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CHAPTER – 05: LAWS OF MOTION, Force: An external agency required to put a stationary body in motion or stop a moving body is called force., Dynamics: The branch of mechanics which deals with the cause for the motion of the objects is called, dynamics., Aristotle’s Fallacy: According to Aristotle an external force is always required to keep a body in motion. This, was falls. An external force is required to overcome the frictional forces which oppose the motion. If there are, no frictional forces, then no force is required to move an object., Example a child has to drag a toy-car on a floor with some force to keep it going. If it releases the string, it, comes to rest. External forces seem to be needed to keep bodies in motion. Left to themselves, all bodies, eventually come to rest. This was Aristotle fallacy., Inertia: The inherent property of a body by virtue of which it remains in the state of rest or uniform motion is, called inertia., Newton’s First Law of Motion (Law of Inertia): It states that ”every body continues to be in its state of rest or, of uniform motion in a straight line unless compelled by an external force to act on it”., Example:, 1. When a stationary bus suddenly starts moving, passengers move backward. In a stationary bus,, passengers are also at rest. As the bus starts to move, the lower part of the body in contact with the, bus starts to move. But upper part remains at rest due to inertia. Thus passenger moves backward., 2. When brakes are applied suddenly to a moving bus, passengers move forward. Because when brakes, are applied, the lower part of the body comes to rest. But upper part still continues to move due to, inertia., 3. A spaceship out in interstellar space with engines switched off continues to move with a uniform, velocity., 4. A book rest on the table is acted by two forces. Force due to gravity is acting downward and normal, reaction on the book by the surface of the table is acting upward. Since the book is observed to be at, rest, the net external force on it must be zero., 5. A ball on a frictionless surface continues to move for ever unless acted by an external force to change, the motion., 6. A moving vehicle gets accelerated to overcome the friction of the road., Momentum (p): The product of mass of a body and its velocity is called momentum. p = mv. It is a vector, quantity. Its S.I unit is kg m/s. Its dimensions are [MLT -1]., 1. Suppose a car and a loaded truck are parked on a horizontal road. A much greater force is needed to, push the truck than the car to bring them to the same speed in same time. Similarly, a greater, opposing force is needed to truck than the car in the same time, if they are moving with the same, speed., 2. If two stones, one light and the other heavy, are dropped from the top of a building, a person on the, ground will find it easier to catch the light stone than the heavy stone. Mass predominates., 3. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. Velocity, predominates., 4. A cricketer catches the ball coming with great speed by drawing his hands backward without any hurt., Thus force not only depends on the change in momentum, but also on how fast the change is brought, about. This leads to the second law of motion., SMGPUC SIRSI, , Page 2

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Second Law of Motion: It states that “the rate of change of momentum of a body is directly proportional to, the applied force and takes place in the direction of the force”., Consider a body of mass m moving with a velocity v Taking the limit, we get F = k, and momentum p = mv. A force F acts on it for a time, interval ∆t. Then velocity changes to v + ∆v and F = k, momentum p + ∆p = m(v + ∆v)., F=k, The change in momentum is ∆p = m∆v., According to Newton’s second law,, F = km, F, , or F = k, , where k is a constant of, , F = kma. For chosen units k = 1., Therefore F = ma., , proportionality., Note:,  1 newton is that force which causes an acceleration of 1 ms-2 to a body of mass 1 kg.,  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. For example, in the motion of a projectile under the vertical gravitational force,, the horizontal component of velocity remains unchanged.,  In the case of system of particles F is the total external force on the system and does not include the, internal forces., Problem: A bullet of mass 0.04 kg moving with a speed of 90 ms-1 enters a heavy wooden block and is stopped, after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?, Solution Given m = 0.04 kg, v0 = 90 ms-1, v = 0, x = 60 cm = 0.6 m, a = ? F = ?, Using the