Lecture Notes in Physics

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PHYSICS 101 AN INTRODUCTION TO PHYSICS This course of 45 video lectures, as well as accompanying notes, have been developed and presented by Dr. Pervez Amirali Hoodbhoy, professor of physics at Quaid-e-Azam University, Islamabad, for the Virtual University of Pakistan, Lahore. GENERAL INFORMATION Purpose: This course aims at providing the student a good understanding of physics at the elementary level. Physics is essential for understanding the modern world, and is a definite part of its culture. Background: It will be assumed that the student has taken physics and mathematics th at the F.Sc level, i.e. the 12 year of schooling. However, B.Sc students are also likely to find the course useful. Calculus is not assumed and some essential concepts will be developed as the course progresses. Algebra and trigonometry are essential. However, for physics, the more mathematics one knows the better. Scope and Duration: The course has 45 lectures, each of somewhat less than one hour duration. All main fields of physics will be covered, together with several applications in each. Language: For ease of communication, all lectures are in Urdu. However, English or Latin technical terms have been used where necessary. The student must remember that further study and research in science is possible only if he or she has an adequate grasp of English. Textbook: There is no prescribed textbook. However, you are strongly recommended to read a book at the level of “College Physics” by Halliday and Resnick (any edition). There are many other such books too, such as “University Physics” by Young and Freedman. Study any book that you are comfortable with, preferably by a well- established foreign author. Avoid local authors because they usually copy. After listening to a lecture, go read the relevant chapter. Please remember that these notes cover only some things that you should know and are not meant to be complete. Assignments: At the end of every lecture summary you will find a few questions that you should answer. The book you choose to consult will have many more. Those students who are seriously interested in the subject are advised to work out several of the questions posed there. In physics you cannot hope to gain mastery of the subject without extensive problem solving. Examinations: Their schedules will be announced from time to time. Tutors: Their duty is to help you, and they will respond to all genuine questions. However, please do not overload them as they have to deal with a large number of students. Happy studying Acknowledgements: I thank the Virtual University team and administration for excellent cooperation, as well as Mansoor Noori and Naeem Shahid, for valuable help. Copyright: Pervez Amirali Hoodbhoy, Professor of Physics, Quaid-e-Azam University, Islamabad. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 3Summary of Lecture 1 – INTRODUCTION TO PHYSICS 1. Physics is a science. Science works according to the scientific method. The scientific method accepts only reason, logic, and experimental evidence to tell between what is scientifically correct and what is not. Scientists do not simply believe – they test, and keep testing until satisfied. Just because some “big scientist” says something is right, that thing does not become a fact of science. Unless a discovery is repeatedly established in different laboratories at different times by different people, or the same theoretical result is derived by clear use of established rules, we do not accept it as a scientific discovery. The real strength of science lies in the fact that it continually keeps challenging itself. 2. It is thought that the laws of physics do not change from place to place. This is why experiments carried out in different countries by different scientists – of any religion or race – have always led to the same results if the experiments have been done honestly and correctly. We also think that the laws of physics today are the same as they were in the past. Evidence, contained in the light that left distant stars billions of years ago, strongly indicates that the laws operating at that time were no different than those today. The spectra of different elements then and now are impossible to tell apart, even though physicists have looked very carefully. 