Newton s Laws of Motion

Transcription

1 Newton s Laws of Motion FIZ101E Kazım Yavuz Ekşi

2 My contact details: Name: Kazım Yavuz Ekşi Notice: Only s from your ITU account are responded. Office hour: Wednesday Website: Website of the course:

3 Chapter 4 4 NEWTON S LAWS OF MOTION 4.1 Force and Interactions 4.2 Newton s First Law 4.3 Newton s Second Law 4.4 Mass and Weight 4.5 Newton s Third Law 4.6 Free-Body Diagrams

4 Force A force is an interaction between two bodies or between a body and its environment. We always refer to the force that one body exerts on a second body. Force is a vector quantity

5 Contact Forces involve direct contact between two bodies Normal Force: always acts perpendicular to the surface of contact, no matter what the angle of that surface. Friction Force: acts parallel to the surface, in the direction that opposes sliding. Tension Force: force exerted by a stretched rope or cord on an object to which it s attached. The push and pull forces in the previous slide are also contact forces.

6 Long Range Forces act even when the bodies are separated by empty space. The force between two magnets. The force of gravity. The gravitational force that the Earth exerts on a body is called weight.

7 Unit of Force To describe a force vector F, we need to describe the direction in which it acts as well as its magnitude. The magnitude of the force describes how much or how hard the force pushes or pulls. The SI unit of the magnitude of force is the newton, abbreviated N.

8 Typical Force Magnitudes

9 Measuring Force A common instrument for measuring force magnitudes is the spring balance. It is a coil spring enclosed in a case with a pointer attached to one end. When forces are applied to the ends of the spring, it stretches by an amount that depends on the force. We can make a scale for the pointer by using a number of identical bodies with weights of exactly 1 N each.

10 Superposition of Forces Experiment shows that when two forces F 1 and F 2 act at the same time at the same point on a body, the effect on the body s motion is the same as if a single force were acting equal to the vector sum of the original forces: R = F 1 + F 2 Any number of forces applied at a point on a body have the same effect as a single force equal to the vector sum of the forces. R = F 1 + F 2 + F 2 + = F

11 Decomposing a Vector Any force can be replaced by its component vectors, acting at the same point. The idea relies on the possibility of superposition: F = F x + F y

12 Decomposing a Vector into Components It s frequently more convenient to describe a force F in terms of it s x and y components F x and F y. The x component of the resultant vector is the sum of the x components of all forces acting. R x = F x R y = F y

13 A body with no net force on it What happens when the net force on a body is zero? The answer is easy if the body is at rest: the body will remain at rest. But what if the body is in motion?

14 Horizontal Motion If we could eliminate friction completely, the puck would never slow down. Once a body has been set in motion, no net force is needed to keep it moving.

15 Newton s First Law of Motion A body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration.

16 Inertia The tendency of a body to keep moving once it is set in motion results from a property called inertia. In Newton s first law, zero net force is equivalent to no force at all. This is just the principle of superposition of forces. When a body is either at rest or moving with constant velocity (in a straight line with constant speed), we say that the body is in equilibrium. F = 0 body in equilibrium In order this to be satisfied Fx = 0, Fy = 0, Fz = 0 body in equilibrium We are assuming that the body can be represented adequately as a point particle. When the body has finite size, we also have to consider where on the body the forces are applied.

17 Exercise: In the movie Star Wars the captain says: We are to hit a asteroid. The ship must be stopped. Shut down the engines! Does stopping the engines really stop the ship? You are driving a Maserati Gran Turismo S on a straight testing track at a constant speed of 250 km/h. You pass a 1971 Volkswagen Beetle doing a constant 75 km/h. On which car is the net force greater?

18 On an Accelerating Frame-I Suppose you are in a bus that is traveling on a straight road and speeding up. If you could stand in the aisle on roller skates, you would start moving backward relative to the bus as the bus gains speed.

19 On an Accelerating Frame-II If instead the bus was slowing to a stop, you would start moving forward down the aisle.

20 On an Accelerating Frame-III In either case, it looks as though Newton s first law is not obeyed; there is no net force acting on you, yet your velocity changes. What s wrong?

21 Inertial Frames of Reference The point is that the bus is accelerating with respect to the earth and is not a suitable frame of reference for Newton s first law. The First Law of Motion is valid in some frames of reference and not valid in others. A frame of reference in which Newton s first law is valid is called an inertial frame of reference. The Earth is at least approximately an inertial frame of reference, but the bus is not.

22 The Earth is not a completely inertial frame owing to the acceleration associated with its rotation. Ex: The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the Earth s equator? Give your answer in m/s 2 and as a fraction of g? (b) If a rad at the equator is greater than g, objects will fly off the Earth s surface and into space. What would the period of the Earth s rotation have to be for this to occur?

