04-1. Newton s First Law Newton s first law states: Sections Covered in the Text: Chapters 4 and 8 F = ( F 1 ) 2 + ( F 2 ) 2.

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1 Force and Motion Sections Covered in the Text: Chapters 4 and 8 Thus far we have studied some attributes of motion. But the cause of the motion, namely force, we have essentially ignored. It is true that we have successfully described an object in freefall and the motion of a projectile, both cases of motion being influenced or driven by the force of gravity. But in both cases the acceleration of the object remained constant. (We described what happened to the object while it was falling, not on what happened to the object once it hit the ground.) This made it unnecessary for us to modify our kinematic definitions or to construct new concepts. However, once we include the concept of force into our study of motion in a general way, by allowing for changes to take place in the magnitude or direction of the force, then we need to anticipate possible changes in the acceleration of the object. The object may speed up, slow down or change direction. The study of motion then expands into dynamics. We shall begin our study of dynamics with the concept of force and the first of the true laws of physics, Newton s Laws of Motion. Later in this and later notes we shall see how these laws can be applied to describe real world problems. The Concept of Force The concept of force is learned early in childhood. We learn that in order to make something move we must push it or pull it; we have to exert a force on it. We also learn that if we jump from a point above the ground then we will drop to the ground (and possibly injure ourselves). In this case something else exerts a force on us. This force we call the force of gravity. In physics, a push or a pull is called a contact force, whereas the force of gravity is called a non-contact force. We shall discuss non-contact forces and the associated concept of a field in a later note. For the moment we focus on contact forces. 1 Force is a Vector If we consider for simplicity a force as a simple pull, then we can quantify the pull in a relative way by observing its effect on something that stretches, like an elastic or spring scale (Figure 4-1). If a spring is 1 A contact force is actually a manifestation of the electromagnetic force and is not a true, distinct force. It just happens to be easier for teaching purposes to continue using the term contact force for the present. pulled it will stretch by an amount that can be measured with a ruler. If we pull on a spring by a certain amount of muscular action then the spring stretches by a certain amount. If two people each pull by the same amount of muscular action then the spring stretches twice as much as before. Figure 4-1. Observations with a spring scale indicate that force is a vector. More importantly, if forces of magnitude F 1 and F 2 are applied to the end of the spring scale in arbitrary perpendicular directions as shown in the figure, then the extension of the spring is found to be proportional to F = ( F 1 ) 2 + ( F 2 ) 2. The conclusion to be drawn is that force is a vector. We now consider Newton s Three laws of motion in modern language and discuss the concepts and definitions associated with each. Newton s First Law Newton s first law states: In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line). 04-1

2 You will be able to witness the effects of Newton s first law for yourself in the experiment Linear Motion. If you place a glider on a horizontal airtrack and release it from rest, it will remain at rest. If you release the glider with some initial velocity, then it will continue to move at a constant velocity (until it encounters the bumper at the end of the track). These observations are consistent with Newton s first law. Law of Inertia Another way of stating the first law is to say if the net force acting on an object is zero, then the acceleration of the object is zero. An object has an attribute that tends to resist a change in its state of motion. This attribute is called inertia. The first law is therefore often called the law of inertia. Inertial Reference Frame In physics, we often make use of the concept of an inertial reference frame. An inertial reference frame is a reference frame in which Newton s first law holds. In other words, an inertial reference frame is one that is not undergoing acceleration itself. A glider, whether stationary or moving with constant velocity on a horizontal airtrack, is an example of an inertial reference frame. Another example of a more-or-less inertial reference frame is a position on the surface of the Earth (say at a worktable in the lab). It is true that the Earth is spinning on its axis and is rotating about the Sun, so a position on the surface of the Earth is therefore undergoing two types of acceleration. But both types are very small in relation to the acceleration of most systems under study; in problems in a first year physics course they can mostly be neglected. Inertial Mass We have stated that an object possesses an attribute that resists a change in its state of motion. By state of motion is meant the following: the state of motion of a ball, for example, can be changed by either throwing it or catching it. A ball made of lead is harder to catch than is a ball made of rubber. This attribute is called inertia as we have seen. A lead ball has more inertia than a rubber ball of the same size. Inertia is more familiarly called mass, and in particular, inertial mass. It may be described in these words: Inertial mass is the measure of an object s resistance to a change in motion in response to an external force. If the same force acts on masses m 1 and m 2 and produces the accelerations a 1 and a 2, respectively, then 04-2 the ratio of the two masses is defined as the inverse ratio of the magnitudes of the accelerations produced by the same force: m 1 = a 2. m 2 a 1 Now if one of these masses happens to be the 1 kg standard mass (described in Note 01) then the mass of the unknown object can be calculated from measurements of the accelerations. 2 Mass is an inherent property of an object, independent of the object s surroundings and of the method used to measure it. Gravitational mass is defined somewhat differently from inertial mass as we shall see. As far as is known an object s inertial and gravitational masses are equal, and are therefore by inference taken to be one and the same property. Newton s Second Law Newton s second law states: The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object s mass. Note that the word object in this statement (and the ones to follow) could be replaced by the word system. Mathematically, the second law means a F m. Writing this expression as an equality and rearranging we have F = ma. [4-1] Thus the vector sum of forces (net force) on an object equals the product of the object s mass and acceleration. The acceleration of an object is in the same direction as the net force on the object. Since eq[4-1] is a vector equation these component equations apply in 3D space: = ma x F y = ma y F z = ma z. 2 Interestingly, though mass is a fundamental quantity, the finding of inertial mass involves calculation not measurement with an instrument. As we shall see in due course, the finding of gravitational mass on the other hand involves measurement with an instrument like a calibrated spring scale or a balance.

