Good quality notes on Physics - Definition of Force, Study notes of Physics

Download Good quality notes on Physics - Definition of Force and more Study notes Physics in PDF only on Docsity!   Digital Pvt. Ltd. [1] Definition of Force According to Aristotle, a constant continuous force is required to keep a body in uniform motion. This is called Aristotle's Fallacy. This cause was first understood in the late 1600s by Sir Isaac Newton. 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 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 and the shape of the body. Units for measurement of force: - Absolute units (i) S.I Newton (N) (ii) C.G.S Dyne Conversion 1N = 510 Dyne Force is a vector quantity, with magnitude and direction. Definition of Force Part - 01   Digital Pvt. Ltd. [2] Laws of Motion Part-01 Force: Push and Pull: - For instance, the force has been defined as an interaction that changes the motion of an object if unopposed. When this statement is examined closely, we see the role of push-pull in this. A force that changes the direction of an object towards you, that would be a pull. On the other hand, if it moves away, it is a push. Sometimes, force is simply defined as a push or pull upon an object resulting from the object’s interaction with another object. Hence, any kind of force is basically a push or a pull. Spring and elastic are also types of forces. The moment you push against it, it tends to resist and react or springs back with the same magnitude. Push and Pull Examples Push is defined as the force that is responsible for an object to move from the state of rest. Examples of push: • Pushing the trolley. • Pushing of the car when it breaks down. • Pushing the table from one place to another. The pull is defined as the force that is responsible for an object to move from the state of rest but in the opposite direction when compared to the push. Examples of pull: • Pulling the curtain. • Dragging the box. • Opening of the door. Whenever we consider a force in a given scenario, it can act as an internal as well as an external force depending on the system we have considered. This is how we have introduced this topic from the basic and it emphasizes on the fact that we have to be careful about the system chosen whenever we are labelling a force as an internal or an external force.   Digital Pvt. Ltd. [2] Laws of Motion Part-02 Inertia of Motion It is the inability of a body to change its state of uniform motion by itself. Example: • 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 then 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. Example: • 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. • When a straight running car turns sharply, the person sitting inside feels a force radially outwards.   Digital Pvt. Ltd. [1] Linear Momentum The total amount 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 instantaneous velocity. Momentum p mv= Momentum = Mass × Velocity  The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.  The direction of the momentum vector is the same as the direction of the velocity vector. SI unit : kg-m/s Dimension : [MLT–1] Illustration 1. Describe the momentum of 5 kg ball moving westward at 2 m/s you must include information about both the magnitude and the direction of the ball? Solution. Linear momentum is P m v=  The magnitude of momentum is |P| = |5| × |2| = 10 kg-m/s. The direction of momentum vector is the same as the direction of the velocity of the ball. As a vector quantity the momentum of an object is fully described by both magnitude and direction is 10 kg-m/s, westward. Illustration 2. A block having mass 2 kg moving with velocity ˆˆ ˆv 2i 3j km/s= + + . Find momentum of the block. Solution. Linear momentum is P m v=  = ˆˆ ˆ2 (2i 3j km/s) + + The magnitude of momentum is 2 2 2P 4 6 2 kg m/s= + + − 2 14kg m/s− Linear Momentum Part - 03   Digital