3. This course will cover the following broad categories: a) Classical Mechanics, which deals with the motion of bodies under the action of forces. This is often called Newtonian mechanics as well. b) Electromagnetism, whose objective is to study how charges behave under the influence of electric and magnetic fields as well as understand how charges can create these fields. c) Thermal Physics, in which one studies the nature of heat and the changes that the addition of heat brings about in matter. d) Quantum Mechanics, which primarily deals with the physics of small objects such as atoms, nuclei, quarks, etc. However, Quantum Mechanics will be treated only briefly for lack of time. 4. Every physical quantity can be expressed in terms of three fundamental dimensions: Mass (M), Length (L), Time (T). Some examples: −1 Speed LT −2 Acceleration LT −2 Force MLT 22− Energy MLT −12− Pressure MLT You cannot add quantities that have different dimensions. So force can be added to force, but force can never be added to energy, etc. A formula is definitely wrong if the dimensions on the left and right sides of the equal sign are different. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 45. Remember that any function f() x takes as input a dimensionless number x and outputs a quantity f (which may, or may not have a dimension). Take, for example, 35 θθ the function f()θ=sinθ. You know the expansion: sinθθ= − + − ⋅⋅⋅ If θ 3 5 had a dimension then you would be adding up quantities of different dimensions, and that is not allowed. 6. Do not confuse units and dimensions. We can use different units to measure the same physical quantity. So, for example, you can measure the mass in units of kilograms, pounds, or even in sair and chatak In this course we shall always use the MKS or Metre-Kilogram-Second system. When you want to convert from one hsystem to another, be methodical as in the example below: mi mi ft11 m hr m 1 =× 1 5280 × × = 0.447 hr hr mi 3.28 ft 3600 s s When you write it out in this manner, note that various quantities cancel out cleanly in the numerator and denominator. So you never make a mistake 7. A good scientist first thinks of the larger picture and then of the finer details. So, estimating orders of magnitude is extremely important. Students often make the mistake of trying to get the decimal points right instead of the first digit – which obviously matters the most So if you are asked to calculate the height of some 6 building using some data and you come up with 0.301219 metres or 4.01219×10 metres, then the answer is plain nonsense even though you may have miraculously got the last six digits right. Physics is commonsense first, so use your intelligence before submitting any answer. 8. Always check your equations to see if they have the same dimensions on the left side as on the right. So, for example, from this principle we can see the equation 22 22 22 v=+ua 2t is clearly wrong, whereas v=+u 13at could possibly be a correct relation. (Here v and u are velocities, a is acceleration, and t is time.) Note here that I use the word possibly because the dimensions on both sides match up in this case. 9. Whenever you derive an equation that is a little complicated, see if you can find a special limit where it becomes simple and transparent. So, sometimes it is helpful to imagine that some quantity in it is very large or very small. Where possible, make a “mental graph” so that you can picture an equation. So, for example, a formula for the distribution of molecular speeds in a 22 -(v-v ) / a 0 gas could look like fe (v)= v . Even without knowing the value of a you can immediately see that a) f (v) goes to zero for large values of v, and v= 0. b) The maximum value of f (v) occurs at v and the 0 function decreases on both side of this value. v 0 Virtual University Physics 101 – Dr. Pervez Hoodbhoy 5QUESTIONS AND EXERCISES – 1 1. Scientists are told to doubt everything and not believe anything that is not provable. Is this a strength or weakness of science? 2. According to the philosopher of science, Sir Karl Popper, even the most well- established and popular scientific theory can never be proved – it can only be disproved. Discuss this strange sounding claim. 3. Suppose we measure time by using hour glasses filled with sand. Discuss the various errors that would exist if we tried to use this as a world standard for time. Give a rough estimate for the error over a 24 hour period. 4. Which of the following equations are definitely wrong: 2 (a) v= at 2 (bx ) =+ 3at vt 2 mc (cE ) = 2 v 1− 2 c E ()dP = 2 v c 1− 2 c In the above xt == distance, time, c and v= velocity, a= acceleration, mE == mass, energy, and P=pressure. 5. Find how much time is taken for a telephone signal to go from your mobile to your friend’s mobile. Assume that the relevant satellite is orbiting the earth at a distance of 250 kilometres, and that the electronic circuits have a delay of 300 microseconds. Give your answer in microseconds. