Example 5.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a. constant rate of<\/b>. What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him.)<\/b><\/p>\n
Answer: <\/b>Because there are no nearby stars exerting gravitational pull on him and the little spaceship exerts minor gravitational attraction on him, the net force acting on the astronaut once he exits the vessel is zero. The astronaut’s acceleration is zero, i.e. Newton’s first rule of motion.<\/span><\/p>\n <\/p>\n Example 5.2A bullet of mass <\/b> moving with a speed of <\/b>enters a heavy wooden block and is stopped after a distance of<\/b>. What is the average resistive force exerted by the block on the bullet?<\/b><\/p>\n Answer: <\/b>The retardation <\/span> of the bullet (assumed constant) is given by<\/span><\/p>\n The retarding force, by the second law of motion, is<\/span><\/p>\n The bullet’s real resistive force, and hence its retardation, may not be uniform. As a result, the response simply reveals the average resistive force.<\/span><\/p>\n <\/p>\n Example 5.3 The motion of a particle of mass <\/b> is described by <\/b>Find the force acting on the particle.<\/b><\/p>\n Answer:<\/b>It is known that,<\/span><\/p>\n Now,<\/span><\/p>\n acceleration, <\/span><\/p>\n Then the force is given by Eq. (5.5)<\/span><\/p>\n As a result, the above equation explains the motion of a particle subjected to gravitational acceleration, with <\/span>representing the position coordinate in the direction of <\/span><\/p>\n <\/p>\n Example 5.4 A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of<\/b>. If the mass of the ball is<\/b>, determine the impulse imparted to the ball. (Assume linear motion of the ball)<\/b><\/p>\n Answer:<\/b>Change in momentum<\/span><\/p>\n Impulse <\/span>when the ball is inlinear motion.<\/span><\/p>\n <\/p>\n Example 5.5 Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in figure.<\/b><\/p>\n What is<\/b><\/p>\n (i) the direction of the force on the wall due to each ball? <\/b><\/p>\n Answer:<\/b>It might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at <\/span>to the normal. This answer is wrong. The force <\/span> the wall is normal to the wall in both cases.<\/span><\/p>\n By using the second law, and then use the third law<\/span><\/p>\n Let <\/span>be the speed <\/span>each ball before and after collision with the wall, and<\/span><\/p>\n the mass of each ball.<\/span><\/p>\n Choose the <\/span>and<\/span>axes as shown in the figure, and consider he change in momentum of the ball in each case:<\/span><\/p>\n case (a),<\/span><\/p>\n Impulse is the change in momentum vector. <\/span><\/p>\n Therefore,<\/span>component of impulse <\/span><\/p>\n SO, the force on the ball due to the wall is normal to the wall, along the negative -direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive <\/span>direction. Because the little time required for the collision has not been indicated in the problem, the quantity of force cannot be determined.<\/span><\/p>\n Case (b)<\/span><\/p>\n Note, while <\/span>changes sign after collision, <\/span>does not. Therefore,<\/span><\/p>\n component of impulse <\/span>component of impulse <\/span><\/p>\n The direction of impulse (and force) is the same as in <\/span>and is normal to the wall along the negative <\/span>direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive<\/span>direction.<\/span><\/p>\n <\/p>\n (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?<\/b><\/p>\n Answer:<\/b>The ratio of the magnitudes of the impulses imparted to the balls in <\/span> and <\/span>is<\/span><\/p>\n <\/p>\n Example 5.6 A mass of <\/b> is suspended by a rope of length <\/b>from the ceiling. A force of <\/b>in the horizontal direction is applied at the midpoint <\/b>of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take <\/b>). Neglect the mass of the rope.<\/b><\/p>\n Answer:<\/b>Figures b and c are known as free-body diagrams. Figure b is the free-body diagram of <\/span>and Fig. c is the free-body diagram of point <\/span>.