Pvt. Ltd. [1] Change in Momentum Change in any physical quantity can be calculated by Change = Final – Initial Similarly change in momentum can be calculated by f ip p p = − f ip mv mv = − p m v =  Illustration 1. A ball of 0.20 kg hits a wall with a velocity of 25 m/s at an angle of 45° with horizontal. if the ball rebounds at 90° to the direction of incidence, calculate the magnitude of change in momentum of the ball. Solution. Initial momentum ( )i i ˆ ˆP m v m v(cos45 )i mvsin45 j=  =   +  − Final momentum ( ) ( )f f ˆ ˆP m v m v(cos45 ) i mvsin45 j=  =   − +  − Change in momentum = (–mv cos45°) – (mv cos45°) = –2mv cos45° 1 p 2mvcos45 2 0.2 25 5 2N s 2  =  =    = − Change in Momentum Part - 04   Digital Pvt. Ltd. [1] Illustration 1. A force ˆˆ ˆF (6i 8j 10k)N= − + produces an acceleration of 22 m/s in a body. Calculate the mass of the body. Solution. From Newton's ndII law |F| |ma| ma= =  Acceleration |F| a m =  2 2 2|F| 6 8 10 m 10kg a 2 + + = = = Illustration 2. A force of 50 N acts on a block 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. Solution. horizontal component of the force 50 3 50sin60 2 =  = acceleration of the block, 2component of force in the direction of acceleration 50 3 1 a 5 3 m/s mass 2 5 = =  = Illustration 3. If force F = 40 – 30t, then impulse in interval [0, t]: - Solution. t2 0 30t I Fdt 40t 2   = = −     = 40t – 15t2 Illustration 4. A force of 40 N acts on a body of mass 10 kg for 10 seconds. The change in its momentum is Solution. F a m = = 4 m/s2 v = u + at v – u = 4(10) v – u = 40 P = m(v–u) = 10(40) P = 400 kg-m/s Basic problems based on Newton's Second Law of Motion Part - 06   Digital Pvt. Ltd. [2] Laws of Motion Part-06 Illustration 5. A player catches a ball of 200 g moving with a speed of 10 m/s. If the time taken to complete the catch is 0.20 s, the force exerted on the player's hand is: - Solution. 3 avg p 200 10 10 F t 0.20 −   = =  Favg = 10N   Digital Pvt. Ltd. [1] 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 F dt   = = =    Unit of impulse = N-s or kg-m/s. Dimension of impulse [F] [t] = [m] [a] [t] = [M1L1T–2T1] = [M1L1T–1] Case-I : If this force is acting from time t1 to t2 , then integrating the above equation, we get – 2 2 1 1 pt t p I Fdt dp= =  Impulse = Change in momentum Case-II : If constant or average force acts on a body, then :- 2 2 1 1 pt avg t p I F dt dp= =   2 2 2 1 1 1 p pt avg t p p I F dt dt dp= = =   2 2 1 1 pt avg t p I F [t] [p]= =  avg 2 12 1I F (t t ) (p p )= − = − avgI F t p=  =  Case-III : Impulse from force time graph : Area under the force time graph represents impulse or change in momentum. e.g. I p Fdt Area under F t graph=  = = − Impulse and Average Force Part - 07   Digital Pvt. Ltd. [1] Impulse-Momentum Theorem dp dI Fdt dp F dt   = = =    Impulse-momentum theorem states that the impulse applied to an object will be equal to the change in its momentum. This theorem can be proven from newton's law. Signifies instantaneous rate of change of momentum hence the impulse is equal to change in momentum. Illustration 1. 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 (b) Average retarding force that stops the hammer Impulse 6 F 60N time 0.1 = = = (c) Average retardation 2F 60 a 60m / s m 1 = = = Illustration 2. 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. Solution. F = ma 6 1 a 10 4 =  2.4 = a (Retardation) v = u + at u = at 12 = 24 t 10   t = 5sec I = F × t = 3 N-s Impulse Momentum Theorem Part - 08   Digital Pvt. Ltd. [2] Laws of Motion Part-08 Illustration 3. A batsman deflects a ball by an angle of 90° without changing its initial speed, which is equal to 54 km/hr. