LAWS OF MOTION

) max s f is independent of the area of contact and varies with the normal force (N) approximately as : ) max f N = µ ( . ) where µ s is a constant of proportionality depending only on the nature of the surfaces in contact. The constant µ s is called the coefficient of static friction. The law of static friction may thus be written as f s ≤ µ s N ( . ) ) max s f the body begins to slide on the surface. It is found experi- mentally that when relative motion has started, the frictional force decreases from the static ) max s f . Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by f k . Kinetic friction, like static fric- tion, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: k k f N ( . ) where µ k ′ the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that µ k is less than µ s . When relative motion has begun, the acceleration of the body according to the second law is ( F – f k )/m. For a body moving with constant velocity, F = f k . If the applied force on the body is removed, its acceleration is – f k /m and it eventually comes to a stop. The laws of friction given above do not have the status of fundamental laws like those for gravitational, electric and magnetic forces. They are empirical relations that are only friction. approximately true. Yet they are very useful in practical calculations in mechanics. Thus, when two bodies are in contact, each experiences a contact force by the other. Friction, by definition, is the component of the contact force parallel to the surfaces in contact, which opposes impending or actual relative motion between the two surfaces. Note that it is not motion, but relative motion that the frictional force opposes. Consider a box lying in the compartment of a train that is accelerating. If the box is stationary relative to the train, it is in fact accelerating along with the train. What forces cause the acceleration of the box? Clearly, the only conceivable force in the horizontal direction is the force of friction. If there were no friction, the floor of the train would slip by and the box would remain at its initial position due to inertia (and hit the back side of the train). This impending relative motion is opposed by the static friction f s . Static friction provides the same acceleration to the box as that of the train, keeping it stationary relative to the train. Example . 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 . . Answer Since the acceleration of the box is due to the static friction, ma = f s ≤ µ s N = µ s m g i.e. a ≤ µ s g ∴ a max = µ s g = . x m s – = . m s – Example . See Fig. . . A mass of kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = ° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ? Fig. . Answer The forces acting on a block of mass m at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force f s opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have m g sin θ = f s , m g cos θ = N As θ increases, the self-adjusting frictional force f s increases until at θ = θ max , f s achieves its ) max s f = µ s N . Therefore, tan θ max = µ s or θ max = tan – µ s When θ becomes just a little more than θ max , there is a small net force on the block and it begins to slide. Note that θ max depends only on µ s and is independent of the mass of the block. For θ max = ° , µ s = tan ° = . Example . 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 . ? What is the tension in the string? (Take g = m s - ). Neglect the mass of the string. (a) (b) (c) Fig. . is the reason why discovery of the wheel has been a major milestone in human history. Rolling friction again has a complex origin, though somewhat different from that of static and sliding 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. We often regard friction as something undesirable. In many situations, like in a machine with different moving parts, friction does have a negative role. It opposes relative motion and thereby dissipates power in the form of heat, etc. Lubricants are a way of reducing kinetic friction in a machine. Another way is to use ball bearings between two moving parts of a machine [Fig. . (a)]. Since the rolling friction between ball bearings and the surfaces in contact is very small, power dissipation is reduced. A thin cushion of air maintained between solid surfaces in relative motion is another effective way of reducing friction (Fig. . (a)). In many practical situations, however, friction is critically needed. Kinetic friction that dissipates power is nevertheless important for quickly stopping relative motion. It is made use of by brakes in machines and automobiles. Similarly, static friction is important in daily life. We are able to walk because of friction. It is impossible for a car to move on a very slippery road. On an ordinary road, the friction between the tyres and the road provides the necessary external force to accelerate the car. Answer As the string is inextensible, and the pully is smooth, the kg block and the kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. . (b)), – T = a Apply the second law to motion of the trolley (Fig. . (c)), T – f k = a . Now f k = µ k N , Here µ k = . , N = x = N. Thus the equation for the motion of the trolley is T – . x = a Or T – = a . These equations give a = m s – = . m s - and T = . N. ⊳ Rolling friction A body like a ring or a sphere rolling without slipping over a horizontal plane will suffer no friction, in principle. At every instant, there is just one point of contact between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. We know, in practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling, some applied force is needed. For the same weight, rolling friction is much smaller (even by or orders of magnitude) than static or sliding friction. This Fig. . Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion. . CIRCULAR MOTION We have seen in Chapter that acceleration of a body moving in a circle of radius R with uniform speed v is v / R directed towards the centre. According to the second law, the force f c providing this acceleration is : c mv f = R ( . ) where m is the mass of the body. This force directed forwards the centre is called the centripetal force. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string. The centripetal force for motion of a planet around the sun is the is the static friction that provides the centripetal acceleration. Static friction opposes the impending motion of the car moving away from the circle. Using equation ( . ) & ( . ) we get the result ≤ mv f N R RN v Rg ≤ [ ∵ N = mg ] which is independent of the mass of the car. This shows that for a given value of µ s and R , there is a maximum speed of circular motion of the car possible, namely max v Rg ( . ) (a) (b) Fig. . Circular motion of a car on (a) a level road, (b) a banked road. gravitational force on the planet due to the sun. For a car taking a circular turn on a horizontal road, the centripetal force is the force of friction. The circular motion of a car on a flat and banked road give interesting application of the laws of motion. Motion of a car on a level road Three forces act on the car (Fig. . (a): (i) The weight of the car, mg (ii) Normal reaction, N (iii) Frictional force, f As there is no acceleration in the vertical direction N – mg = N = mg ( . ) The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This by definition is the frictional force. Note that it Motion of a car on a banked road We can reduce the contribution of friction to the circular motion of the car if the road is banked (Fig. . (b)). Since there is no acceleration along the vertical direction, the net force along this direction must be zero. Hence, N cos θ = mg + f sin θ ( .19a) The centripetal force is provided by the horizontal components of N and f . N sin θ + f cos θ = mv R ( .19b) But f s N ≤ ) max f f R ≤ = µ k f R k = µ µ s (co-efficient of static friction) and µ k (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that µ k is less than µ s . POINTS TO PONDER . Force is not always in the direction of motion. Depending on the situation, F may be along v , opposite to v , normal to v or may make some other angle with v . In every case, it is parallel to acceleration. . If v = at an instant, i.e. if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v = but the force continues to be its weight mg and the acceleration is not zero but g . . Force on a body at a given time is determined by the situation at the location of the body at that time. Force is not ‘carried’ by the body from its earlier history of motion. The moment after a stone is released out of an accelerated train, there is no horizontal force (or acceleration) on the stone, if the effects of the surrounding air are neglected. The stone then has only the vertical force of gravity. . In the second law of motion F = m a , F stands for the net force due to all material agencies external to the body. a is the effect of the force. m a should not be regarded as yet another force, besides F . . The centripetal force should not be regarded as yet another kind of force. It is simply a name given to the force that provides inward radial acceleration to a body in circular motion. We should always look for some material force like tension, gravitational force, electrical force, friction, etc as the centripetal force in any circular motion. . Static friction is a self-adjusting force up to its limit µ s N (f s ≤ µ s N) . Do not put f s = µ s N without being sure that the maximum value of static friction is coming into play. . The familiar equation mg = R for a body on a table is true only if the body is in equilibrium. The two forces mg and R can be different (e.g. a body in an accelerated lift). The equality of mg and R has no connection with the third law. . The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for simultaneous mutual forces between a pair of bodies. Unlike their meaning in ordinary language, action does not precede or cause reaction. Action and reaction act on different bodies. . The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’, ‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’ in different contexts. For clarity, every force and its equivalent terms encountered in mechanics should be reduced to the phrase ‘force on A by B ’. . For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground. . The objective concept of force in physics should not be confused with the subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of our body are subject to an inward force, but we have a feeling of being pushed outward – the direction of impending motion. EXERCISES (For simplicity in numerical calculations, take g = m s - ) . Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. . A pebble of mass . kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of ° with the horizontal direction? Ignore air resistance. . Give the magnitude and direction of the net force acting on a stone of mass . kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of km/h, (c ) just after it is dropped from the window of a train accelerating with m s - , (d) lying on the floor of a train which is accelerating with m s - , the stone being at rest relative to the train. Neglect air resistance throughout. . One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is : (i) T, (ii) l mv T − , (iii) l mv + T , (iv) T is the tension in the string. [Choose the correct alternative]. . A constant retarding force of N is applied to a body of mass kg moving initially with a speed of m s - . How long does the body take to stop ? . A constant force acting on a body of mass . kg changes its speed from . m s - to . m s - in s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ? . A body of mass kg is acted upon by two perpendicular forces N and N. Give the magnitude and direction of the acceleration of the body. . The driver of a three-wheeler moving with a speed of km/h sees a child standing in the middle of the road and brings his vehicle to rest in . s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is kg and the mass of the driver is kg. . A rocket with a lift-off mass , kg is blasted upwards with an initial acceleration of . m s - . Calculate the initial thrust (force) of the blast. . A body of mass . kg moving initially with a constant speed of m s - to the north is subject to a constant force of . N directed towards the south for s. Take the instant the force is applied to be t = , the position of the body at that time to be x = , and predict its position at t = – s, s, s. . A truck starts from rest and accelerates uniformly at . m s - . At t = s, a stone is dropped by a person standing on the top of the truck ( m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.) . A bob of mass . kg hung from the ceiling of a room by a string m long is set into oscillation. The speed of the bob at its mean position is m s - . What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position. . A man of mass kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of m s - , (b) downwards with a uniform acceleration of m s - , (c) upwards with a uniform acceleration of m s - . What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ? . Figure . shows the position-time graph of a particle of mass kg. What is the (a) force on the particle for t < , t > s, < t < s? (b) impulse at t = and t = s ? (Consider one-dimensional motion only). Fig. . . Two bodies of masses kg and kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case? . Two masses kg and kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released. . A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions. . Two billiard balls each of mass . kg moving in opposite directions with speed m s - collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ? . A shell of mass . kg is fired by a gun of mass kg. If the muzzle speed of the shell is m s - , what is the recoil speed of the gun ? . A batsman deflects a ball by an angle of ° without changing its initial speed which is equal to km/h. What is the impulse imparted to the ball ? (Mass of the ball is . kg.) . A stone of mass . kg tied to the end of a string is whirled round in a circle of radius . m with a speed of rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of N ? . If, in Exercise . , the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks : (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ? . Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.