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Chapter 3 NEWTON’S LAWS OF MOTION

Umm Al- Qura University جا معة أم القرى

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Chapter 3 NEWTON’S LAWS OF MOTION

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  1. Umm Al-Qura University جا معة أم القرى Faculty of Scienceكـلية العلوم Department of Physics قسم الفيزياء General Physics (For medical students): 403104-3 403104-3 : المدخل للفيزياء الطبية Chapter 3 NEWTON’S LAWS OF MOTION A. Prof. Hamid NEBDI hbnebdi@uqu.edu.sa Faculty of Science. Department of Physics. Room: 315 second floor. Phone: 3192

  2. Course’s Topics 3.1- Force, Weight & Gravitational Mass 3.2- Density 3.3- Newton’s 1st law 3.4- Equilibrium 3.5- Newton’s 3rd law 3.6- Newton’s 2nd law 3.8- Some Examples of Newton’s Laws 3.9- Gravitational Forces 3.12- Friction

  3. INTRODUCTION - Having learned how to describe motion, we can now turn to the more fundamental question of what causes motion. - Although there are many kinds of forces in nature, the effects of any force are described accurately by three general laws first stated by IsaacNewton. - Even though twentieth century advances have shown that Newton’s laws are inadequate at the atomic scale and at velocities comparable to the speed of light (3x108 m/s). - But these laws are fully adequate for must applications in fields such as astronomy, biomechanics, geology and engineering.

  4. 3.1- FORCE, WEIGHT, AND GRAVITATIONAL MASS - If we push or pull an object, we are exerting a force on it. Forces have both magnitudes and directions, so they are vector quantities. - It is found that the net or total force on an object is the vector sum of all the forces acting on the object. For example, if two forces equal in magnitude but opposite in direction act on an object, there is no net force. - Forces that are exerted only when two objects are in contact are referred to as contact forces. - Other forces, including gravitational, magnetic, and electric forces, can be exerted between objects that are not in contact. • The S.I. force unit is the newton (N), and the British unit is the pound (lb): 1 N = 0.225 lb

  5. WEIGHT, AND GRAVITATIONAL MASS - A particularly important force is the gravitational force on an object, which is referred to as its weight w. - Closely related to the weight is the gravitational mass mof on an object, which is defined as the weight divided the gravitational acceleration g at the location of the object: m = w/g (3.1) - The gravitational force pulls an object downward along the direction it will fall, if it is not supported. - Thus, the weight w is parallel to the gravitational acceleration g, and Eq. 3.1 can be written in vector form as : w = m g (3.2) - The S.I. unit of the gravitational mass is the kilogram (kg). With Eq. 3.2, the force and mass units are related by: 1 N = (1 kg) (1m s-2) = 1kg m s-2 - For Example, A man who weights 1000 N on the earth has a gravitational mass of: m = w /g = 1000 N / 9.8 m s-2 = 102 kg

  6. 3.2- DENSITY - When we talk about the properties of materials in a general way, rather than about specific objects it is often convenient to refer to, the mass per unit volume or the density. - If a sample of a material has a mass m and a volume V, the density is given by the ratio: (The symbol is the Greek letter “rho”) - In S.I. units, densities are measured in kilograms per cubic metre (kg m-3). - For example, if a block of wood has a mass of 50 kg and a volume of 0.1 m3, its density is: - Note that the density varies with the temperature and the pressure, especially for gases. - Closely related is the concept of the relative density or specific gravity, which is defined as the ratio of the density of a substance to that of water of 0C (Celsius).

  7. Example 3.1: A cylindrical aluminum rod has a radius of 1.2 cm and a length of 2 m. What is its mass? Solution 3.1: Table 3.2 lists the density of aluminum (2700 kg m-3). Thus, with the definition of the density,  = m/V , we can find the mass of the rod once we have its volume. For a cynlinder : Converting the radius to metres, the mass is:

  8. 3.3- NEWTON’S FIRST LAW - Newton’s first law states that: Every object continues in a state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces acting upon it. - An equivalent statement of the first law is that if there is no force on an object, or if there is no net force when two or more forces act on the object, then: (1) an object at rest remains at rest, and (2) an object in motion continues to move with constant velocity. - Newton’s first law as stated does not hold true for someone who is accelerating.

  9. 3.4- EQUILIBRIUM • Newton’s first law tells us that the state of an object remains unchanged whenever the net force on the object is zero. • This can happen if no forces or if two forces act on an object. More commonly, it occurs because two or more forces acting on an object add to zero or “balance.” • When the state of motion of an object remains unchanged even though two or more forces act upon it, the object is said to be in equilibrium. • For example, assuming you are sitting at rest in your chair, the net force on you is zero. However, your weight - the gravitational force exerted by the earth- is pulling you downward. This force is offset by upward forces exerted on you by the chair and the floor. This is an illustration of the equilibrium. • The first law applies to objects in uniform motion as well as objects at rest. • The simplest kind of equilibrium situation is one where two forces act on an object. When you stand motionless, you experience the downward gravitational pull of the earth, your weight w. The weight is balanced by an upward force extended on you by the floor. This force is perpendicular or normal to the floor, so it is called a normal force.