23 The earth is not a completely inertial frame also because of its motion around the sun. Ex: The radius of the Earth s orbit around the Sun (assumed to be circular) is km and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s 2?

24 Law of Inertia The Earth is not a completely inertial frame, owing to the acceleration associated with its rotation and its motion around the Sun. But we saw that these effects are quite small. Because Newton s first law is used to define what we mean by an inertial frame of reference, it is sometimes called the law of inertia. If we have an inertial frame of reference A, in which Newton s first law is obeyed, then any second frame of reference B will also be inertial if it moves relative to A with constant velocity. Consider now a third object moving with constant velocity with respect to A. Observers in frames A and B will disagree about the velocity of P, but they will agree that P has a constant velocity (zero acceleration) and has zero net force acting on it.

25 Law of Inertia There no single inertial frame of reference is preferred over all others for formulating Newton s laws. All inertial frames are equivalent. The state of rest and the state of motion with constant velocity are not very different; both occur when the vector sum of forces acting on the body is zero.

26 Test Your Understanding In which of the following situations is there zero net force on the body? (i) an airplane flying due north at a steady 120 m/s and at a constant altitude; (ii) a car driving straight up a hill with a slope 3 at a constant 90 km/h (iii) a hawk circling at a constant 20 km/h at a constant height of 15 m above an open field; (iv) a box with slick, frictionless surfaces in the back of a truck as the truck accelerates forward on a level road at 5 m/s 2.

27 Newton s Second Law We have seen that when the net force is zero the object keeps its velocity (zero acceleration). Second law is about what happens when the net force is not zero? Experiment shows that acceleration is proportional to the net force. We conclude that a net force acting on a body causes the body to accelerate in the same direction as the net force.

28 Circular Motion That the acceleration is proportional to the net force also apply to a body moving along a curved path. Hockey puck moving in a horizontal circle on an ice surface of negligible friction. A rope is attached to the puck and to a stick in the ice The rope exerts an inward tension force of constant magnitude on the puck. The net force and acceleration are both constant in magnitude and directed toward the center of the circle.

29 Mass and Force Experiments show that for any given body, the magnitude of the acceleration is directly proportional to the magnitude of the net force acting on the body. F a The proportionality constant is called the inertial mass, or simply the mass, of the body and denote it by m. Thus F = m a

30 Mass and Inertia Mass is a quantitative measure of inertia. The greater its mass, the more a body resists being accelerated. It takes a smaller engine to accelerate a car than a truck. A car is easier to slow down than a truck.

31 Unit of Mass The SI unit of mass is the kilogram. The kilogram is officially defined to be the mass of a cylinder of platinum-iridium alloy kept in a vault near Paris. We can use this standard kilogram, along with F = m a, to define the newton. One newton is the amount of net force that gives an acceleration of 1 meter per second squared to a body with a mass of 1 kilogram. 1 newton = (1 kilogram)(1 meter per second squared) 1 newton = 1 kg m/s 2

32 Measuring Mass We can also use F = m a to measure mass. Suppose we apply a constant net force F to a body having a known mass m 1 and we find an acceleration of magnitude a 1. We then apply the same force to another body having an unknown mass m 2 and we find an acceleration of magnitude a 2. According to F = m a we have m 1 a 1 = m 2 a 2 m 2 m 1 = a 1 a 2 For the same net force, the ratio of the masses of two bodies is the inverse of the ratio of their accelerations. In principle we could use this idea to measure an unknown mass but it is usually easier to determine mass indirectly by measuring the body s weight.

33 Stating Newton s Second Law The principle of super-position of forces also holds true when the net force is not zero and the body is accelerating. Newton s second law of motion: If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The mass of the body times the acceleration of the body equals the net force vector. F = m a Newton s second law is a fundamental law of nature, the basic relationship between force and motion.

34 Remarks on Newton s Second Law F = m a is a vector relation: Fx = ma x, Fy = ma y, Fz = ma z The statement of Newton s second law refers to external forces. F = m a is valid only when m is constant. Newton s second law is valid only in inertial frames of reference, just like the first law.

35 Using Newton s Second Law Ex: A worker applies a constant horizontal force with magnitude 20 N to a box with mass 40 kg resting on a level floor with negligible friction. What is the acceleration of the box?

36 Using Newton s Second Law First choose a coordinate system and to identify all of the forces acting on the body in question. It s usually convenient to take one axis either along or opposite the direction of the body s acceleration, which in this case is horizontal. The forces acting on the box are (i) the horizontal force F exerted by the worker, of magnitude 20 N; (ii) the weight of the box; and (iii) the upward supporting force n exerted by the floor.