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 6Summary of Lecture 2 – KINEMATICS I 1. x(t) is called displacement and it denotes the position of a body at time. If the displacement is positive then that body is to the right of the chosen origin and if negative, then it is to the left. 2. If a body is moving with average speed v then in time t it will cover a distance d=vt. But, in fact, the speed of a car changes from time to time and so one should limit the use of this formula to small time differences only. So, more accurately, one defines an average speed over the small time interval ∆t as: distance travelled in time ∆t average speed = ∆t 3. We define instantaneous velocity at any time t as: xt()−xt( ) ∆x 21 v =≡ . tt−∆t 21 Here ∆∆ xt and are both very small quantities that tend to zero but their ratio v does not. x 2 ∆x x 1 t∆t t 1 2 4. Just as we have defined velocity as the rate of change of distance, similarly we can define instantaneous acceleration at any time t as: v(tt )− v( )∆v 21 a=≡ . tt−∆t 21 Here ∆∆ v and t are both very small quantities that tend to zero but their ratio a is not zero, in general. Negative acceleration is called deceleration. The speed of a decelerating body decreases with time. 5. Some students are puzzled by the fact that a body can have a very large acceleration but can be standing still at a given time. In fact, it can be moving in the opposite direction to its acceleration. There is actually nothing strange here because position, velocity, and acceleration are independent quantities. This means that specifying one does not specify the other. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 76. For constant speed and a body that is at x=0 at time t=0, x increases linearly with time, x∝= tx (or vt). If the body is at position xt at time = 0, then at 0 time t, x=+xt v . 0 7. For constant acceleration and a body that starts from rest at t= 0, v increases linearly with time, v∝=ta (or vt). If the body has speed v at t= 0, then at 0 time ,ta v=+t v . 0 8. We know in (6) above how far a body moving at constant speed moves in time t. But what if the body is changing its speed? If the speed is increasing linearly (i.e. constant acceleration), then the answer is particularly simple: just use the same formula as in (6) but use the average speed: (v++ v at) / 2 . So 00 2 1 we get x=+xa (v+ v+t)t / 2=+x vt+at . This formula tells you how 00 0 0 0 2 far a body moves in time t if it moves with constant acceleration a, and if started at position x at t=0 with speed v . 0 0 9. We can eliminate the time using (7), and arrive at another useful formula that tells us what the final speed will be after the body has traveled a distance 22 equal to x−= xt after time , v v+ 2a(x−x ). 00 0 10. Vectors: a quantity that has a size as well as direction is called a vector. So, for example, the wind blows with some speed and in some direction. So the wind velocity is a vector. 11. If we choose axes, then a vector is fixed by its components along those axes. In one dimension, a vector has only one component (call it the x-component). In two dimensions, a vector has both x and y components. In three dimensions, the components are along the x,y,z axes. r 12. If we denote a vector rx= ( ,y) then, r ==x r cosθ , andr =y=r sinθ. xy 22 2 Note that x+=yr . Also, that tanθ=y /x. 13. Two vectors can be added together geometrically. We take any one vector, move it without changing its direction so that both vectors start from the same point, and then make a parallelogram. The diagonal of the parallelogram is the resultant. rrr CA=+B 14. Two vectors can also be added algebraically. In this case, we simply add the components of the two vectors along each axis separately. So, for example, Two vectors can be put together as (1.5,2.4)+− (1, 1)= (2.5,1.4). Virtual University Physics 101 – Dr. Pervez Hoodbhoy 8QUESTIONS AND EXERCISES – 2 1. A train starts at rest, accelerates with constant acceleration a for 5 minutes, then travels at constant speed for another 5 minutes, and then decelerates with –a. Suppose it travels a distance of 10km in all. Find a. 2. A ball is dropped from a height of 10 m. At the same time, another ball is thrown vertically upwards at an initial speed of 10 m/sec. How high above the ground will the two balls collide? 3. A train follows the speed-time graph as in the figure below: v 0 4T T T t→ 5 5 a) Find the total distance travelled during its journey. b) Find the average speed. 4. Find the resultant of the two velocity vectors v== (3,2) and v (4,6) and the angle 12 that the resultant makes with the first vector. 5. a) Is it always true that the resultant of two vectors is bigger than either one of the vectors? b) Is it possible to define a function of two vectors that is a scalar? Give an example. c) Is it possible to define a function of two vectors that is a vector? Give an example. 6. Show that the magnitude of the resultant of two vectors is always less than or equal to the sum of the magnitudes of the two vectors. In what situation are the two equal? Virtual University Physics 101 – Dr. Pervez Hoodbhoy 9Summary of Lecture 3 – KINEMATICS II 1. The concept of the derivative of a function is exceedingly important. The derivative shows how fast a function changes when its argument is changed. (Remember that for fx ( ) we say that f is a function that depends upon the argument xf . You should think of as a machine that gives you the value fx when you input .) 