Mechanics Newton s Laws (cont d)

Transcription

1 Mechanics Newton s Laws (cont d) Lana heridan De Anza College Oct 16, 2018

2 Last time net force example Newton s first law Newton s second law mass vs weight force diagrams

3 Overview Newton s second law examples Newton s third law action-reaction pairs of forces some types of forces

4 Diagrams of Forces: Free-Body Diagram N Physical picture This is a free-body diagram. We represent the chair as a point-particle with force vectors pointing outward. interest (c) Choose a convenient coordinate system (d) Resolve fo y N N N x = 0 N y = N W F W W x = 0 W y = O x Free-body diagram We also picked a coordinate system (x, y axes).

5 object, which we will model as a particle. T us isolate only those forces on the object and We can choose our system analysis. to be more than one object. This is three interacting objects, a monitor sitting on a table, on the Earth: Diagrams of Forces n F tm n F tm F mt F g F me F Em F g F Em 1 Figure from erway & Jewett. a b

6 ly those forces on the object and eliminate the Force Diagrams We could later refine our system into pieces. Here is a depiction of the forces that act on a single object, the monitor. tm n F tm n F tm F g F Em F g F Em F g F Em

7 Clearly, we would like to use Newton s second law gas that can be released through varying combinati around the unit, producing a force of about 10 pou enough propellant for a six-hour EVA (extra-vehicular We show the physical situation in Figure 5 7 (a), w An astronaut useson a jet a 655-kg pack tosatellite. push on The a 655-kg corresponding satellite. If free-body the d satellite starts at rest shown and in moves Figure m(b). after Note 5.00that seconds we have of chosen pushing, what is the direction force, F, of exerted the push. on itnow, by theif astronaut? the satellite starts at after 5.00 seconds of pushing, what is the force, F, exer Force Diagrams, Newton s econd Law, and Kinematics astronaut using a jet llite ation. (b) The freehe satellite. Only one ellite, and it is in the. y F x (a) Physical picture (b) Free-body

8 be released through varying combinations of 24 nozzles spaced nit, Force producing Diagrams, a force of Newton s about 10 pounds. econd The Law, MMUs and contain llant Kinematics for a six-hour EVA (extra-vehicular activity). he physical An astronaut situation uses in Figure 5 7 a jet pack to (a), push where on a an 655-kg astronaut satellite. pushes If the atellite. The corresponding free-body diagram for the satellite is satellite starts at rest and moves m after 5.00 seconds of re 5 7 (b). Note that we have chosen the x axis to point in the pushing, what is the force, F, exerted on it by the astronaut? he push. Now, if the satellite starts at rest and moves m nds of ketch: pushing, what is the force, F, exerted on it by the astronaut? y F x picture (b) Free-body diagram!!

9 Force Diagrams, Newton s econd Law, and Kinematics An astronaut uses a jet pack to push on a 655-kg satellite. If the satellite starts at rest and moves m after 5.00 seconds of pushing, what is the force, F, exerted on it by the astronaut? Given: x, t, m Want: F

10 Force Diagrams, Newton s econd Law, and Kinematics An astronaut uses a jet pack to push on a 655-kg satellite. If the satellite starts at rest and moves m after 5.00 seconds of pushing, what is the force, F, exerted on it by the astronaut? Given: x, t, m Want: F trategy: to find the force we must find the acceleration. x = v 0x t a xt 2

11 Force Diagrams, Newton s econd Law, and Kinematics

12 Newton s econd Law Implications Quick Quiz You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v? A 4 t B 2 t C t 2 D t 4 4 &J page 116.

13 , and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 F 1 = 5.0 N F = 8.0 N 2 60 oving s suro two F 1 x hapter 3, predict the approximate n the same direction. s problem is categorized as one that

14 , and time: Example s 2 (5.4) e is the pound (lb). A force of 1 lb is oduces an acceleration of 1 ft/s 2 : /s 2 Consider a 0.3 kg hockey puck on frictionless ice. Find its acceleration. y F 2 F 1 = 5.0 N F = 8.0 N 2 60 oving s suro two F 1 x hapter 3, predict the approximate n the same direction. s problem is categorized as one that

15 Newton s Third Law Newton s Third Law is commonly stated as For every action, there is an equal and opposite reaction. However it is more precisely stated: Newton III If two objects (1 and 2) interact the force that object 1 exerts on object 2 is equal in magnitude and opposite in direction to the force that object 2 exerts on object 1. F 1 2 = F 2 1

16 Newton s Third Law Main idea: you cannot push on something, without having it push back on you. If object 1 pushes on (or interacts with) object 2, then the force that object 1 exerts on object 2, and the force that object 2 exerts on object 1 form an action reaction pair.