  10. Example 3.2: A woman has a mass of 60 kg. She is standing on a floor and remains at rest. Find the normal force exerted on her by the floor. Solution 3.2: The net force w + Non the woman must be zero . Thus the upward normal force must have the same magnitude as her weight, which acts downward. Symbolically: w + N = 0 and w = -N Her weight is: Hence the normal force has a magnitude of N= 588 N

  11. TYPES OF EQUILIBRIUM - There are three types of equilibrium: unstable, stable, and neutral 1- Unstable equilibrium: a small displacement leads to an unbalanced force that further increase the displacement from the equilibrium location (ball in position A). 2- Stable equilibrium: a small displacement leads to an unbalanced force that tends to restore the object to the equilibrium location (ball in position B). 3- Neutral equilibrium: it is in equilibrium at any location near C.

  12. 3.5- NEWTON’S THIRD LAW • The third law relates the forces that two objects exert on each other, and it’s familiar in a general way from our everyday experiences. • For example: suppose you are at rest in a swimming pool. If you push a wall with your legs, the wall exert a force that propels you further into the pool. The reaction force the wall exerts on you is opposite in direction to the (action) force you exert on the pool. - Newton observed that whenever a person exerts a force on an object, the object exerts a force on the person that is equal in magnitude and opposite in direction. These two forces are called action and reaction • The general statement of the third law is: If one object exerts a force F on a second, then the second object exerts an equal but opposite force -F on the first.

  13. 3.5- NEWTON’S SECOND LAW • When, there is a net force acting on an object , the object undergoes an acceleration in the same direction as the force. • Thus, we can relate the net force F and the acceleration a by Newton’s second law: F = m a The proportionality constant m is called the inertial mass of the object. • The mass of an object is a measure of the amount of matter it comprises or, in other words, its inertia.

  14. Example 3.6: A child pushes a sled across a frozen pond with a horizontal force of 20 N. Assume friction is negligible. a) If the sled accelerates at 0.5 ms-2, what is its mass? b) Another child with a mass of 60 kg sits on the sled. What acceleration, the same force produce now? Solution 3.6: a) The sled does not move vertically, so its weight and the normal force exerted by the ice cancel. Thus the net force on the sled is the 20 N exerted by the child. From the second law: F = m a The mass m is: b) The total mass of the passager and the sled is 60 kg+40 kg=100 kg Thus the acceleration is now: The same force produces a small acceleration when applied to a more massive object.

  15. 3.8- SOME EXAMPLES OF NEWTON’S LAWS We employ a systematic procedure for relating the acceleration of an object or objects to the forces present as follows: 1- For each object, we draw a careful sketch showing the forces acting on that object. 2- We then apply Newton’s second law F = m ato each object separately . If there are n forces F1, F2, … Fn, acting on an object, the net force Fis the sum of the forces and we have: F = F1 + F2 + . . . + Fn = m a In component form, this is Fx = F1x + F2x + …+ Fnx = m ax (3.5) Fy = F1y + F2y + …+ Fny = m ay (3.6) 3- We usually don’t substitute numerical values immediately. Instead we solve the equations for unknown quantities symbolically and then substitute the numbers if they are given. This procedure facilitates checking for algebra and physics errors and often reduces the amount of arithmetic required. 4- We include the units of the numerical quantities and see that the final answer has the correct dimensions.

  16. Example 3.9: A child pulls a train of two cars with a horizontal force F of 10 N. Car 1 has a mass m1 = 3 kg and car 2 has a mass m2 = 1 kg. The mass of the string connecting the cars is small enough so it can be set equal to zero, and friction can be neglected. a) Find the normal forces exerted on each car by the floor b) What is the tension in the string? c) What is the acceleration of the train?

  17. Solution 3.9: In this problem, there is a system of several objects. The three objects of interest are the two cars and the string connecting them. When Newton’s second law F = ma is applied to each object, a set of equations results that must be solved simultaneously. The fact that the string is massless means that the net force F = ma on the string must be 0. a) Since ay = 0 for each car, so Fy = m ay = 0, so N2 – w2 = 0 and N1 – w1 = 0 Thus the normal forces exerted by the floor are:

  18. Solution 3.9: b) For both cars, ax = a , so for car 1, Fx = m1 ax becomes F – T = m1 a (i) Both T and a are unknowns, so we cannot find them from this one equation alone; as many equations as there are unknowns are needed to solve a problem. The second equation relating T and a is obtained by applying Fx = m2axto car 2: T = m2 a (ii) This gives a = T/m2. Substituting this for a in Eq. i, F – T = m1 (T/m2) Solving for T, T = F / (1 + (m1/m2)) = 10 N / ( 1 + (3 kg /1 kg)) = 2.5 N c) Now we can find a from Eq. ii: a = T/m2= 2.5 N / 1 kg = 2.5 m.s-2 Note that if we had considered the train as a single object of mass m = m1 + m2 = 4 kg, we could immediately have found the acceleration from a = F / m = 10 N / 4 kg = 2.5 m.s-2. However, the normal and tension forces can only be found by considering the individual cars and not the system as whole.