formula v2 = + 2ax we have 0 = 902 + 2a(0.6),, 1.2a = -8100, a =, , = -6750 ms-2., , Resistive (retarding) force F = ma = 0.04 (-6750) = -270 N., Problem: The motion of a particle of mass m is described by y = v0t + gt2. Find the force acting on the, particle., Solution, , Given y = v0t + gt2,, then velocity v =, , =, , acceleration a =, , =, , (v0t + gt2) = v0(1)+ g(2t) = v0 + gt,, (v0 + gt) = 0 + g(1) = g., , Then force is given by F = ma = mg., Impulsive Force: A large force acting for a short interval of time to produce a finite change in momentum is, called impulsive force. Example kicking a football, hitting a cricket ball with a bat, hitting a volley ball., Impulse: The product of large force and time of action is called impulse. Impulse = force x time. S.I unit of, impulse is N s. Its dimensions are [M L T-1]., Now impulse = F.t = (ma) t = m(at)., We know that v = v0 + at or at = (v – v0)., Substituting this in the above equation,, we get F = m (v – v0) = mv – mv0 i.e change in momentum., Therefore impulse is equal to the change in momentum., SMGPUC SIRSI, , Page 3

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Problem: A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of, 12 ms–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion, of the ball) Solution: Given m = 0.15 kg, v0 = -12 ms-1, v = +12 ms-1, impulse = ?, Impulse = change in momentum = m(v – v0) = 0.15[12 – (-12)] = 0.15 x 24 = 3.6 Ns in the direction from, the batsman to the bowler., Newton’s third law of motion: It states that “to every action, there is always an equal and opposite reaction”., Consider to objects A and B. If A exerts a force FAB on B, then simultaneously object B exerts a force FBA on the, object A. These two forces are equal but opposite in direction. ⃗ AB = - ⃗ BA., Example,  A swimmer pushes the water in the backward direction and water pushes him in the forward direction.,  In rocket motion, the exhaust gases push the atmosphere backward with greater speed. Meanwhile, atmosphere pushes the rocket in the forward direction., Conservation or Momentum: It states that “ the total momentum of an isolated system of interacting, particles is conserved”. In other words the total momentum before interaction (collision) is equal to the total, momentum after interaction (collision)., Consider two bodies A and B undergoing collision. Let initial momenta be PA and PB. After the collision,, the momenta become, and, respectively. According to Newton’s second law, force of A due to B is the, rate of change of momentum of the body A., FAB =, , or FAB, , =, , ----(1)., , Similarly force on B due to A is the rate of change of momentum of the body B., FBA =, , or FBA, , =, , ----(2)., , According to Newton’s third law, these two forces are equal and opposite,, FAB = - FBA,, = -(, )=, =, ., Thus momentum after collision is equal to the momentum before collision., Equilibrium of a Particle: A particle is said to be in equilibrium, of the, vector sum of the forces acting on it is zero. i.e net force on the particle is, zero. Consider three forces ⃗ , ⃗ and ⃗ are acting on a particle as shown, in the figure. If the resultant of ⃗ and ⃗ is equal and opposite to ⃗ , then, the particle is at equilibrium., Problem: A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the, horizontal direction is applied at the midpoint P of the rope, as shown in the figure. What is the angle the rope, makes with the vertical in equilibrium? (Take g = 10 m s-2). Neglect the mass of the rope., Solution, of three forces - the tensions, Figures (b) and (c) are known T1 and T2, and the horizontal, as free-body diagrams. Fig. force 50 N. Resolving T1 we, (b) is the free-body diagram, get, of W and Fig. (c) is the freebody diagram of point P. T1 cos θ = T2 = 60 N, T1 sin θ = 50 N,, Clearly,, tan θ = 5/6,, T2 = 6 × 10 = 60 N., Consider the equilibrium of θ =, = 400., the point P under the action, SMGPUC SIRSI, , Page 4

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Common Forces in Mechanics:, Contact Forces: The force between two objects which are in contact is called contact force., The component of contact force normal to the surfaces in contact is called normal reaction., The component of contact force parallel to the surfaces in contact is called friction., Bouyant force, viscous force, air resistance, tension in a string and force due to spring etc are examples, of contact forces., Friction: Static Friction ( fs ): The friction which opposes the impending motion is called static friction. It is a, self-adjusting force. Consider a body of mass m at rest on a horizontal table. The force of gravity (mg) is, cancelled by the