3 Unit of Force From eq[4-1] the dimension of force equals the product of the dimensions of mass and acceleration, namely M.L.T 2. Its units are kg.m.s 2. 1 kg.m.s 2 is called a newton. 1 newton (abbreviated N) may be defined in words as the force that produces an acceleration of 1 m.s 2 on a mass of 1 kg y x Once the resultant force on an object of known mass is known then the object s acceleration can be calculated. Let us consider an example. a = 3.66 m/s/s Example Problem 4-1 Finding the Acceleration of an Object Three forces of magnitude 10.0 N act on a ball of mass 2.00 kg in the directions shown in Figure 4-2. Calculate the resultant acceleration of the ball. F2 = 10.0 N F3 = 10.0 N 30.0 y 2.00 kg F1 = 10.0 N 30.0 Figure 4-2. Three forces of magnitude F1, F2 and F3 act on a ball. Solution: This problem is an extension of Example Problem 2-3. It was shown in that problem that the resultant force vector is of magnitude 7.32 N acting in a direction 30.0 wrt to the ve x-axis (Figure 2-13). Since we have a = F m a = 7.32(N) 2.00(kg) = 3.66 m.s 2. The acceleration vector points in the same direction as the resultant force vector (Figure 4-3). x Figure 4-3. The acceleration of the ball in Figure 4-2 during the elapsed time the three forces act. WARNING! Force and Acceleration An object undergoes an acceleration only so long as the resultant or net force on the object is also non-zero. In the event that the force is zero the acceleration is also zero. Gravitational Force and Weight We have seen that mass is a fundamental property of an object (being the sum total of the masses of its constituent atoms and molecules). Under ordinary conditions (non-relativistic speeds) the mass of an object remains constant. An object has the same mass whether measured on the Earth or on the Moon. The weight of an object is different from mass. It is defined in these words: In physics, the weight of an object is defined as the magnitude of the resultant force of gravity on the object. We have seen that when an object is released near the surface of the Earth it always falls downwards with the same acceleration, namely g. In this situation there is only one force acting on the object, the force of gravity. Thus F = Fg. It follows from the second law that Fg = ma = mg, and so the weight of any object of mass m is 04-3

