NEWTONS LAWS OF MOTION AND FRICTIONS STRAIGHT LINES

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1 EWTOS LAWS O OTIO AD RICTIOS STRAIGHT LIES ITRODUCTIO In this chapter, we shall study the motion o bodies along with the causes o their motion assuming that mass is constant. In addition, we are going to treat the translational motion o rigid bodies. Beore we start doing anything else and since this chapter is about the Laws o otion, we are going to state these laws irst. Later we are going to discuss its applications in Statics and Dynamics. ORCE A pull or push which changes or tends to change the state o rest or o uniorm motion or direction o motion o any object is called orce. orce is the interaction between the object and the source (providing the pull or push). It is a vector quantity. Eect o resultant orce : (1) may change only speed () may change only direction o motion. (3) may change both the speed and direction o motion. (4) may change size and shape o a body kgm g cm Unit o orce : newton and (S System)dyne and s s 1 newton = 10 5 dyne Dimensional ormula o orce : [ L T ] (CGS System) undamental orces All the orces observed in nature such as muscular orce, tension, reaction, riction, elastic, weight, electric, magnetic, nuclear, etc., can be explained in terms o only ollowing our basic interactions: Gravitational orce The orce o interaction which exists between two particles o masses m 1 and m, due to their masses is called gravitational orce. Electromagnetic orce orce exerted by one particle on the other because o the electric charge on the particles is called electromagnetic orce. ollowing are the main characteristics o electromagnetic orce (a) These can be attractive or repulsive. (b) These are long range orces (c) These depend on the nature o medium between the charged particles. (d) All macroscopic orces (except gravitational) which we experience as push or pull or by contact are electromagnetic, i.e., tension in a rope, the orce o riction, normal reaction, muscular orce, and orce experienced by a deormed spring are electromagnetic orces. These are maniestations o the electromagnetic attractions and repulsions between atoms/molecules. uclear orce It is the strongest orce. It keeps nucleons (neutrons and protons) together inside the nucleus inspite o large electric repulsion between protons. Radioactivity, ission, and usion, etc. result because o unbalancing o nuclear orces. It acts within the nucleus that too upto a very small distance. Weak orce It acts between any two elementary particles. Under its action a neutron can change into a proton emitting an electron and a particle called antineutrino. The range o weak orce is very small, in act much smaller than the size o a proton or a neutron. 1

2 IDS O ORCES orce System Wise Source (Origin) Reerence rame ature (Workdone) ield Contact Real Pseudo External Internal Conservative on-conservative The irst classiication is sel explanatory. All orces in case o dynamics o a particle can be classiied in two ways w.r.t. its origin as: (a) contact orces, and (b) non-contact orces. on-contact orce Those orces which do not require contact between the bodies to act, or example gravitational orce, electromagnetic orce etc., are known as non-contact orces. Contact orce () orces which act when bodies are in contact are known as contact orces. It is usually convenient to resolve contact orces into components, one parallel to the surace o contact, the other perpendicular to the surace o contact. ormal orce or ormal Reaction (): The component o the contact orce that is perpendicular to the surace o contact is known as the ormal reaction orce. It measures how strongly the suraces in contact are pressed together. riction orce (): It is the component o the contact orce parallel to the surace o contact. The direction o the orce o riction is opposite to the direction o relative motion between the suraces, or is such as to oppose any tendency o relative motion between the suraces. Example 1 : Two blocks are kept in contact on a smooth surace as shown in igure. Draw normal orce exerted by A on B. Block A does not push block B, so there is no molecular interaction between A and B. Hence normal orce exerted by A on B is zero. TESIO: When a body is connected by means o a string or a rope, a orce may be exerted on the body by the string or the rope. This orce is called tension. (i) I a string is ideal (inextensible and massless), then the magnitude o the accelerations o any number o masses connected through this string is always same. (ii) a a a m a m I a string is massless, the tension in it is same everywhere; on the other hand i a string is not massless (i.e. its mass is not negligible), tension at dierent points may be dierent.