<\/span><\/p>\n Consider the equilibrium of the weight <\/span><\/p>\n Consider the equilibrium of the point <\/span>under the action of three forces – the tensions <\/span>and<\/span>, and the horizontal force<\/span>. The resulting force’s horizontal and vertical components must each disappear separately:<\/span><\/p>\n which gives that<\/span><\/p>\n It’s worth noting that the solution is independent of the rope’s length (which is assumed to be massless) or the point at which the horizontal force is applied.<\/span><\/p>\n <\/p>\n Example 5.7 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 <\/b>.<\/b><\/p>\n Answer:<\/b>Because the box’s acceleration is caused by static friction,<\/span><\/p>\n <\/p>\n Example 5.8 A mass of <\/b> rests on a horizontal plane. The plane is gradually inclined until at an angle <\/b>with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?<\/b><\/p>\n Answer:<\/b>The forces acting on a block of mass <\/span>at rest on an inclined plane are:<\/span><\/p>\n (i) the weight <\/span>acting vertically downwards<\/span><\/p>\n (ii) the normal force <\/span>of the plane on the block, and<\/span><\/p>\n (iii) the static frictional force <\/span>opposing the impending motion. <\/span><\/p>\n The resultant of these forces must be zero in equilibrium. We can find the weight <\/span> by resolving it in the two ways illustrated.<\/span><\/p>\n As <\/span>increases, the self-adjusting frictional force, <\/span> increases until at <\/span>achieves its maximum value, <\/span><\/p>\n Therefore,<\/span><\/p>\n When <\/span>becomes just a little more than<\/span>. <\/span><\/p>\n There is a small net force on the block and it begins to slide. Note that <\/span>depends only on <\/span>and is independent of the mass of the block.<\/span><\/p>\n <\/p>\n Example 5.9 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 <\/b> ? What is the tension in the string? (Take<\/b>). Neglect the mass of the string.<\/b><\/p>\n Answer: <\/b>As the string is inextensible, and the pully is smooth, the <\/span> block and the <\/span>trolley both have same magnitude of acceleration. According to the second law to motion of the block ,<\/span><\/p>\n According to the second law to motion<\/span><\/p>\n Now,<\/span><\/p>\n Thus the equation for the motion of the trolley is <\/span><\/p>\n These equations give,<\/span><\/p>\n <\/p>\n Example 5.10 A cyclist speeding at <\/b> on a level road takes a sharp circular turn of radius <\/b> without reducing the speed. The co-efficient of static friction between the tyres and the road is <\/b>Will the cyclist slip while taking the turn?<\/b><\/p>\n Answer:<\/b>Frictional force alone can supply the centripetal force required to keep a bike going in a circular circle without skidding on an unbanked road. The frictional force is insufficient to deliver the requisite centripetal force if the cyclist’s speed is too high, or if the turn is too acute (i.e. of too short a radius), or both.<\/span><\/p>\n The condition for the cyclist not to slip is given by<\/span><\/p>\n Now, <\/span><\/p>\n The condition is disregarded. During the circular turn, the rider will slip.<\/span><\/p>\n <\/p>\n Example 5.11 Acircular racetrack of radius <\/b> is banked at an angle of<\/b>. If the coefficient of friction between the wheels of a race-car and the road is<\/b>, what is the <\/b><\/p>\n (a) optimum speed of the racecar to avoid wear and tear on its tyres, and <\/b><\/p>\n Answer:<\/b>On a banked road, the frictional force and the horizontal component of the normal force combine to generate centrifugal force, which keeps the automobile going in a circular turn without skidding. The normal response component is sufficient to give the required centripetal force at the optimal speed, and the frictional force is not required.<\/span><\/p>\n The optimum speed <\/span> is given as<\/span><\/p>\n Here <\/span> we have<\/span><\/p>\n (b) maximum permissible speed to avoid slipping?<\/b><\/p>\n Answer:<\/b>The maximum permissible speed <\/span>is given by <\/span><\/p>\n <\/p>\n Example 5.12 See A wooden block of mass <\/b>rests on a soft horizontal floor. When an iron cylinder of mass <\/b>is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of<\/b>. <\/b><\/p>\n <\/p>\n What is the action of the block on the floor (a) before and (b) after the floor yields ? Take<\/b>. Identify the action-reaction pairs in the problem.