  19. 3.9- GRAVITATIONAL FORCES • Newton’s study of planetary motion led him to infer the formula for the gravitational force between two masses. • This formula is referred to as the law of universal gravitation and is regarded as a fundamental law of nature. • The law of universal gravitation states that all objects in the universe attract each other. • For two objects of ay shape that are so small compared to their separation that may be considered as points particles, the law has a simple form: (3.7) where G is the gravitational constant and has a measured value of: G = 6.67 x 10-11 N m2 kg-2 and m, m’ are the masses of the particles, r is the distance separated their centers. • The gravitational forces are directed along the line connecting the centers of the two spheres. • The magnitude of the gravitational force varies as 1/r2, so Eq. 3.7 is referred to as an inverse square law.

  20. 3.12- FRICTION (a) - Friction is a force that always act to resist the motion of one object on another. - Frictional forces are very important, since they make it possible for a us to walk, use wheeled vehicles and hold books. - Frictional forces in fluids are called viscous forces. They are often quite small compared to the friction between solid surfaces. Thus the use of lubricating liquid such as oil, which clings to the surface of metals, greatly reduces friction. - When we walk or run, we are not conscious of any friction in our knees or other leg joints. These joint, really, are lubricated by synovial fluid, which is squeezed through the cartilage lining the joints when they move. This lubricant tends to be absorbed when the joint is stationary, increasing the friction and making it easier to maintain a fixed position. This is an excellent example of theclever biological engineering that GOD has employed.

  21. 3.12- FRICTION (b) - We consider a block at rest on a horizontal surface (Fig. 3.19 a, b). Since the block is at rest, the first law requires that the net force on the block be zero. The vertical forces are the weight w and the normal force N, so we must have : N = w. In the horizontal direction, there is no applied force and no motion, so the frictional force must also be zero, according to the first law. - If we apply a small horizontal force T to the rights and if the block remains at rest, the friction force fscan no longer be zero, since the first law requires that the net force on the block be zero, or fs = T. - If T is gradually increased fs increases also. Eventually when T become large enough, the block begins to slide. Thus there is a maximum possible static friction force fs(max).

  22. 3.12- FRICTION (c) - Experimentally it is found that fs(max) has the following properties: 1-fs(max)is independent of the contact area. 2-For a given pair of surfaces fs(max) is proportional to the normal force N. 3- The number relating fs(max) and N called the coefficient of static friction, s, is defined by: fs(max) = sN (3.12) The coefficient s depends on: - the nature of the two surfaces, - their cleanliness and smoothness, - the amount of moisture present, … For metal on metal: sis between 0.3 and 1 When lubricating oils are used s is about 0.1 For Teflon on metals, s0.04 In a healthy hip joint, the synovial fluid reduces s to the remarkably low value 0.003. 4-The force necessary to keep an object sliding at constant velocity is smaller than that required to start it moving. Thus the sliding or kinetic friction fkis less than fs(max). It is independent of the contact area, and is satisfies: fk= kN (3.13) Here kis the coefficient of kinetic friction and is determined by nature of the two surfaces. 5- k is nearly independent of the velocity, and, since fk fs(max) : k  s (3.14)

  23. Example 3.17: A 50-N block is on a flat, horizontal surface. a) If a horizontal force T=20 N is applied and the block remains at rest; what is the frictional force? b) The block starts to slide when T is increased to 40 N. What is s ? c) The block continues to move at constant velocity if T is reduced to 32 N. What is K? Solution 3.17: a) Since the block remains at rest when the force T is applied, the frictional force fs must fs = T = 20 N :be equal but opposite to T. consequently b) Since the block just begins to slide when the applied force is increased to 40 N, the maximum frictional force must be: fs(max) = 40 N The vertical forces must add to zero, so the normal force N is equal to the weight, w, or 40 N /30 N = 0.8 s =fs(max) / N = :50 N. Hence c ) Since the block moves with constant velocity when a 32 N force is applied, the net force must be zero. Consequently the frictional force fk must equal the applied force, or: fk = 32 N Again the normal force must equal the weight, or 50 N so: As expected, the coefficient of kinetic friction k is less than the coefficient of static friction s.

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