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg) Solution. vf = vi = 15 m/s I = m f i(v v )− 2 2 f iI m v (v )= + 15 I 15 2 100 =  = 3.1815 (N-s) Illustration 4. The magnitude of the force (in newton) acting on a body varies with time t (in microsecond) as shown in fig. AB, BC and CD are straight line segments. Find the magnitude of the total impulse of the force (in N–s) on the body from t = 4 µs to t = 16 µs. Solution. Impulse = F.dt = Area of F – t graph between 4 µs to 16 µs = 6 61 1 (200 800) 2 10 800 10 10 2 2 − −+   +    = 10–3 + 4 × 10–3 = 5 × 10–3 Ns   Digital Pvt. Ltd. [1] Conservation of Linear Momentum According to Newton’s Second law, if net external force on the system is zero, then linear momentum of the system remains conserved. According to Newton's Second Law. ext dp F dt = If extF 0= dp 0 p constant dt =  = in finalp p= for two particles system 1 2p p constant+ = 1 2p p 0 + = 1 2p p = − (change in momentum of Ist particle = –change in momentum of IInd particle) Examples of Law of Conservation of Momentum Following are the examples of law of conservation of momentum: 1. Air-filled balloons 2. System of gun and bullet 3. Motion of rockets Illustration 1. There are cars with masses 5 kg and 10 kg respectively that are at rest. A car having the mass 10 kg moves towards the east with a velocity of –15ms . Find the velocity of the car with mass 4 kg with respect to ground. Solution. Given, 1m 5kg= 2m 10kg= 1v ?= 2v 5m/s= We know from the law of conservation of momentum that, Conservation of Linear Momentum Part - 09   Digital Pvt. Ltd. [1] Rocket Propulsion m = Mass of rocket relv = Velocity of exhaust gases w.r.t rocket dm dt =Rate of burning of fuel Case-I : If rocket is accelerating upwards, then - Net upwards force on rocket = ma rel dm v mg ma dt − = Where relv is the relative velocity of the ejected mass w.r.t rocket The acceleration of the rocket is given as: relv dm a g m dt = − Case-II : If rocket is moving with constant velocity, then a = 0 rel dm v mg dt = Types of Rocket Propulsion The following are the different types of rocket propulsion: • Liquid-fuel chemical propulsion • Solid fuel chemical propulsion • Cold-gas chemical propulsion Variable Mass Problems Part - 11   Digital Pvt. Ltd. [2] Laws of Motion Part-11 Illustration 1. A 800 kg rocket is set for a vertical firing. If the exhaust speed of gases is 2000 m/s w.r.t rocket, then calculate the mass of gas ejected per second to supply the thrust needed to overcome the weight of rocket. Solution. Force required to overcome the weight of rocket F = mg and thrust needed = rel dm v dt so rel rel dm dm mg 800 9.8 v mg 3.92 kg /s dt dt v 2000  =  = = = Illustration 2. 62.80 10 kg is the mass at liftoff of a Saturn. The fuel is burnt at the rate of 41.40 10 kg/s and the exhaust velocity is 32.40 10 m/s w.r.t rocket. What is the initial acceleration? Solution. Given: Exhaust velocity, 3 rev 2.40 10=  m/s Mass of the rocket, 6m 2.80 10=  kg Rate of fuel burnt, 4dm 1.40 10 dt =  kg/s Acceleration due to gravity, 2g 9.80 m/s= Substituting the above, in the formula, we get relv dm a g m dt = − ( ) 3 4 2 6 2.40 10 m /s a 1.40 10 kg /s 9.80 m /s 2.80 10 kg  =  −  2a 2.20 m/s= Illustration 3. If the force on a rocket moving in force free space with an exhaust velocity of gases 600 m/sec w.r.t rocket is 500 N, then the rate of combustion of the fuel, is :- Solution. dm 500 0.833kg / s dt 600 = = Illustration 4. For a Rocket propulsion velocity of exhaust gases relative to rocket is 1.8 km/s. If mass of rocket system is 600 kg, then the rate of fuel consumption for a rocket to rise up with an acceleration 2.4 m/s2 will be :- Solution. dm 600 9.8 600 2.4 1800 dt  +  =  dm 7320 4.067kg / s dt 1800 = =   Digital Pvt. Ltd. [3] Laws of Motion Part-11 Illustration 5. A rocket of mass 600 kg is to be projected vertically upwards. The gases are exhausted vertically downwards with velocity 30 m/s with respect to the rocket. What is the minimum rate of burning fuel, so as to just lift the rocket upwards against the gravitational attraction? (Take g = 10 m/s2) Solution. Minimum rate of burning fuel is dm dt  mg = u dm dt or, dm mg 600 10 dt u 30  = = = 200 kg/s.   