<\/b><\/p>\n Answer:<\/b><\/p>\n (a) The block has come to a halt on the floor<\/span>.<\/span>Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to<\/span>; and the normal force <\/span> of the floor on the block. According to the First Law of motion,the net force on the block must be zero i.e.,<\/span>.By applying third law of the motion on the the action of the block (i.e. the force exerted on the floor by the block) is equal to <\/span> and directed vertically downwards.<\/span><\/p>\n (b) The system (block + cylinder) accelerates downwards with<\/span>. The free-body diagram of the system shows two forces on the system : the force of gravity due to the earth<\/span>; and the normal force <\/span>by the floor. The internal forces between the block and the cylinder are not shown in the system’s free-body diagram.Applying the second law to the system,<\/span><\/p>\n By the third law, the action of the system on the floor is equal to <\/span>vertically downward. Action-reaction pairs:<\/span><\/p>\n For (a): (i) the force of gravity <\/span>on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to <\/span>directed upwards (not shown in the figure).<\/span><\/p>\n (ii) the force on the floor by the block (action); the force on the block by the floor (reaction).<\/span><\/p>\n For (b): (i) the force of gravity <\/span> on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to <\/span>,<\/span><\/p>\n directed upwards (not shown in the figure).<\/span><\/p>\n (ii) the system’s force on the floor (activity); the system’s force on the floor (reaction). <\/span><\/p>\n In addition, for (b), the cylinder’s force on the block and the block’s force on the cylinder form an action-reaction pair:<\/span><\/p>\n The crucial thing to remember is that an action-reaction pair is made up of reciprocal forces between two bodies that are always equal and opposing. <\/span><\/p>\n An action-reaction pair can never be formed by two equal and opposite forces acting on the same body. <\/span><\/p>\n The normal force on the mass by the floor and the force of gravity on the mass in (a) or (b) are not action-reaction pairings.These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is<\/span>, while the normal force <\/span> is<\/span><\/p>\n <\/p>\n Exercises<\/b><\/p>\n <\/p>\n 5.1 Give the magnitude and direction of the net force acting on<\/b><\/p>\n Answer:<\/b>Net force is zero.The raindrops are falling at a steady rate.As a result, the acceleration is zero. <\/span><\/p>\n The net force exerted on the rain drop is zero, according to Newton’s second law of motion.<\/span><\/p>\n Answer: <\/b>Net force is zero. The cork’s weight is acting downward. <\/span><\/p>\n It is counterbalanced by the water’s buoyant force in the upward direction, <\/span><\/p>\n As a result, there is no net force acting on the floating cork.<\/span><\/p>\n Answer:<\/b>Net force is zero.In the sky, the kite is stationary, i.e. it is not moving at all. <\/span><\/p>\n As a result, according to Newton’s first rule of motion, there is no net force acting on the kite.<\/span><\/p>\n <\/p>\n Answer:<\/b>Net force is zero.The automobile is going at a consistent speed down a bumpy route. <\/span><\/p>\n As a result, it has no acceleration. There is no net force operating on the automobile, according to Newton’s second law of motion.<\/span><\/p>\n <\/p>\n Answer: <\/b>Net force is zero.All fields have no affect on the high-speed electron. <\/span><\/p>\n As a result, there is no net force acting on the electron.<\/span><\/p>\n <\/p>\n 5.2 A pebble of mass <\/b> is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,<\/b><\/p>\n Ignore air resistance.<\/b><\/p>\n Answer:<\/b>, in vertically downward direction, in all cases.<\/span><\/p>\n Gravitational acceleration always operates downward, regardless of the direction of motion of an item. <\/span><\/p>\n\n
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