3 C B C B m (iii) A String is massless i.e TA TB TC A String is not massless i.e. TA TB TC and TA TB TC I riction is present between the pulley and the string, tension is dierent on the two sides o the pulley But i the riction is absent between the pulley and the string, tension will be same on both sides o the pulley, provided the pulley is massless. T T 1 There is no riction between pulley and string i.e., T T 1 T 1 There is riction between pulley and string i.e., T T 1 T ELASTIC ORCE An ideal spring ollows Hooke s law which says that the orce applied by a springs on bodies connected to it is proportional to its extension or compression (change in length rom its natural length). Consider a light spring which is connected to a vertical wall as shown in the igure. Suppose that it is pulled to the right by means o a orce, which causes the spring to get elongated by x over its natural length instantaneous rest to ring orce will be equal to applied orce and opposite direction.. Then, x = kx where k is a constant that is characteristic o the spring also known as the spring constant. EWTO S LAWS O OTIO irst Law o otion: Every body continues in its state o rest, or o uniorm motion in a straight line, unless it is compelled by an external impressed orce to change that state. Second Law o otion: The rate o change o momentum o a body is proportional to the impressed orce, and takes place in the direction in which the orce acts. dp dp dv m ma, i m is a constant xz dv dm or, m v, i m is not constant. REARS 1. The direction o motion o a particle does not in general coincide with the direction o the orce acting on it. It is the rate o change o velocity, which is related to the orce.. The acceleration must be measured with respect to an inertial reerence rame. x Example : A body o mass m = 1 kg alls rom a height h = 0 m rom the ground level. What is the magnitude o total change in momentum o the body beore it strikes the ground? Since the body alls rom rest (u = 0) through a distance h beore striking the ground, the speed v o the body is given by kinematical equation. 3

4 v = u + as; v = gh Putting a = g and s = h, we obtain The magnitude o total change in momentum o the body p = mv 0 = mv, where v = gh p = m gh = (1) (10 0) kg-m/sec p = 0 kg-m/sec. Third Law o otion: To every action there is an equal and opposite reaction. Here by action and reaction, one means orce only, i an object exerts a orce on a second, then the second object exerts an equal but opposite orce in the opposite direction. man(action) man ground. Ground(Reaction) Signiicance o ewton s Laws: (a) (b) (c) The irst law tells us about the natural state o a body, which is in motion along a straight line with constant speed or rest. It is also known the law o inertia. The second law tells us that i a body does not ollow its natural state o motion then it is under the inluence o other bodies, that is, a net unbalanced orce must be acting on it. The third law tell us about the nature o orce, that is, orce exist in pairs. REARS 1. The acceleration must be measured with respect to an inertial reerence rame.. The direction o motion o a particle does not in general coincide with the direction o the orce acting on it. It is the rate o change o velocity, which is related to the orce. Example 3 : W A horse reuses to pull a cart. The horse reasons, according to ewton s third law, whatever orce I exert on the cart, the cart will exert an equal and opposite orce on me so the resultant orce will be zero and I will have no chance o accelerating the cart. What is wrong with this reasoning? T T igure Horse pulling a cart. The cart will accelerate to the right i the orce T exerted on it by the horse is greater than the rictional orce exerted on the cart by the ground. The orce Tis equal and opposite to T, but because it is exerted on the horse it has no eect on the motion o the cart. igure is a sketch o a horse pulling the cart. Since we are interested in the motion o the cart, we have circled it and indicated the orces acting on it. The orce exerted by the horse is labelled T. Other orces on the cart are its weight W, the vertical support orce o ground, and the horizontal orce exerted by the ground labeled (or riction). The vertical orces W and balance each other. The horizontal orces are T to the right and to the let. The cart will accelerate i T is greater than. T is the reaction orce exerted on the horse, not the cart. It has no eect on the motion o the cart. It does eect the motion o the horse. I the horse is to accelerate to the right, there must be a orce (to the right) exerted by the ground on the horse s eet that is greater than T. This example illustrates the importance o a simple diagram (ree body diagram) in solving mechanics problems. Had the horse drawn a simple diagram, he would have seen that he need only push back hard against the ground so that the ground would push him orward. REE BODY DIAGRA In this diagram the object o interest is isolated rom its surroundings and the interactions between the object and the surroundings are represented in terms o orces. 4