Digital Pvt. Ltd. [1] Normal Reaction The contact force by which two surfaces in contact push each other perpendicular to the contacting surfaces, is known as normal reaction. • Electromagnetic in nature • The pushing force acting towards the body. • Contact force in a direction perpendicular to surfaces in contact • Always produced in action reaction pair • Normal Reaction cannot pull an object. • Self-adjusting in nature • Normal Reaction can never be negative, its minimum value is ZERO • N = 0 implies two conditions • Bodies are not in physical contact • Bodies are in physical contact but not pushing each other Illustration 1. Show the normal reaction forces in following figures ? Solution. Here N1 and N2 normal reaction forces perpendicular to the well define surface towards the centre of the body. Normal Reaction Force Part - 13   Digital Pvt. Ltd. [1] Two Bodies in Contact Free body diagrams : 1F N m a− = …..(1) 2N m a= …..(2) On adding the above equations 1 2F m a m a= + ( )1 2F a m m= +  1 2 F a m m = + ….(3) OR net total F a m = putting the value of 'a' from equation (3) in (2), we get 2 1 2 m F N m m = + (here N = contact force) Three bodies in contact : 1 1F N m a− = …..(1) 1 2 2N N m a− = …..(2) 2 3N m a= …..(3) Problems on Normal Reaction Part - 14   Digital Pvt. Ltd. [2] Laws of Motion Part-14 On adding the above equations 1 2 3F m a m a m a= + + ( )1 2 3F a m m m= + +  1 2 3 F a m m m = + + …..(4) OR Net total F a m = Putting the value of 'a' from equation (4) in (3), we get 3 2 1 2 3 Fm N m m m = + + Putting the value of 'a' from (4) in (1); 1 1F m a N− = 1 1 1 2 3 Fm N F m m m = − + +  2 3 1 1 2 3 (m m )F N m m m + = + + Illustration 1. 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. 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 the action of a force F F 35 a 5 m / s m M 2 5 = = = + + Force of contact MN Ma 5 5 25N= =  = If the force is applied to M then its action on m will be MN ma 2 5 10N= =  = 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.   Digital Pvt. Ltd. [2] Laws of Motion Part-15 (b) String is assumed to be inextensible so that the magnitude of accelerations of the blocks tied to the strings are always same. (c) (i) If the string is massless and frictionless, tension throughout the string remains same. (ii) If the string is massless but not frictionless, at every contact tension changes. (iii) If the string is not light, tension at each point of the string will be different depending on the acceleration. (d) If a force is directly applied to a string, say a man is pulling a string from the other end with some force, then tension will be equal to the applied force irrespective of the motion of the pulling agent, irrespective of whether the box moves or not, man moves or not.   Digital Pvt. Ltd. [3] Laws of Motion Part-15 (e) String is assumed to be massless unless stated, hence 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 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.   Digital Pvt. Ltd. [1] Motion of Connected Bodies • Two Connected Bodies : • Free body diagram : 1F T m a− = …..(1) 2T m a= …..(2) On adding the above equations 1 2F m a m a= + ( )1 2F a m m= + 1 2 F a m m = + …..(3) Or net total F a m = Putting the value of 'a' from equation (3) in (2), we get – 2 1 2 m F T m m = + • Three Connected Bodies : Free body diagram: - Problems based on Tension Force Part - 16   Digital Pvt. Ltd. [4] Laws of Motion Part-16 For 1m 1 1 1m g T m a− =  ( )1 1T m g a= − …….(1) For 2m 2 1 2 2m g T T m a+ − =  2 2 2 1T m g m a T= − +  ( ) ( )2 2 1T m g a m g a= − + − (from equation 1)  ( )( )2 1 2T m m g a= + − …….(2) For 3m 3 3 2 3 3F T and m g T T m a= + − =  ( ) ( )( )3 3 1 2T m g a m m g a= − + + − (from equation 2)  ( )( )3 1 2 3T m m m g a= + + − …….