4 Fg = F g = mg (N). [4-2] Weight is a positive scalar with the unit newton. Newton s Third Law Newton s Third law states: If two objects interact, the force F 12 exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force F 21 exerted by object 2 on object 1, that is, F12 = F21. The idea is illustrated in Figure 4-4. [4-3] Figure 4-4. To every force there is an equal and opposite reaction force. Shown here are the gravitational forces two objects exert on one another. The book has a mass m and is subject to the gravitational force of the Earth. But the book is in contact with the table. Since the Earth pulls downwards on the book and the book is in contact with the table, then the book pushes downwards on the table with the same magnitude of force. Since the book pushes downwards on the table, the table reacts (third law) by pushing upwards on the book. This reaction force (denoted n ) is called a normal force when it is perpendicular (or normal) to a surface. The upshot is that the sum of forces on the book is F = FTB + FEB = n + mg = 0. (Here the notation FTB means the force the table exerts on the book.) Figure 4-5 is the free-body diagram of the book. A free-body diagram is a diagram showing just the forces acting on the object of interest. The Idea of Tension Tension is a word used to describe the contact force exerted by a rope or a string on whatever it is attched to. A problem involving tension is illustrated in Figure 4-6. A rope is attached to point W on an immovable wall. A boy at position B pulls on the rope with force FBR in a direction towards the right. Neither wall nor boy actually move. In the process of being pulled, the rope becomes taut and is subject to what is called tension. To explore this idea further with more familiar contact forces consider an object like a book lying at rest on a table (Figure 4-5). Since the book is stationary, and in an inertial reference frame, it is described by the first law. FWR FRW Boy FTB = n W FRB B FBR 04-4 FEB = mg book m table Figure 4-5. Free-body diagram of a book lying at rest on a table in the laboratory. Wall Figure 4-6. The forces involved when a boy pulls on a rope attached to an immovable wall. Two free-body diagrams are shown here for points W and B. You should be able to reason that the forces shown, namely FBR, FRB, FRW and FWR are all of equal magnitude. F RW and F RB both refer to the same tension in the rope. Tension may be defined in these words:

5 The tension in a rope is the magnitude of the force the rope exerts on whatever it is attached to. Note 04 It is important to note that Figure 4-6 shows two freebody diagrams for the points W and B. The two forces in the left half of the figure are the forces acting only on W. The two forces on the right half of the figure act only on the point B. An Object in Translational Equilibrium An object at rest or moving with constant velocity is said to be in a state of translational equilibrium. Translational equilibrium is the state of an object that satisfies Newton s first law. 3 The condition of translational equilibrium may be stated mathematically as F = 0. Figure 4-7a. A traffic light at rest. This means that in 3D space we have the component equations = 0 F y = 0 F z = 0. [4-4] The book shown in Figure 4-5 is an example of an object in translational equilibrium. More precisely, since the object is in a state of translational equilibrium and is at rest, it is said to be in a state of static equilibrium. Applications of Newton s Laws In an introductory physics course there are a number of classical problem types. These include the problem of tension in a cable pulling or supporting an object, the working of the Atwood machine and the analysis of objects pushing on one another. We consider these three types in examples. Example Problem 4-2 A Traffic Light at Rest A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support as in Figure 4-7a. The upper cables make angles of 37.0 and 53.0 with the horizontal. These upper cables are not as strong as the vertical cable, and will break if the tension in them exceeds 100. N. Does the traffic light remain in this situation, or will one of the cables break? Figure 4-7b. The free-body diagram of the traffic light. Solution: Two points on this diagram are of special interest: the light itself and the knot at which the three cables come together. Both points are in states of static equilibrium (at least they are prior to the instant when the ropes break, if they do break). The free-body diagram of the traffic light itself is drawn in Figure 4-7b. Figure 4-7b shows only two forces: the force of gravity acting downwards and the tension T 3 acting upwards. Applying the equilibrium condition to the traffic light in the y-direction F y = 0 gives T 3 F g = 0. This means that 3 In later notes we shall extend this condition to include rotational motion. T 3 = F g = 122 N. 04-5