5 m mg Earth (A).B.D. o a particle in gravitational ield. The earth pulls the particle o mass m by a orce mg. a g (B).B.D. o a block placed on a horizontal surace. Two vertical orces act on the block : The earth pulls the block downward by mg, and the surace pushes the block upward by. g (D).B.D. o a block placed on a rough surace being pulled by an external orce. There are our orces acting on the block: the gravitational pull g; the normal reaction ; the external orce ; and the tangential orce o riction. m mg (C).B.D. o a ball suspended by a string. Two vertical orces act on the ball : The earth pulls the ball downward by mg and the string pulls the ball upwards by T. x T kx g (E).B.D. o a block supported by a spring o stiness k. Two vertical orces act on the block: the gravitational pull, and the spring orce kx. EQUILIBRIU A system is said to be in equilibrium i it does not tend to undergo any urther change o its own, i.e. any urther change must be produced by external means (e.g. orces) i.e., orces which have zero linear resultant and zero torque. Such orces (and the object) are said to be in equilibrium. Resolution o a orce: When a orce is replaced by an equivalent set o components, it is said to be resolved. H = cos V = sin REARS athematically a body is said to be in equilibrium i (a) net orce acting on it is zero i.e. net 0 (b) net moments o all the orces acting on it about any axis is also zero. I or a body 0 i.e. the body is said to be in translational equilibrium 0 i.e., ma 0 dv m. 0 (as m is assumed to be a constant). Or v const. i.e., i a body is in translational equilibrium it will be either at rest or in uniorm motion in a straight line. I it is at rest, the equilibrium is called static. STEPS TO SOLVE THE PROBLE 1. ake a simple sketch showing the body under consideration.. Identiy the orces acting on the body, draw arrows on your sketch to show the direction o each orce acting on the body. 3. Choose a coordinate system and resolve the orces into components that are parallel to the coordinate axes. 4. Write the equation or equilibrium along each axis o the co-ordinate system. 5. Solve the equation or the required unknown(s). 5

6 Example 4 : A block o mass 10 kg is suspended with two strings, as shown in the igure. ind the tension in each string. (g = 10 m/s ) T 1 30 o T O The ree body diagram o the joint O is drawn as shown in the igure T cos 30 o 30 o y T 10 kg x T 1 O T sin 30 o Example 5 : 100.B.D. o the joint O Applying equations or equilibrium. x = 0 T sin 30 T 1 = 0..(i) y = 0 T cos = 0..(ii) Thus, T = 00 0 cos 30 3 Substituting the value o T in equation (i), we get T 1 = T sin 30 = Two blocks o mass m 1 and m are attached at the ends o an inextensible string, which passes over a smooth massless pulley. I m 1 > m, ind (a) the acceleration o each block (b) the tension in the string. The ree body diagram o each block is shown in the igure. T T m m1 a m m 1 a 1 m g m 1 g ote The block m 1 is assumed to be moving downward and the block m is assumed to be moving upward. It is merely an assumption and it does not imply the real direction o motion. I the values o a 1 and a come out to be positive then only the assumed directions are correct; otherwise the bodies move in the opposite directions. Since the pulley is smooth and massless, thereore, the tension on each side o the pulley is same. Applying ewton s second Law on Block m 1 m 1 g T = m 1 a 1..(i) Block m m g + T = m a..(ii) umber o unknowns : T, a 1 and a (three) umber o equations : only two Obviously, we require one more equation to solve the problem. ote that the whenever one inds the number o equations less than the number o unknowns, one must think about the constraint relation. In the previous problems we have obtained the constraint relation by experience and judgment. ow we are going to explain the mathematical procedure or this. 6