2. Functions do not always have to be written as fx ( ). x(t) is also a function. It tells us where a body is at different times t. 3. The derivative of xt ( ) at time t is defined as: dx∆x ≡ lim ∆→ t 0 dt∆t xt() +∆t −x(t) = lim . ∆→ t 0 ∆t ∆x ∆ t 2 4. Let's see how to calculate the derivative of a simple function like xt ( )=t . We must first calculate the difference in xt at two slightly different values, and t +∆t,while remembering that we choose ∆t to be extremely small: 2 2 ∆=xt+∆t −t () 2 22 =+tt () ∆ + 2t∆t−t ∆∆ xxdx =∆tt + 2 ⇒ lim = = 2 ∆→ t 0 ∆∆ ttdt n 5. In exactly the same way you can show that if xt ( )=t then: dx∆x n−1 ≡= lim nt ∆→ t 0 dt∆t This is an extremely useful result. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 106. Let us apply the above to the function x(t) which represents the distance moved by a body with constant acceleration (see lecture 2): 1 2 ( ) v xt=+ x t+ at 00 2 dx 1 =+ 0 v + at (2 )= v +at 00 dt 2 dv This clearly shows that =+ 0 aa = (acceleration is constant) dt 7. A stone dropped from rest increases its speed in the downward direction according dv to =≈ g 9.8 m/sec. This is true provided we are fairly close to the earth, otherwise dt the value of g decreases as we go further away from the earth. Also, note that if we measured distances from the ground up, then the acceleration would be negative. 22 dv d dx dx dx ⎛⎞ 8. A useful notation: write == . We call the second derivative ⎜⎟ 22 dt dt dt dt dt ⎝⎠ of xt with respect to , or the rate of rate of change of x with respect to t. 9. It is easy to extend these ideas to a body moving in both the x and y directions. The position and velocity in 2 dimensions are: r ˆˆ rx= (t)( iy+ t)j r r dr dx dy ˆˆ v== ij + dt dt dt ˆˆ =+ vij v xy ˆˆ Here the unit vectors ij and are fixed, meaning that they do not depend upon time. rr 10. The scalar product of two vectors AB and is defined as: r r AB⋅=ABcosθ You can think of: rr AB⋅= (A)(Bcosθ ) rrr =× (length of AB ) (projection of on A) OR, rr AB⋅= (B)(Acosθ ) r rr =× (length of BA ) (projection of on ) B. ˆˆ ˆ ˆ ˆ ˆ Remember that for unit vectors ii⋅ =j⋅= j 1 and i⋅= j 0. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 11QUESTIONS AND EXERCISES – 3 1. Apply the definition of derivative (see point 3 above) to calculate: d 1 a) 2 dt t+1 d 1 b) 3 dt t+ 2 d c) Show that sintt= cos dt d d) Show that costt =−sin dt To do this, carefully study the example worked out in point 4 above. 2. Suppose the position of a particle changes with respect to time according to the relation 3 bb+ t 2 01 xt ( )=+ a at+at + +csindt 01 2 t a) What are the dimensions of aa , ,a ,b ,b ,c,d ? 01 2 0 1 d b) Find xt ( ). dt 2 d c) Find xt ( ). 2 dt 3. We shall find the derivative of a product of two functions in this problem. Suppose that df g() t +∆t h() t +∆t − g(t)h(t) ft ()=g()t h()t . Start from the definition, = lim . Now ∆→ t 0 dt∆t dg write gt ( +∆t) =g(t) + ∆t and a similar equation for f (t + ∆t). Hence show that: dt df dg dh a) =+hg dt dt dt dg (1/ ) 1dg b) =− 2 dt g dt d(/ f g) f dg 1 df c) =− + 2 dt g dt g dt Virtual University Physics 101 – Dr. Pervez Hoodbhoy 12Summary of Lecture 4 – FORCE AND NEWTON’S LAWS 1. Ancient view: objects tend to stop if they are in motion; force is required to keep something moving. This was a natural thing to believe in because we see objects stop moving after some time; frictionless motion is possible to see only in rather special circumstances. 2. Modern view: objects tend to remain in their initial state; force is required to change motion. Resistance to changes in motion is called inertia. More inertia means it is harder to make a body accelerate or decelerate. 3. Newton’s First Law: An object will remain at rest or move with constant velocity unless acted upon by a net external force. (A non-accelerating reference frame is called an inertial frame; Newton’s First Law holds only in inertial frames.) 4. More force leads to more acceleration: ⇒∝aF 5. The greater the mass of a body, the harder it is to change its state of motion. More mass means more inertia. In other words, more mass leads to less acceleration: 1 ⇒∝ a m Combine both the above observations to conclude that: F a∝ m F 6. Newton's Second Law: a= (or, if you prefer, write as Fm= a). m 7. Fm= a is one relation between three independent quantities (m,a,F). For it to be useful, we must have separate ways of measuring mass, acceleration, and force. Acceleration is measured from observing the rate of change of velocity; mass is a measure of the amount of matter in a body (e.g. two identical cars have twice the mass of a single one). Forces (due to gravity, a stretched spring, repulsion of two like charges, etc) will be discussed later. -2 8. Force has dimensions of mass ×= acceleration MLT . In the MKS system the unit of force is the Newton. It has the symbol N where: 2 1 Newton = 1 kilogram.metre/second . 