(3)   Digital Pvt. Ltd. [1] Tension in rod Given Mass of Rod = M Length of Rod = L F – T = ma  T = F – ma  T = F – m F M F a M   =    Mass of length 'L' = M  Mass of unit length = M L  Mass (m) of length 'x' = M x L Put this value of 'm' in equation (1) M F x T F x T F 1 L M L     = −  = −        Illustration 1. 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. Tension in rod and heavy strings Part - 17   Digital Pvt. Ltd. [2] Laws of Motion Part-17 Solution. (i) Acceleration F a (m M) = + (ii) Force exerted by the rope on the block is M.F T Ma (m M) = = + (iii) 1 m m 2M F T M a 2 2 m M +      = + =      +      Tension in rope at midpoint is 1 (m 2M)F T 2(m M) + = +   Digital Pvt. Ltd. [3] Laws of Motion Part-18 Illustration 1. A block of mass 25 kg is raised in two different ways by a 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 ? Solution. Mass of the block, m = 25 kg ; mass of the man, M = 50 kg 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 2. 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 (b) the tensions in the strings (Neglect friction). (g = 10 m/s2)   Digital Pvt. Ltd. [4] Laws of Motion Part-18 Solution. (a) In this case net pulling force = mAg sin60° + mBg sin60° – mCg sin30° = (mA +mB)g sin60° – mCg sin30° = (1 + 3) × 10 × 3 2 – 2 × 10 × 1 2 = 20 3 – 10 = 20 × 1.732 – 10 = 24.64 N Total mass being pulled = 1 + 3 + 2 = 6 kg  Acceleration of the system 24.64 a 6 = = 4.1 m/s2 (b) For the tension in the string between A and B. mAg sin60° – T1 = (mA)(a)  T1 = mAg sin60° – mAa = mA(g sin60° – a)  T1 = (1) 3 10 4.1 2    −     = 4.56 N For the tension in the string between B and C. T2 – mCg sin30° = mCa  T2 = mC(a + g sin30°) = 2 1 4.1 10 18.2N 2    + =      Illustration 3. In the figure blocks A, B and C have acceleration a1, 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. 1 2mg mg a g m − = = ; 2 2m m g a g 2m m 3 − = = + 3 mg mg mg g a 2m 2 + − = =   Digital Pvt. Ltd. [5] Laws of Motion Part-18 Illustration 4. 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 160 – T = N Condition for just loosing the contact is N = 0 160 – T = 0  T = 160 …(ii) from equation (i) & (ii) 160 – 120 = 12 × a  a = 3.33 m/s2   Digital Pvt. Ltd. [3] Laws of Motion Part-19 Solution. (a) FBD of box x B AF 0 T cos60 T 0 =   − = ....(i) y BF 0 T sin60 10 3 0 =  − = ...(ii) Solving equations (i) and (ii) We have, TA = 10 N and TB = 20 N (b) FBD of box x B AF 0 T cos60 T sin 0 =   −  = ...(i) y A BF 0 T cos T sin60 10 3 0 =  + − = ....(ii) From equation (i) and (ii) We have A 10 3 T 3sin cos = +  TA is minimum when ( 3sin cos ) +  is maximum. Now, 2 2 max( 3sin cos ) ( 3) (1) 2 +  = + = Therefore (TA)min = 10 3 N 5 3 N 2 = Illustration 3. A 70 kg man standing on a weighing machine in a 50 kg lift pulls on the rope, which supports the lift as shown in the figure. Find the force with which the man should pull the rope to keep the lift stationary. Also, find the weight of the man as shown by the weighing machine. Solution. Magnitude of tension everywhere in the string is same. For equilibrium of the lift. yF 0 500 N 2T=  + = ...(i) To analyse the equilibrium of the man let us assume him as a block   Digital Pvt. Ltd. [4] Laws of Motion Part-19 yF 0 N T 700=  + = ...(ii) From equations (i) & (ii), we have T = 400 N and N = 300 N Here, T is the pull of mass and N is the reading of the weighing machine.   Digital Pvt. Ltd. [1] Spring Force A linear spring is generally a helical metallic wire. When the two free ends of the spring are pulled away or pushed towards each other, the length of the spring is changed. The spring has a tendency to come back to its original length and it develops an opposition to change in its length. This opposing force (F) is the restoring force. This force is electromagnetic in nature. Hooke’s Law Under small extension/compression, The opposing force is directly proportional to the change in length and opposite to direction of pull/push. = − spring F kx Spring force