m2 T2 m3 T3 m1 Solution F 120 (i) Acceleration of = the system a = = 4 m/s² m1 + m2 + m3 5 + 10 + 15 (ii) T1 = m1a = 5 × 4 = 20 N T2 = (m1 + m2)a = (5 + 10) 4 = 60 N Illustration 14. ® A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m as shown in fig. A horizontal force F is applied to one end of the rope. Find the (i) Acceleration of the rope and the block (ii) Force that the rope exerts on the block. (iii) Tension in the rope at its mid point. M F m Solution F (i) Acceleration a = (m + M) a (ii) Force exerted by the rope on the block is a M.F M T T F = = T Ma L (m + M) FBD of block FBD of rope m   m + 2M   F  (iii) T1  + M  a =  =  m+M a 2   2    T1 L/2 T1 L/2 f (m + 2M)F M Tension in rope at midpiont is T1 = m/2 m/2 2(m + M) Illustration 15. The system shown in fig. is in equilibrium. If the spring balance is calibrated in newtons, what does it record in each case? (g = 10 m/s²) Solution T T T T one weight acts as and the other as weight 30° so tension T = 10 g (C) 10 kg = 100 N 10 kg T =10 × 10 × sin30° (A) 10 kg 10 kg = 10 × 10 × ½ = 50 N (B) T = 2 × 10 × g = 2 × 10 × 10 = 200 N 182  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical BEGINNER'S BOX-3 1. In the given figure determine T1 : T2 : T3 T3 5kg T2 3kg T1 2kg 2. Find the tension in the chain at a distance Y from the support. Mass of chain is M. Y L ® 3. Calculate TP and TQ (P is mid point) a0=0.2m/s2 Q g=9.8m/s2 1.9kg P 0.2 kg 2.9 M 4. A uniform rope of mass M and length L is placed on a smooth T F horizontal surface. A horizontal force F is acting at one end of rope. x Calculate the tension in the rope at a distance x as shown. L 5. Calculate the tension T in the rope at a distance x from end A in the B M A F2 F1 x following diagram assuming F1 > F2. L 6. A man in a lift carrying a 5 kg bag. If the lift moves vertically downwards with g/2 acceleration. Find the tension in the handle of the bag. 7. Calculate T for the following diagram : a P T T m1 m2 8. Calculate TA, TB, TC for the following diagram : (B is the mid point of rope) 24N 60° 2 kg 5 kg C A 5 kg B 9. A dynamometer is attached to two blocks of masses 6 kg and 4 kg. 6 kg 4 kg F = 20 N F = 10 N Forces of 20 N and 10 N are applied on the blocks as shown in figure. Find the dynamometer reading in the steady state. 183 ®  TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical 13. PULLEY SYSTEMS The only function of pulley (which has  Ideal pulley is considered massless and frictionless. no friction on its axle to retard rotation)  Ideal string is massless and inextensible. is to change the direction of force  A pulley may change the direction of force in the through the cord that joins the two string but not the tension. blocks. Some Cases of Pulley Case I Case II m1 = m2 = m m1 > m2 Tension in the string T now for mass m1, T a T = mg m m1 g – T = m1a.........(i) T a Acceleration 'a' = zero m for mass m2 T m2 Reaction at the point of suspension of T – m2 g = m2 a.........(ii) m1 the pulley or thrust on pulley. R = 2T = 2 mg. adding (i) and (ii) ® (m1 − m2 ) 2m1m2 2W1 W2 a= = g and T = g (m1 + m2 ) (m1 + m2 ) W1 + W2 net pulling force acceleration = total mass to be pulled 2 × Product of masses Tension = g Sum of masses Reaction at the suspension point of pulley (or Thrust on pulley) 4m1m2g 4W1W2 R = 2T = = (m1 + m2) W1+W2 Case III : For mass m1 : T = m1 a a For mass m2 : m2g – T = m2 a m1 T m2 g m1m2 acceleration a = and T = g (m1 + m2 ) (m1 + m2 ) R = 2T a Reaction at suspension point of pulley R = 2T m2 Case IV : (m1 > m2) a m1g – T1 = m1a (i) T2 – m2g = m2a (ii) T1 M T2 T1 – T2 = Ma (iii) adding (i), (ii) and (iii) T1 T2 (m1 − m2 ) a= g m1 m2 (m1 + m2 + M) Case V : Mass suspended over a pulley along with m1g m2g another on an inclined plane. For mass m1 : m1g – T = m1 a For mass m2 : T – m2g sinθ = m2 a (m − m2 sin θ) acceleration a = 1 g  (m1 + m2 ) T m1 a m1m2 (1 + sin θ)   T= g (m1 + m2 ) θ m2 g m1 g 184  Physics : Newton's laws of motion & Friction TG: @Chalnaayaaar ® Pre-Medical Case VI : Masses m1 and m2 are connected by a string passing over a pulley m1sinα > m2sinβ m1gsinα – T = m1a.......