6 The force T 3 exerted by the vertical cable balances the force of gravity on the light. As for the knot, its free-body diagram is drawn in Figure 4-7c, with a set of arbitrarily-chosen x-and y- axes. T 2 = 1.33T 1 = 97.4 N. Both of these values are less than 100. N so the cables will not break. Example Problem 4-3 The Atwood Machine Figure 4-7c. The tensions in the ropes act at a common point. The Atwood machine, an example of which is shown in Figure 4-8, is an apparatus commonly used in introductory physics labs for studying Newton s laws. The version shown here has masses m 1 and m 2 connected together with a rope that passes over a pulley whose friction is assumed to be negligible. 4 We shall assume here that m 1 is less than m 2 so the former should be expected to accelerate upward, the latter downward. Calculate an expression for the magnitude of the acceleration a of the masses and the tension T in the rope. The forces in the ropes can be resolved into their x- and y-components as listed in the following table: Force x-component y-component T1 T 1 cos37.0 T 1 sin37.0 T 2 T 2 cos53.0 T 2 sin53.0 T N Eqs[4-4} applied to the knot give = T 2 cos 53.0 o T 1 cos37.0 o = 0, F y = T 1 sin37.0 o + T 2 sin 53.0 o 122N = 0. Solving the first equation for T 2 gives T 2 in terms of T 1 : cos37.0 o T 2 = T 1 cos53.0 o =1.33T 1. Substituting T 2 into the second of the above equations gives T 1 sin 37.0 o + (1.33T 1 )sin 53.0 o 122N = 0 Figure 4-8. An Atwood machine. In this note we assume up is positive, down is negative. Solution: We take up as positive, down as negative. Applying Newton s second law to m 1 we have F y = T m 1 g = m 1 a. [4-5] Applying the second law to m 2 we have so that T 1 = 73.4 N. Finally we substitute back to get T 2 : We assume we can also neglect the moment of inertia of the pulley. In this course, because of a lack of time, we shall have to omit the concept of moment of inertia.

7 F y = T m 2 g = m 2 a. [4-6] Subtracting eq[4-6] from eq[4-5] gives m 1 g g = m 1 a a from which we can solve for the magnitude a: a = m m 2 1 g. m 1 The acceleration of the two-block system is a = F m 1. Note 04 (b) We now treat each block individually. The freebody diagrams of m 1 and m 2 are drawn in Figures 4-9b and c where the contact force is denoted P. This expression is general and includes two special cases. If m 2 > m 1 then a is positive and m 1 accelerates upwards and m 2 downwards. If m 1 > m 2 then a is negative and m 1 accelerates downwards, m 2 upwards. Substituting the above expression for a into eq[4-5] yields T: T = 2m m 1 2 g. m 1 Figure 4-9b. The free-body diagram of m 1. This expression is general and applies for any value of m 1 and m 2. For example, if m 1 = m 2 then T = mg and a = 0, that is, neither object moves. Example Problem 4-4 One Block Pushes Another Two blocks of mass m 1 and m 2, with m 1 > m 2, are placed in contact with each other on a surface assumed to be frictionless (Figure 4-9a). A constant horizontal force F is applied to m 1 as shown. Find (a) the magnitude of the acceleration of the system of two blocks and (b) the magnitude of the contact force between the two blocks. Figure 4-9c. The free-body diagram of m 2. From Figure 4-9c the only horizontal force acting on m 2 is the contact force P12 (the force exerted by m 1 on m 2 ), which is directed to the right. Applying the second law to m 2 gives = P 12 = m 2 a. Substituting the expression for a from above gives the magnitude of the contact force requested: Figure 4-9a. A force is applied to two blocks in contact. Solution: (a) The blocks are in contact and therefore experience the same acceleration produced by F. Thus (system) = F = (m 1 )a. m P 12 = m 2 a = 2 F. m 1 Note that the force on the system as a whole (and on m 1 ) is F but the force on m 2 is less than F. 04-7

8 To Be Mastered Definitions: force, Newton s Laws of Motion, inertial reference frame Definitions: mass, inertial mass, gravitational mass, weight, tension Definition: weight Definition: tension Definition: free-body diagram 1. State Newton s Laws. Typical Quiz/Test/Exam Questions 2. A puck of mass g rests on a frictionless ice surface. There are two ropes attached to the puck by a pin at the puck s center. Two people pull as hard as they can in a horizontal direction, one person on each rope. If they pull in the same direction as shown, the puck has an acceleration of 1.52 m.s 2 to the right. If the direction of rope 1 is reversed and the people pull in opposite directions, then the puck has an acceleration of m.s 2 to the left. rope 1 puck rope 2 Answer the following questions: (a) What magnitude of force does each rope exert on the puck? (b) What are the corresponding tensions in the ropes? (c) What would be the acceleration of the puck if each person pulls with forces of equal magnitudes in opposite directions? (d) What law(s) of motion do these questions involve? 3. An object of mass 1.0 kg is hanging motionless by a cable from a fixed support (see the figure). Answer the following questions: support cable object (a) What is the weight of the object? (b) What force does the cable exert on the support? (c) What force does the cable exert on the object? (d) What is the acceleration of the object? 04-8

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