9. Forces can be internal or external. For example the mutual attraction of atoms within a block of wood are called internal forces. Something pushing the wood Virtual University Physics 101 – Dr. Pervez Hoodbhoy 13 is an external force. In the application of Fm= a, remember that F stands for the total external force upon the body. 10. Forces are vectors, and so they must be added vectorially: rr r r FF = +F +F + ⋅⋅⋅⋅ 12 3 This means that the components in the x ˆ direction must be added separately, those in the y ˆ direction separately, etc. 11. Gravity acts directly on the mass of a body - this is a very important experimental observation due to Newton and does not follow from Fm= a. So a body of mass mF experiences a force =mg while a body of mass m experiences a force 111 2 Fm= g, where g is the acceleration with which any body (big or small) falls under 22 the influence of gravity. (Galileo had established this important fact when he dropped different masses from the famous leaning tower of Pisa) 12. The weight of a body WW is the force which gravity exerts upon it, =mg. Mass and weight are two completely different quantities. So, for example, if you used a spring balance to weigh a kilo of grapes on earth, the same grapes would weigh only 1/7 kilo on the moon. 13. Newton's Third Law: for every action there is an equal and opposite reaction. More precisely, FF =− , where F is the force exerted by body B upon A whereas F AB BA AB BA is the force exerted by body AB upon . Ask yourself what would happen if this was not true. In that case, a system of two bodies, even if it is completely isolated from the surroundings, would have a net force acting upon it because the net force acting upon both bodies would be FF+≠ 0. AB BA 14. If action and reaction are always equal, then why does a body accelerate at all? Students are often confused by this. The answer: in considering the acceleration of a body you must consider only the (net) force acting upon that body. So, for example, the earth pulls a stone towards it and causes it to accelerate because there is a net force acting upon the stone. On the other hand, by the Third Law, the stone also pulls the earth towards it and this causes the earth to accelerate towards the stone. However, because the mass of the earth is so large, we are only able to see the acceleration of the stone and not that of the earth. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 14QUESTIONS AND EXERCISES – 4 Q.1 The picture shown here makes the fairly obvious statement that action and reaction forces in this situation are equal. However, why do you feel pain if there is no net force acting upon your hand? Q.2 Two astronauts are in outer space connected with a rope in a state of complete weightlessness. Each pulls one end of a rope with his hands. Describe what will happen as time goes on. r ˆˆ ˆ ˆ Q.3 Add together the forces Fx =+ 3 5y−zˆˆ and F=−2x− 3y+ 2z and obtain the 12 magnitude of the resultant force. Q.4 A cube of constant density ρ is pushed with a pressure P from one side. The cube is place on a smooth level surface. Find the acceleration of the cube, and the distance covered after time t. Q.5 Describe all the forces acting upon a ladder that is leaning against a wall. Both the floor and the wall are rough. Q.6 In the above situation, suppose the floor and wall suddenly become perfectly smooth. What will the net force on the ladder become, and what will be the acceleration in the horizontal and vertical directions? Virtual University Physics 101 – Dr. Pervez Hoodbhoy 15Summary of Lecture 5 – APPLICATIONS OF NEWTON’S LAWS – I 1. An obvious conclusion from F= ma is that if F== 0 then a 0 How simple, yet how powerful This says that for any body that is not accelerating the sum of all the forces acting upon it must vanish. 2. Examples of systems in equilibrium: a stone resting on the ground; a pencil balanced on your finger; a ladder placed against the wall, an aircraft flying at a constant speed and constant height. 3. Examples of systems out of equilibrium: a stone thrown upwards that is at its highest point; a plane diving downwards; a car at rest whose driver has just stepped on the car's accelerator. 4. If you know the acceleration of a body, it is easy to find the force that causes it to accelerate. Example: An aircraft of mass m has position vector, r 32 4 ˆˆ r=+() at bt i+(ct+ dt) j What force is acting upon it? 22 r dx d y ˆˆ SOLUTION: Fm=+i m j 22 dt dt 2 ˆˆ =+ 6bmti m(2c+12dt ) j 5. The other way around is not so simple: suppose that you know F and you want to find x. For this you must solve the equation, 2 dx F = 2 dt m This may or may not be easy, depending upon Fx (which may depend upon both as well as t if the force is not constant). 