(i) T – m2g sinβ = m2a.......(ii) After solving equation (i) and (ii) (m1 sin α − m2 sin β) Acceleration a = g (m1 + m2 ) α m1 g m2 g β m1m2 (sin α + sin β)  Tension T= g (m1 + m2 ) Case VII : For mass m1 : T1 – m1g = m1a T1 For mass m2 : m2g + T2 – T1 = m2a T a a m1 A For mass m3 : m3g – T2 = m3a m2 ® m1 g m2 g (m2 + m3 − m1 ) ⇒ a= g T (m1 + m2 + m3 ) m3 we can calculate tensions T1 and T2 from above equations m3 g Illustrations Illustration 16. A block of mass 25 kg is raised in two different ways by a F 50 kg man as shown in fig. What is the action in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the yielding of the floor ? Mg Solution F Mg Mass of the block, m = 25 kg ; mass of the man, M = 50 kg mg mg (a) (b) Force applied to lift the block F = mg = 25 × 9.8 = 245 N Weight of the man, Mg = 50 × 9.8 = 490 N (a) When the block is raised by the man by applying a force F in the upward direction, reaction being equal and opposite to F will act on the floor in addition to the weight of the man. ∴ Action on the floor Mg + F = 490 + 245 = 735 N (b) When the block is raised by the man applying force F over the rope (passing over the pulley) in the downward direction, reaction being equal and opposite to F will act on the floor against the weight of the man. ∴ Action on the floor Mg – F = 490 – 245 = 245 N since floor yields to a normal force of 700 N, mode (b) should be adopted by the man to lift the block. Illustration 17. In the adjacent figure, masses of A, B and C are 1 kg, 3 kg and 2 kg respectively. Find (a) the acceleration of the system and 2 60° 30° (b) the tensions in the strings (Neglect friction). (g = 10 m/s ) Solution 185  ® TG: @Chalnaayaaar Physics : Newton's laws of motion & Friction Pre-Medical (a) In this case net pulling force = mAg sin60° + mBg sin60° – mCg sin30° = (mA + mB) g sin 60° – mC g sin 30° 3 1 = (1 + 3) × 10 × – 2 × 10 × = 20 3 − 10 = 20 × 1.732 – 10 2 2 = 24.64 N F.B.D. of A Total mass being pulled = 1 + 3 + 2 = 6 kg T1 24.64 ∴ Acceleration of the system a = 2 = 4.1 m/s 6 (b) For the tension in the string between A and B. mAgsin60° mAg sin60° – T1 = (mA) (a) ∴ T1 = mA g sin60° – mA a = mA (g sin60° – a)  3  ∴ T1 = (1)  10 × − 4.1  = 4.56 N  2    ® For the tension in the string between B and C. T2 – mC g sin30° = mCa T2   1  ∴ T2 = mC (a + g sin30°) = 2 4.1 + 10    = 18.2 N   2  Illustration 18. In the figure blocks A, B and C have accelerations a1, A B C a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2mg and mg respectively. Find the value of a1, a2 and a3. Solution m m m 2mg − mg 2m − m g =a1 = g ;= a2 = g 2m m m 2m + m 3 F1=2mg F2=mg mg + mg − mg g =a3 = 2m 2 Illustration 19. A 12 kg monkey climbs a light rope as shown in fig. The rope passes over a pulley and is attached to a 16 kg bunch of bananas. Mass and friction in the pulley are negligible so that the effect of pulley is only to reverse the direction of force of the rope. What maximum acceleration can the monkey have without lifting the bananas? (Take g = 10 m/s2) Solution For Monkey T – 120 = 12 × a....(i) For Bananas T 160 – T = N N T Condition for just loosing the