normal reaction force (N) of the table. Now suppose a force F is applied horizontally to the, body. Static friction arises in the opposite direction. As the force increases, friction also increases. After a, certain limit the body starts to move. The limiting value of static friction (fs)max is independent of the area of, contact and varies with the normal force(N) approximately as (fs)max = 𝜇sN where 𝜇s is a constant of, proportionality depending only on the nature of the surfaces in contact and is called coefficient of static, friction., The law of static friction states that the static friction is directly proportional to the normal reaction force up to, a certain limit., Kinetic or sliding friction ( fk ): Frictional force that opposes relative motion between surfaces in contact is, called kinetic or sliding friction and is denoted by fk. fk = 𝜇kN where 𝜇k is a constant of proportionality, depending only on the nature of the surfaces in contact and is called coefficient of kinetic friction., The law of kinetic friction states that the kinetic friction is directly proportional to the normal reaction force., (𝜇k < 𝜇s)., Problem Determine the maximum acceleration of the train in which a box lying on its floor will remain, stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15., Solution Given 𝜇s = 0.15, g = 10 ms-2, since the acceleration of the box is due to the static friction, ma = fs, 𝜇sN = 𝜇smg i.e. a 𝜇sg , amax = 𝜇sg = 0.15 x 10 = 1.5 ms-2., Problem: A mass of 4 kg rests on a horizontal plane as shown in the figure. The plane is gradually inclined until, at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction, between the block and the surface?, Solution, θmax depends only on μs and is, Resolving the weight mg, we have independent of the mass of the, mg sin θ = fs ,, block., mg cos θ = N, For θmax = 15°,, For θ = θmax, fs achieves its μs = tan 15°, maximum value, (fs)max = 𝜇sN., μs = 0.27, tan θmax = 𝜇s or θmax =, 𝜇, Problem: What is the acceleration of the block and trolley system shown in a Fig.(a), 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 ms-2). Neglect mass of the string., Solution Applying second law to motion of the, block [Fig.(b)],, 30 – T = ma, 30 – T = 3a ----(1)., Apply the second law to motion of the trolley, [Fig.(c)], T – fk = ma, T – fk = 20a., SMGPUC SIRSI, , Page 5

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Now fk = μkN, Here μk = 0.04, N = 20 x 10 = 200 N. Thus the equation for the motion of the trolley is, T – 0.04 x 200 = 20a or T – 8 = 20a -----(2). Adding eqn. (1) and (2) we get 22 = 23a or a =, , = 0.96 ms-2 and, , T = 27.1 N. Rolling Friction: Contact force which opposes the rolling of an object is called rolling friction., During rolling, the surfaces in contact get momentarily deformed a little, and this results in a finite area (not a, point) of the body being in contact with the surface. The net effect is that the component of the contact force, parallel to the surface opposes motion. Rolling friction < kinetic friction < static friction., Advantages of friction:, 1. We are able to walk due to friction., 2. We are able to write on the paper or board due to friction., 3. We are able to stop machines and automobiles by applying brakes due to friction., Disadvantages of friction:, 1. Friction opposes relative motion between the objects., 2. Friction causes heating of machines and automobile wheels., 3. It causes wearing and tearing of tyres of an automobile., Ways to reduce friction:, 1. Friction can be reduced by applying lubricants to machinery parts., 2. It can be reduced by polishing the surface., 3. It can be reduced by using ball bearings in the wheels., 4. It can be reduced by maintaining a thin cushion of air between the solid surfaces., Circular motion:, Motion of a car on a level road: Consider circular motion of a car on a, level road as shown in the figure. The three forces acting on the car are, (i) weight of the car mg downwards, (ii) normal reaction N upwards, (iii), frictional force f towards the centre of the circular path. As there is no, acceleration in the vertical direction N–mg = 0, N = mg ------(1), The centripetal force required for circular motion is along the surface of, the road, and is provided by the static friction between road and the car, tyres along the surface. fs, , 𝜇sN =, , or v2, , =, , =𝜇, , Motion of a car on a level road, , which is independent of the mass of the car., This shows that for a given value of μs and R, there is a maximum speed of circular motion of the car possible,, namely vmax = √, Motion of a car on a banked road: In a curved road the outer edge is raised to a certain height with respect to, the inner edge to avoid skidding is called banking of road., Consider circular motion of a car on a banked, road as shown in the figure. The three forces, acting on the car are (i) weight of the car mg, downwards, (ii) normal reaction N normal to the, road, (iii) frictional force f towards the centre of, the circular path., Motion of a car on a banked road, As there is no acceleration in the vertical, direction, N cos θ = mg + f sin θ ------(1), The centripetal force is provided by the horizontal components of N and f., SMGPUC SIRSI, , Page 6