17 us isolate only those forces Newton s Third Law: Action Reaction analysis. Pairs ewton s Third Law 119 n F tm s 2 F 12 F 21 r - t s - F 12 F 21 Figure 5.5 Newton s third law. The force F 12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force F 21 exerted by object 2 1 a F mt F g F me F Em

18 5.6a. The gravita- Pitfall Prevention 5.6 Defining a ystem Consider5.6 these Newton s particles Third which Law exert a force119 on each other: n two objects, we ted by a on b. The erts on object 2 is ect 1 is called the rthermore, either e terms for convet objects and must, the force acting by the Earth on 2 Figure 5.5 Newton s third law. agnitude They are of attracted. this The Each force will F12 accelerate exerted by object toward 1 the other. xerted by the procelerate the Earth on object 2 is equal in magnitude and opposite in direction to the force F21 exerted by object 2 projectile toward on object 1. s acceleration due F 12 F 12 F 21 F 21 1

19 Defining a ystem Consider5.6 these Newton s particles Third which Law exert a force119 on each other: n two objects, we ted by a on b. The erts on object 2 is ect 1 is called the rthermore, either e terms for convet objects and must, the force acting by the Earth on agnitude of this xerted by the procelerate the Earth projectile toward s acceleration due 2 F 12 F 12 F 21 F 21 Figure 5.5 Newton s third law. They are attracted. The Each force will F12 accelerate exerted by object toward 1 the other. on object 2 is equal in magnitude But wait: do the forces and opposite cancel? in direction to the force F21 exerted by object 2 F on object = F 2 1 F F 2 1 = 0 Is the net force zero? How can they each accelerate? 5.6a. The gravita- Pitfall Prevention 5.6 1

20 Defining a ystem Consider these 5.6 particles Newton s Third which Law exert a force 119 on each other: between two objects, we e exerted by a on b. The t 1 exerts on object 2 is on object 1 is called the ms; furthermore, either se these terms for conveifferent objects and must ample, the force acting xerted by the Earth on Is the net force zero? the magnitude of this orce exerted by the proust accelerate the Earth tes the projectile toward ver, its acceleration due 2 F 12 F 12 igure The 5.6a. only The force gravitao this accelerates. force is the force n Does Not Always Equal mg In on particle Pitfall Prevention 1 is F , so the net force is not zero: it onitor does not acceler- the situation shown in Figure 5.6 F 21 Figure 5.5 Newton s third law. The force F12 exerted by object 1 on object 2 is equal in magnitude No! The forces act on and different opposite in objects. direction to To find if particle 1 accelerates, we find the force net Fforce 21 exerted onby particle object 2 1. We do not on object 1. consider forces on particle 2. F 21 1

21 Action and Reaction Why when we fire a cannon does the cannon ball move much faster forward than the cannon does backwards? Why when we drop an object does it race downwards much faster than the Earth comes up to meet it?

22 Action and Reaction Why when we fire a cannon does the cannon ball move much faster forward than the cannon does backwards? Why when we drop an object does it race downwards much faster than the Earth comes up to meet it? The masses of each object are very different! From Newton s second law a = F m If m is smaller, a is bigger. If m is very, very big (like the Earth), the acceleration is incredibly small.

23 ject, which we will model as a particle. Therefore, a free-body dia isolate Forceonly Diagrams those forces on the object and eliminate the other forc alysis. Question. Do the two forces shown in the diagram that act on the monitor form an action-reaction pair under Newton s third law? n F tm n F tm n F tm F mt F g F me (A) Yes. (B) No. F Em Figure 5 F g F Em F g F Em b c the force the gravi exerted b force F m diagram s shows th

24 ome types of forces: Gravitation and Weight For the moment, we only care about this force in that it gives objects weight, W. F g = W = mg The force F g or W, acts downwards towards the center of the Earth.

25 ome types of forces The Normal Force The normal force supports and object sitting on a surface. It acts in a direction perpendicular to the surface. For an object of weight W sitting on a level surface (that is not accelerating), the normal force, N is N = W Be careful! There are many cases in which the above equation is not true! 1 Figure from

26 The Normal Force The normal force supports an object that sits on a surface, but its magnitude is different in different circumstances. In general, one needs to work out what it will be in each problem. ome cases where the normal force is different than the weight of an object are: the object is in an accelerating elevator. the object sits on an incline.