Part - 20   Digital Pvt. Ltd. [4] Laws of Motion Part-20 Solution. (a) 3g – kx = 3a …(i) 2g + kx – T = 2a …(ii) T = a …(iii)  Acceleration of the system is 250 a m / s 6 = (b) Free body diagram of 1 kg block gives T = ma = (1) 50 50 N N 6 6   =    (c) Free body diagram of 3 kg block gives 30 – kx = ma but ma = 3 × 50 6 = 25 N 30 25 5 x 0.1m 10cm k 50 − = = = = Illustration 4. Find initial accelerations of A and B respectively, when string is cut Solution. Before cutting the string T = mg kx = T+ mg, kx = 2mg After cutting the string   Digital Pvt. Ltd. [5] Laws of Motion Part-20 A mg a g m = =  B kx a m =  B 2mg a 2g m = =    Digital Pvt. Ltd. [1] Frame of Reference A system with respect to which position or motion of a particle is described is known as frame of reference. We can classify frame of reference in two categories :- 1. Inertial Frame of Reference 2. Non inertial frame of Reference Inertial Frame of Reference The frame for which law of inertia is applicable is known as inertial frame of reference. All the frames which are at rest or moving uniformly with respect to an inertial frame, are also inertial frame. Non-Inertial Frame of Reference The frame for which law of inertia is not applicable is known as non-inertial frame of reference. All the frames which are accelerating or rotating with respect to an inertial frame will be non-inertial frames. Pseudo Force To apply Newton’s law of motion in non-inertial reference frame we need to apply pseudo force. It is an imaginary force which is used to explain the motion of objects from non-inertial reference frames. Magnitude : =  − pseudo F Mass of body Acceleration of non inertial frame (w.r.t. Observer) Direction : Opposite to the direction of acceleration of non-inertial frame(observer) = − pseudo obsF m a Pseudo force does not follow action reaction law. Illustration 1. A pendulum of mass m is suspended from the ceiling of a train moving with an acceleration 'a' as shown in figure. Find the angle  in which pendulum is in equilibrium w.r.t. train. Also calculate tension in the string. Frame of Reference and Pseudo Force Part - 21   Digital Pvt. Ltd. [2] Laws of Motion Part-22 Case-III : If the lift is accelerated downwards, then – mg – N = ma N = mg – ma Wapp or N=m(g–a) So, Wapp < Wactual If the lift is under free fall, it implies that its acceleration is equal to the acceleration due to gravity. i.e. a = g, then Wapp = 0 It means that person in lift will feel weightless The apparent weight of any body falling freely is zero   Digital Pvt. Ltd. [1] Friction Friction is the opposing force that is set up between the surfaces of contact, when one body slides or tends to do so on the surface of another body. Friction does not oppose motion. Rather it opposes relative motion between two bodies. Cause of Friction Old view: When two surfaces are in contact with each other, irregularities of one body get interlocked with the irregularities in the surface of the other. This interlocking opposes the tendency of relative motion. Modern view : Friction arises on account of Intermolecular forces of attraction between the two surfaces at the point of actual contact. Friction depends on the following factors : 1. Friction force depends only on the area covered by contact particles of contact surfaces (actual contact area) it does not depend on the area covered by the body (apparent contact area) 2. Except static friction (fs), friction force depends on normal reaction (N) f  N (A) (B) (C) (D) (D) Introduction to Friction Part - 23   Digital Pvt. Ltd. [2] Laws of Motion Part-23 Illustration 1. Why do we slip on a muddy road? Solution. Water on a muddy road provides a thin layer in between our feet and road. This layer breaks the interlocking and decreases the friction. Illustration 2. State whether the following statement is true or false: When a person walks on a rough surface, the frictional force exerted by the surface on the person is opposite to the direction of his motion. Solution. False, when a person walks on a rough surface the man exerts a backward frictional force on the surface. As a result, the surface exerts a forward frictional force, according to Newton's III law.   