37 Using Newton s Second Law The box doesn t move vertically, so the y-acceleration is zero: a y = 0 Our target variable is the x-acceleration, a x. F x = F = 20 N = ma x a x = 20 N = 0.50 m/s2 40 kg

38 Mass and Weight Weight of a body is the gravitational force that the Earth exerts on the body. The terms mass and weight are often misused and interchanged in everyday conversation. Mass is a scalar quantitiy characterizing the inertial properties of a body. Weight is a vector quantity. Mass and weight are related: Bodies having large mass also have large weight. w = m g where g is the acceleration due to gravity. A large stone is hard to throw because of its large mass, and hard to lift off the ground because of its large weight. Remember that g is the magnitude of g, so g is always a positive number, by definition. Thus w = mg, is the magnitude of the weight and is also always positive.

39 Mass and Weight The concept of mass plays two rather different roles in mechanics: The weight of a body (the gravitational force acting on it) is proportional to its mass; we call the property related to gravitational interactions gravitational mass. On the other hand, we call the inertial property that appears in Newton s second law the inertial mass.

40 Newton s Third Law A force acting on a body is always the result of its interaction with another body, so forces always come in pairs. Experiments show that whenever two bodies interact, the two forces that they exert on each other are always equal in magnitude and opposite in direction. Newton s third law of motion: If body A exerts a force on body B (an action ), then body B exerts a force on body A (a reaction ). These two forces have the same magnitude but are opposite in direction. These two forces act on different bodies. F A on B = F B on A The two forces in an action-reaction pair never act on the same body.

41 Exercise: Three blocks on a plane Three blocks connected by strings are pulled by a force F. What is the acceleration of the system? What are the tensions in the strings?

42 Exercise: Atwood Machine Two blocks connected by a string along a pulley. What is the acceleration of the system? What is the tension in the string?

43 Exercise: Atwood Machine T m 2 g = m 2 a m 1 g T = m 1 a (m 1 m 2 )g = (m 1 + m 2 )a a = m 1 m 2 m 1 + m 2 g T = m 2 (a + g)

44 Exercise: Inclined Plane What is the acceleration of the system? What is the tension in the string?

45 Exercise: Inclined Plane What is the acceleration of the system? What is the tension in the string?

46 Exercise: Dangling Rope A uniform rope of mass M and length L hangs from the limb of a tree. Find the tension a distance x from the bottom.

47 Exercise: Dangling Rope The force diagram for the lower section of the rope is shown in the sketch. The section is pulled up by a force of magnitude T (x), where T (x) is the tension at x. The downward force on the rope is its weight T (x) = Mg(x/L). The total force on the section is zero since it is at rest. Hence T (x) = Mg L x At the bottom of the rope the tension is zero (T (0) = 0), while at the top the tension equals the total weight of the rope (T (L) = Mg).

48 Exercise: Whirling Rope A uniform rope of mass M and length L is pivoted at one end and whirls with uniform angular velocity ω. What is the tension in the rope at distance r from the pivot? Neglect gravity.

49 Exercise: Whirling Rope This example cannot be solved by direct application of Newton s second law. However, by treating each small section of the system as a particle, and taking the limit using calculus, we can obtain a differential equation which leads to the solution. Consider the small section of rope between r and r + r. The length of the section is r and its mass is m = M r/l. Because of its circular motion, the section has a radial acceleration. Therefore, the forces pulling either end of the section cannot be equal, and we conclude that the tension must vary with r. The inward force on the section is T (r), the tension at r, and the outward force is T (r + r). Treating the section as a particle, its inward radial acceleration is rω 2. [This point can be confusing; it is just as reasonable to take the acceleration to be (r + r)ω 2. However, we shall shortly take the limit r 0, and in this limit the two expressions give the same result.]

50 Exercise: Whirling Rope The equation of motion for the section is T (r + r) T (r) = mrω 2 = Mrω2 r L The problem is to find T (r), but we are not yet ready to do this. However, by dividing the last equation by r and taking the limit r 0, we can find an exact expression for dt /dr. dt dr T (r + r) T (r) = lim r 0 r = Mrω2 L To find the tension, we integrate. dt = Mω2 L rdr

51 Exercise: Whirling Rope Integrating we obtain dt = Mω2 L rdr T (r) = C Mω2 2L r 2 where C is an integration constant to be determined by a boundary condition. Since the end of the rope at r = L is free, the tension there must be zero, T (L) = 0. T (r) = Mω2 2L (L2 r 2 )

52 A ball thrown straight up has zero velocity at its highest point. Is the ball in equilibrium at this point? If the two ends of a rope in equilibrium are pulled with forces of equal magnitude and opposite direction, why is the total tension in the rope not zero?