7 How to determine Constraint Relation? 1. Assume the direction o acceleration o each block, e.g. a 1 (downward) and a (upward) in this case.. Locate the position o each block rom a ixed point (depending on convenience), e.g. centre o the pulley in this case. 3. Identiy the constraint and write down the equation o constraint in terms o the distance assumed. a x x 1 m m 1 Position o each block is located w.r.t. centre o the pulley. or example, in the chosen problem, the length o string remains constant is the constraint or restriction. Thus, x 1 + x = constant Dierentiating both the sides w.r.t. time, we get dx1 dx 0 Each term on the let side represents velocity o the block. Since we have to ind a relation between accelerations, thereore, we dierentiate it once again w.r.t. time. Thus d x1 d x 0 Since the block m 1 is assumed to be moving downward (x 1 is increasing with time) dx1 a, 1 a 1 > 0 and block m is assumed to be moving upward (x is decreasing with time) dx a, a > 0 Thus a 1 a = 0 or a 1 = a = a Alternatively, one can also assume the distance o each block rom the ground as shown in the igure. One can easily see that an expression or length o the string can not be written unless we locate the centre o the pulley w.r.t. ground. I y 0 be the distance o the ground rom the centre o the pulley then the length o string is (y 0 y 1 ) + (y 0 y ) = constant y 0 y 1 y = constant Dierentiating twice w.r.t. time, we get ow, 0 dy1 d y d y 0 1 a and 1 dy a Thus a 1 a = 0 a 1 = a = a Substituting a 1 = a = a in equations (i) and (ii) and ater solving them, we get a = m m1 m g m 1 a y m m1m also T = g m1 m m 1 y 1 a 1 a 1 y o 7

8 Example 6 : The igure shows one end o a string being pulled down at constant velocity v. ind the velocity o mass m as a unction o x. Using constraint equation x b + y = length o string = constant dx dy Dierentiating w.r.t. time :. x x b = 0 dy = v dx = v x b x REERECE RAES A rame o reerence is basically a coordinate system in which motion o object is analyzed. There are two types o reerence rames. (a) Inertial reerence rame: rame o reerence moving with constant velocity. (b) on-inertial reerence rame: A rame o reerence moving with non-zero acceleration. m ma where a is the acceleration o the body relative to the non-inertial rame. The term m is an example o a pseudo-orce. In general, the Second Law takes the orm: P ma, in a non-inertial rame. Example 7 : All suraces are smooth in the adjoining igure. ind such that block remains stationary with respect to wedge. Acceleration o (block + wedge) is a = m Let us solve the problem by using both rames. rom inertial rame o reerence (Ground).B.D. o block w.r.t. ground (Apply real orces): with respect to ground block is moving with an acceleration a. y = 0 cos = mg...(i) and x = ma sin = ma...(ii) rom Eqs. (i) and (ii) a = g tan = ( + m) a = ( + m) g tan rom non inertial rame o reerence (Wedge) :.B.D. o block w.r.t. wedge (real orces + pseudo orce) w.r.t. wedge, block is stationary y = 0 cos = mg...(iii) x = 0 sin = ma...(iv) rom Eqs. (iii) and (iv), we will get the same result i.e. = ( + m) g tan. 8

9 WEIGHIG ACHIE : A weighing machine does not measure the weight but measures the orce exerted by object on its upper surace. Example 7 : A man o mass 60 kg is standing on a weighing machine () o mass 5kg placed on ground. Another same weighing machine is placed over man s head. A block o mass 50kg is put on the weighing machine (1). Calculate the readings o weighing machines (1) and (). Answer : 500, RICTIO Whenever the surace o a body slides over that o another, each body exerts a orce o riction on the other parallel to the suraces. The orce o riction on each body is in a direction opposite to its motion relative to the other body. The orce o riction comes into action only when there is a relative motion between the two contact suraces or when an attempt is made to have it. It is a sel adjusting orce, it can adjust its magnitude to any value between zero and the limiting (maximum) value i.e 0 max The rictional orce acting between any two suraces at rest with respect to each other is called the orce o static riction. And the rictional orce acting between suraces in relative motion with respect to each other is called the orce o kinetic riction or sliding riction. TYPES O ORCE O RICTIO riction acting between two bodies the relative motion is known as kinetic riction while the orce o riction acting between bodies, opposing the onset o relative motion is known as static riction. The orce o riction acting between layer o luids in known as viscous orce. It is worth noting that : (i) (ii) I a body is at rest on a rigid horizontal surace and no pulling orce is acting on it, orce o riction on it is zero. ow i a orce is applied to pull the body and it does not move, the orce o riction which acts on it is equal in magnitude and opposite in direction to the applied orce i.e. riction is a sel-adjusting orce. Since the body is at rest, the riction is called static riction. ma a mg =0 (iii) (iv) mg =ma mg = I the applied orce is increased, the orce o static riction also increases. I the applied orce exceeds a certain (maximum) value, the body starts moving. This maximum orce o static riction upto which body does not move is called limiting riction. Thus, the orce o static riction is a sel-adjusting orce with an upper limit which is known as the limiting orce o static riction. This limiting orce o riction ( L ) is ound experimentally to proportional to the orce o normal reaction i.e. = dynamic riction L or, L s where s is a dimensionless constant called coeicient o static riction which depends upon the nature o suraces in contact. riction = s Applied orce 9