Digital Pvt. Ltd. [3] Laws of Motion Part-24 Solution. N = mg – F sin30° Fcos30° = Frmax = N (for just move the block) Fcos30°= ( )mg Fsin30 −  3 1 1 F mg F 2 23    = −     3 F F mg 2F mg 2 2  = −  = mg 50 F 25N 2 2 = = = Illustration 3. A body of mass 0.1 kg is pressed against a wall with a horizontal force 5 N. If coefficient of friction is 0.5 then find force of friction. Solution. extN F 5N= = mg = 0.1 × 10 = 1N ( ) max Fr N= ( ) max Fr 0.5 5 2.5N=  = ( ) max mg Fr So Fr = mg = 1N Illustration 4. What is the acceleration of the block and trolley system shown in fig. if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m/s2). Neglect the mass of the string. 3 kg 20 kg   Digital Pvt. Ltd. [4] Laws of Motion Part-24 Solution. N = mg = 200N Fr N 0.04 200 8N= =  = T – Fr = 20a (for trolley) ……(1) T – 8 = 20a 3g – T = 3a (for block) ……(2) From (1) and (2) 3g – 8 = 23a 222 a m / s 23 = T = 30 – 3a = 30 – 30 × 22 23 = 27.13N   Digital Pvt. Ltd. [1] Illustration 1. Length of a chain is L and coefficient of static friction is  . Calculate the maximum length of the chain which can hang from the table without sliding. Solution. Let y be the maximum length of the chain that can be held outside the table without sliding. Length of chain on the table = (L – y) Weight of the part of the chain on table ( ) M W' L y g L = − Weight of hanging part of the chain M W yg L = For equilibrium: limiting force of friction on (L–y) length = weight of hanging part of the chain of y length ( ) M M L N W W' W L y g yg L y y y L L 1   =  =  − =  − =  = + Illustration 2. A block of mass m rests on a rough horizontal surface as shown in figure (a) and (b). Coefficient of friction between the block and surface is  . A force F = mg act at an angle  with the vertical side of the block. Find the condition for which the block will move along the surface. Motion and Equilibrium on rough horizontal surface Part - 25   Digital Pvt. Ltd. [1] Contact Force Let f be the force of friction and N the normal reaction, then the net contact force by the surface on the object is 2 2 surfaceF N f= + . Its minimum value (when f = 0) is N and maximum value (when f = µN) is 2N 1 µ+ Therefore 2 surfaceN F N 1 µ  + Angle of Friction() The angle which the resultant of the force of limiting friction fL and normal reaction N makes with the direction of normal reaction N.  = L f tan N  = µN tan N  = S tan µ Illustration 1. A horizontal force of 49 N is just able to move a block of wood weighing 10 kg on a rough horizontal surface. Calculate the coefficient of friction and angle of friction. Contact Force and Angle of Friction Part - 26   Digital Pvt. Ltd. [2] Laws of Motion Part-26 Solution. Here, F = 49 N, N = w = mg = 10 × 9.8N so s F 49 0.5 N 10 9.8  = = =   tan = µ = 0.5  Angle of friction  = tan–1(0.5)   Digital Pvt. Ltd. [1] Angle of Repose or Angle of Sliding It is defined as the minimum angle of inclination of a plane with the horizontal at which a body placed on it just begins to slide down or equivalently the maximum angle of inclination of plane with the horizontal at which a body placed on it down not slide. LF mgsin=  ……….(i) N = mg cos LF N mgcos= =  ……….