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N sin θ + f cos θ =, , ------(2) But f, , 𝜇sN. To obtain maximum velocity, f = 𝜇sN. Substituting this in eqn. (1), , we get, N cos θ = mg + 𝜇sN sin θ, N(cos θ - 𝜇s sin θ) = mg or N =, 𝜇sN cos θ =, , or N (sin θ + 𝜇s cos θ) =, , ,, , cos θ) =, , =, , ------(3) From eqn. (2) N sin θ +, (sin θ + 𝜇s, , substituting eqn. (3) here we get,, (, , =, , (, , ), ), , =, , or, , =√, , (, (, , ), ), , This is the expression for maximum possible speed of a car on a banked road. This is greater than for a flat, road., Optimum speed: The speed of a vehicle at which frictional force is not needed at all to provide the necessary, centripetal force is called optimum speed. For 𝜇s= 0, v0 = √, . Driving at this speed on a banked road, will cause little wear and tear of the tyres., Problem: A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without, reducing the speed. The co-efficient of static friction between the tyres and the road is 0.1. Will the cyclist slip, while taking the turn?, Solution Given R = 3 m, g = 9.8 ms-2, 𝜇 = 0.1, v = 18 km/h = 18 x 5/18 = 5 m/s. The condition for the cyclist not, to slip is given by v2 𝜇, . Now 𝜇, = 0.1 x 3 x 9.8 = 2.94 (ms-1)2, v2 = 52 = 25 (ms-1)2. The condition is not, obeyed. The cyclist will slip while taking the circular turn., Problem: A circular racetrack of radius 300 m is banked at an angle of 150. If the coefficient of friction between, the wheels of a race-car and road is 0.2, what is the (a) optimum speed of the race-car to avoid wear and tear, on its tyres, and (b) maximum permissible speed to avoid slipping?, Solution At the optimum speed, the normal reaction’s component is enough to provide the needed, centripetal force, and the frictional force is not needed. Here R = 300 m, θ = 15°, g = 9.8 ms-2,, we have The optimum speed, v0 = √, = √, The maximum permissible speed vmax is given by, =√, , (𝜇, (, , 𝜇, , ), ), , =√, , =√, , = 28.06 ms-1., , = √, , =√, , =√, , = 38.12 ms-1., , Problem: A wooden block of mass 2 kg rests on a soft horizontal floor as shown in the figure. When an iron, cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder, together go down with an acceleration of 0.1 ms–2. What is the action of the block on the floor (a) before and, (b) after the floor yields? Take g = 10 ms–2. Identify the action-reaction pairs in the problem., Solution (a) The block is at rest on the floor. Its free-body diagram, shows two forces on the block, the force of gravitational attraction, by the earth equal to 2 × 10 = 20 N; and the normal force R of the, floor on the block. By the First Law, the net force on the block, must be zero i.e., R = 20 N. Using third law the action of the block, (i.e. the force exerted on the floor by the block) is equal to 20 N, and directed vertically downwards., (b) The system (block + cylinder) accelerates downwards with 0.1, ms-2. The free-body diagram of the system shows two forces on, the system, the force of gravity due to the earth (270 N) and the, normal force R′ by the floor. Applying the second law to the system,, SMGPUC SIRSI, , Page 7

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270 – R′ = 27 × 0.1N ie. R′ = 267.3 N. By the third law, the action of the system on the floor is equal to 267.3 N, vertically downward., Action-reaction pairs: For (a): (i) the force of gravity (20 N) on the block by the earth (say, action); the force of, gravity on the earth by the block (reaction) equal to 20 N directed upwards. (ii) the force on the floor by the, block (action); the force on the block by the floor (reaction)., For (b): (i) the force of gravity (270 N) on the system by the earth (say, action); the force of gravity on the, earth by the system (reaction), equal to 270 N, directed upwards. (ii) the force on the floor by the system, (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the, cylinder and the force on the cylinder by the block also constitute an action-reaction pair., Note: The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not, action-reaction pairs., , SMGPUC SIRSI, , Page 8