6. Ropes are useful because you can pull from a distance to change the direction of a force. The tension, often denoted by T, is the force you would feel if you cut the rope and grabbed the ends. For a massless rope (which may be a very good approximation in many situations) the tension is the same at every point along the rope. Why? Because if you take any small slice of the rope it weighs nothing (or very little). So if the force on one side of the slice was any different from the force force on the other side, it would be accelerating hugely. All this was for the "ideal rope" which has no mass and never breaks. But this idealization if often good enough. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 16 7. We are all familiar with frictional force. When two bodies rub against each other, the frictional force acts upon each body separately opposite to its direction of motion (i.e it acts to slow down the motion). The harder you press two bodies against each rr r other, the greater the friction. Mathematically, FN=µ , where N is the force with which you press the two bodies against each other (normal force). The quantity µ is called the coefficient of friction (obviously). It is large for rough surfaces, and small for smooth rr ones. Remember that FN=µ is an empirical relation and holds only approximately. This is obviously true: if you put a large enough mass on a table, the table will start to bend and will eventually break. 8. Friction is caused by roughness at a microscopic level - if you look at any surface with a powerful microscope you will see unevenness and jaggedness. If these big bumps are levelled somehow, friction will still not disappear because there will still be little bumps due to atoms. More precisely, atoms from the two bodies will interact each other because of the electrostatic interaction between their charges. Even if an atom is neutral, it can still exchange electrons and there will be a force because of surrounding atoms. 9. Consider the two blocks below on a frictionless surface: T T F m m 1 2 We want to find the tension and acceleration: The total force on the first mass is FT− and so FT −=ma. The force on the second mass is simply T and so T=ma. Solving the 1 2 F mF 2 above, we get: T= and a= . mm+ mm+ 1212 10. There is a general principle by which you solve equilibrium problems. For equilibrium, the sum of forces in every direction must vanish. So FF==F= 0. You may always xy z choose the xy , , z directions according to your convenience. So, for example, as in the lecture problem dealing with a body sliding down an inclined plane, you can choose the directions to be along and perpendicular to the surface of the plane. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 17 QUESTIONS AND EXERCISES – 5 Q.1 The relation FN=µ is an approximate one only. Discuss various circumstances in which a)it would be correct to use the relation, b)where it would not be correct. Q.2 Force is a vector quantity but FN=µ is written as a relation between the magnitudes of two forces. So what is the direction of the frictional force ? Q.3 In (9), a frictionless surface was considered. Now assume that the coefficient of friction is µ. What will be the acceleration and the tension now. Assume that the body is pulled with sufficient force so that it moves. Q.4 For a body resting on an inclined plane as below: N Mg sinα Mg cosα α Mg a) Find the minimum angle at which the body starts to slide. b) If the angle exceeds the above, find the acceleration. c) Find the time taken to slide a length d down the slope if the body starts at rest. Q5. Consider the system of two masses shown. Find the acceleration of the two masses (are they the same?) for the following two situations: a) The table is frictionless. b) The coefficient of friction µ= 2. Virtual University Physics 101 – Dr. Pervez Hoodbhoy 18Summary of Lecture 6 – APPLICATIONS OF NEWTON’S LAWS – II 1. As a body moves through a body it displaces the fluid. it has to exert a force on the fluid to push it out of the way. By Newton's third law, the fluid pushes back on the body with an equal and opposite force. The direction of the fluid resistance force on a body is always opposite to the direction of the body's velocity relative to the fluid. 