(ii) From (i) and (ii) so 1L s s f mgsin µ tan tan µ N mgcos − = = = =  This is fact is used for finding the coefficient of static friction in the laboratory. Angle of repose () = Angle of friction () Illustration 1. Check in which case, extra force is needed for block to slide down. Solution. Case-I : - Angle of repose = 1tan− (0.5) ( )1tan 0.5 37−   No extra force needed for block two slide down. Case-II : - Angle of repose = ( )1tan 0.9− ( )1tan 0.9 37−   Extra force needed for block two slide down. Angle of Repose Part - 27   Digital Pvt. Ltd. [2] Laws of Motion Part-28 Note : (i) If we want to prevent the downward slipping of body then minimum upward force required is = mg sin – µk mg cos (ii) If a body is projected in upward direction along the inclined plane then retardation of body is a = g [sin + µkcos] retardation of a body up a rough inclined plane may be greater than 'g' Pulley with friction between block and surface For mass m1 : m1g – T = m1a and N = m2g cos  µN = µm2 g cos q For mass m2 : T – µm2g cos – m2g sin = m2a on solving, Acceleration 1 2 1 2 m m (sin cos ) a g (m m )  −  +   =   +  Tension 1 2 1 2 m m (1 sin cos )g T (m m ) +  +   = + Illustration 1. A block of mass 2 kg slides down an inclined plane which makes an angle 30° with the horizontal. The coefficient of friction between the block and the surface is 3 2 . (i) What force must be applied to the block so that it moves down the plane without acceleration ? (ii) What force should be applied to the block so that it moves up without any acceleration ? Solution. Make a 'free–body' diagram of the block. Take the force of friction opposite to the direction of motion. (i) Project forces along and perpendicular to the plane. Perpendicular to plane N = mg cos Along the plane F + mg sin – f = 0 ( there is no acceleration along the plane) F + mg sin – µN = 0  F + mg sin = µmg cos F = mg (µ cos – sin) = 2 × 9.8 ( 3 2 cos 30° – sin 30°) = 19.6 ( 3 2 × 3 2 – 1 2 ) = 19.6 ( 3 1 4 2 − ) = 4.9 N (ii) This time the direction of F is reversed and that of the frictional force is also reversed.  N = mg cos ; F = mg sin + f  F = mg (µcos + sin) = 19.6 3 1 4 2   +    = 24.5 N   Digital Pvt. Ltd. [3] Laws of Motion Part-28 Illustration 2. A block of mass 1 kg rests on an incline as shown in figure. (a) What must be the frictional force between the block and the incline if the block is not to slide along the incline when the incline is accelerating to the right at 3 m/s2 ? (b) What is the least value of µs which can have for this to happen ? Solution. N = m (g cos37° + a sin37°) = 1(9.8 × 0.8 + 3 × 0.6) = 9.64 N mg sin 37° = ma cos 37° + f (a) f = 1(9.8 × 0.6 – 3 × 0.8) = 3.48 (b) f = µN  f 3.48 0.36 N 9.64  = = =   Digital Pvt. Ltd. [1] Two Block System in Friction Consider two blocks A and B placed one above the other, resting on a horizontal surface. A horizontal force F applied on either blocks, tend to move the system of blocks. Problems involving such situations can conveniently be solving by following mentioned step. Step 1 : Draw the FBD of the combined block system. If friction appears in the FBD, then take its limiting value (maximum static friction) If applied force > limiting friction then motion is possible, otherwise not. If movement occurs then, either the blocks move together or separately depending on the fact that whether frictional forces are able to support the combined motion or not. Step 2 : Assuming combined motion, find the common acceleration aC. Draw the FBD of the body on which external force is not applied. Find the frictional force f required to make it move combinedly with the other block. Compare the above calculated force with the limiting value fL (maximum static friction). If f  fL, then both move together with common acceleration ac. Otherwise, they move separately. Step 3 : For separate motion, draw the individual FBD’s of both blocks with kinetic friction forces acting wherever applicable. Find the individual accelerations of the two blocks using Newton’s second law. Illustration 1. Calculate the accelerations of the blocks and the force of friction between them. Solution. Step 1 : Draw FBD of the combined block system. Obviously, there is an unbalanced force (F = 50N) in the horizontal direction.  Movement occurs. Two Block Problems Part - 29