10 (v) I the applied orce is increased above its limiting value, it is observed that relative motion occurs. The orce o riction opposing motion is known as kinetic riction (sliding riction). where is coeicient o kinetic riction and is less than According to laws o riction, the orce o kinetic riction is independent o the area o contact and independent o the relative velocities o the bodies. s. 1. In problems i s and REARS are separately not given but only a single is given then use and L S. I more than two blocks are placed one over the other on a the ground then normal reaction between the two blocks will be equal to the weight o the blocks over the common surace. or example 1 mag (ormal reaction between A and B) = (mb m A)g (ormal reaction between B and C) Example 8 : A block o mass = 10 kg is placed at rest on a horizontal surace as shown in the igure. The coeicient o riction between the block and the surace is µ s = 0.3 and k = 0.. It is pulled with a horizontal orce. ind the magnitude o the riction i (a) = 0 (b) = 40 The maximum value o riction orce is s(max) =µ s = µ s g or s(max) = (0.3) (10) (10) = 30 (a) To keep the block stationary the magnitude o riction orce should be = = 0 since < max. Thereore the orce o riction is =0 (b) s(max) < so the body moves and now orce o riction = k = 0.(10)(g) = 0 ote that in this case riction orce is unable to keep the block stationary and the block accelerates with a = 40 0 k m/s 10 A B C g = 0 = 40 ote that riction orce is not always equal to µ s. It is the limiting or maximum value o static riction. At any stage riction may attain any value between 0 and µ s. 0 µ s AGLE O RICTIO Suppose a body is placed on an inclined surace whose angle o inclination varies between 0 to /. The coeicient o riction between the body and the surace is µ s. Let the initial value o be zero and i we slowly start increasing the value o, then at a particular value o = the block just starts to move. This value o = is called the angle o riction. athematically, i the block is just about to move, then mg sin = When =, mg sin = max Or mg sin = µ s = µ s mg cos or tan = µ s Thus = tan 1 µ s mg sin g mg cos A block o mass m is placed on an incline whose inclination may be varied between 0 to /. When = the riction orce is maximum and block just starts sliding 10

11 The angle o riction is that minimum angle o inclination o the inclined plane at which a body placed at rest on the inclined plane is about to slide down. When (or tan -1 µ s ) the body is in equilibrium. When the angle o inclination is more than the angle o riction ( > ) the block starts sliding down with acceleration. And, i we wish to keep it in equilibrium an external orce has to be applied. Example 9 : A block o mass m is placed at rest on an inclined plane whose angle with the horizontal is more than the angle o riction ( > ). An external orce parallel to the inclined plane is applied in the upward direction to keep it in equilibrium. ind the magnitude o the orce. m > s Solution Since the block has a tendency to move downward, the orce o riction acts upward on the block as shown in its ree body diagram. Applying equation o equilibrium, = mg cos mg sin mg cos ree Body Diagram o the block when riction orce is acting at its maximum value + max = mg sin or = mg sin max = mg sin µ s mg cos or = mg (sin µ s cos ) ote that this is the minimum magnitude o orce required to keep it in equilibrium. I we apply a orce slightly more than this the block does not start moving up but the magnitude o the riction orce gets reduced. It becomes equal to zero when the external orce attain a value equal to = mg sin, as shown in igure. I the magnitude o is urther increased then the block has a tendency to move upward; the direction o riction orce gets reversed. The block will not start moving up unless the external orce attains the maximum value. The ree body diagram o the body is shown in igure Applying the equations o equilibrium = mg cos mgsin =mgsin = 0 mgcos When = mg sin, = 0 max max = mg sin + max or max = mg (sin + µ s cos ) Conclusion The block remains stationary i min max or mg(sin µ s cos ) mg(sin + µ s cos ) mg sin max mg cos The body has a tendency to move upward. The riction orce acts downward. 11