2. The magnitude of the fluid resistance force usually increases with the speed of the body through the fluid. Typically, fk= v (an empirical law). Imagine that you drop a ball bearing into a deep container filled with oil. After a while the ball bearing will approach its maximum (terminal) speed when the forces of gravity and friction balance each other: mg== kv from which v mg / k. final 3. The above was a simple example of equilibrium under two forces. In general, while solving problems you should a)draw a diagram, b)define an origin for a system of coordinates, c)identify all forces (tension, normal, friction, weight, etc) and their x and yx components, d)Apply Newton's law separately along the and y axes. e) find the accelerations, then velocities, then displacements. This sounds very cook-book, and in fact it will occur to you naturally how to do this while solving actual problems. 4. Your weight in a lift: suppose you are in a lift that is at rest or moving at constant velocity. In either case a=0 and the normal force N and the force due to gravity are exactly equal, NM−= g0. ⇒N=Mg But if the lift is accelerating downwards then MgN −=Ma or N=M (g−a) . So now the normal force (i.e. the force with which the floor of the lift is pushing on you) is decreased. Note that if the lift is accelerating downwards with acceleration a (which it will if the cable breaks) then N=0 and you will experience weightlessness just like astronauts in space do. Finally, if the lift is accelerating upwards then a is negative and you will feel heavier. 5. Imagine that you are in a railway wagon and want to know how much you are accelerating. You are not able to look out of the windows. A mass is hung from the roof. Find the acceleration of the car from the angle made by the mass. a T Virtual University Physics 101 – Dr. Pervez Hoodbhoy 19T We first balance the forces vertically: Tm cosθ=g θ And then horizontally: Tm sinθ=a a From these two equations we find that: tanθ= g Note that the mass m doesn't matter - it cancels out mg 6. Friction is a funny kind of force. It doe not make up its mind which way to act until some other force compels it to decide. Imagine a block lying on the floor. If you push it forward, friction will act backward. And if you push it to the left, friction will act to the right. In other words, the direction of the frictional force is always in the opposite direction to the applied force. 7. Let us solve the following problem: a rope of total length L and mass per unit length m is put on a table with a length l hanging from one edge. What should be l such that the rope just begins to slip? N () Ll− µN ml g l mL()−l g To solve this, look at the balance of forces in the diagram below: in the vertical direction, the normal force balances the weight of that part of the rope that lies on the table: Nm=− () L lg. In the horizontal direction, the rope exerts a force mlg to the right, which is counteracted by the friction that acts to the left. Therefore µNm= lg. Substituting N µL from the first equation we find that l= . Note that if µ isvery small then even a small µ+1 piece of string that hangs over the edge will cause the entire string to slip down. 9. In this problem, we would like to calculate the minimum force F such that the small block does not slip downwards. Clearly, M F m since the 2 bodies move together, Fm =+ ( M )a . This gives F a= . We want the friction µNm to be at least as large as the downwards force, g. () mM+ ⎛⎞ F So, we put Nm==a m from which the minimum horizontal force needed to ⎜⎟ () mM+ ⎝⎠ () mM+ g prevent slippage is F= . µ Virtual University Physics 101 – Dr. Pervez Hoodbhoy 20QUESTIONS AND EXERCISES – 6 Q.1 a) Why is it necessary to choose a reference frame? b) If you used spherical coordinates to solve a problem instead of rectangular cartesian coordinates, would it make a difference? c) Why do we pretend that the earth is an inertial frame when we know it is not? d) Without making measurements by looking outside, can you know whether or not your frame is inertial? e) Look up the meaning of "inertial guidance". f) Newton thought that one frame that was fixed to the distant stars was the best (or most preferred frame). Was this a sensible thought at that time? Discuss. Q.2 For a particular liquid and a certain shape of body the relation between the frictional 2 resistance and velocity is of the form Fa=+ v bv . a) What are the dimensions of ab and ? b) If a body of mass M is dropped into a deep vessel containing the liquid, what is the maximum speed that it will attain? c) What will be the initial acceleration just as it enters the liquid? Q.3 A rope is pulled with force F on the floor. (a) Find its acceleration and the tension Tl at distance . (b) Repeat if the coefficient of friction is µ. () Ll− l T A F T B Solution to part (a): Take a small piece of the rope and look at the forces acting to the left and the right. The net force on part A of the rope gives its acceleration: FT −=mla Similarly, the net force on part B of the rope gives its acceleration: Tm=− (L l)a l ⎛⎞ From these two equations it follows that Tl depends on : T=−F 1 ⎜⎟ L ⎝⎠ Virtual University Physics 101 – Dr. Pervez Hoodbhoy 21

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