27 Elevator Problems a = 0 Elevator is at rest or moving with constant velocity. You feel the same as you normally do. Your weight and normal force are both of magnitude mg.

28 Elevator Problems a = +a j (a is a positive number) Elevator could be moving upward increasing speed or downward decreasing speed. You feel as if your weight has increased. Your weight is mg j, but the normal force is n = m(g + a) j.

29 Elevator Problems a = a j (a is a positive number) Elevator could be moving upward and slowing down or moving downward increasing speed. You feel as if your weight has decreased. Your weight is mg j, but the normal force is n = m(g a) j.

30 The Normal Force The normal force supports an object that sits on a surface, but its magnitude is different in different circumstances. In general, one needs to work out what it will be in each problem. ome cases where the normal force is different than the weight of an object are: the object is in an accelerating elevator. the object sits on an incline.

31 Object on an Incline Problems with an object placed on an incline often require us to find the net force on the object or its acceleration. 1 Figures from erway & Jewett

32 Object on an Incline d, the force again decreases from front to back. The coupler connecting the loco ge force Problems to slow with down anthe object rest of placed the cars, on an but incline the final often coupler requiremust to apply a for last car. find the net force on the object or its acceleration. Consider a car on a frictionless driveway. 2 (Or free to roll, with frictionless, massless wheels.) naway Car AM y riveway inclined. n he car, assuming mg sin 11a to conceptuyday experience, ncline will accelsame thing 1 Figures hapfrom erway a & Jewett u x b mg cos u u F g = m

33 Object on an Incline AM y d n g mg sin u u- e, l- p- le r- a u x b mg cos u u F g = m g The forces acting on the car: weight and normal force. Figure 5.11 (Example 5.6) (a) A car on a frictionless incline. (b) The free- thisdiagram case, it for isthe useful car. The to pick black adot coordinate represents the system position that of is the rotated: center Inbody the of mass x axis of the points car. We along will learn slope, about the center y axisof perpendicular mass in Chapter to 9. the slope. x

34 car. Object on an Incline ay Car AM y ay inclined n r, assuming mg sin u o conceptuexperience, e will accelthing hapes not set.) u The forces a acting on the car: weight and b normal force. Figure 5.11 (Example 5.6) (a) A car on a frictionless incline. (b) The freebody diagram acts downwards for the car. The black anddot the represents normal the position force acts of the center in a as a particle The weight of mass of the car. We will learn about center of mass in Chapter 9. tes. Furtherry common category of problems in which an object moves under the influence direction perpendicular to the inclined surface. of e-body diagram for the car. The only forces acting on the car are the normal force hich acts perpendicular to the plane, and the gravitational force Fg 5 mg, which lems involving inclined planes, it is convenient to choose the coordinate axes with x x mg cos u u F g = m g The normal force has a magnitude such that it cancels out the component of the weight perpendicular to the surface. x

35 Object on an Incline away Car AM y iveway inclined n e car, assuming mg sin u a to conceptuay experience, cline will accelme thing haprakes not set.) u o, the forces a in the (tilted) y-directionb cancel: Figure 5.11 (Example 5.6) (a) A car on a frictionless incline. (b) The free- car as a particle body diagram for the car. The black dot represents the position of the center of mass of the car. We Fwill y = learn n about mg center cos of mass θ = in 0Chapter 9. erates. Furthera very common category of problems in which an object moves under the influence of Rearranging: free-body diagram for the car. The only forces acting on the car are the normal force e, which acts perpendicular to the plane, and the gravitational force Fg 5 mg, which oblems involving inclined planes, it is convenient to choose the coordinate axes with x icular to beit as less in Figure than5.11b. the weight, With these mg. axes, we represent the gravitational force by sin u along the positive x axis and one of magnitude mg cos u along the negative y axis. x n = mg cos θ mg cos u u Fg = m g If θ > 0 (it is an incline, not a flat surface), the normal force will x

36 st car. Object on an Incline away Car AM y iveway inclined n e car, assuming mg sin u a to conceptuay experience, cline will accelme thing haprakes not set.) u weight: In the (tilted) a x-direction, there is only ba component from the Figure 5.11 (Example 5.6) (a) A car on a frictionless incline. (b) The free- car as a particle body diagram for the car. The black dot represents the position of the center of mass of the car. We will learn F erates. Furthera very common category of problems in which an object moves under the influence x about = mg center sin of mass θ in Chapter 9. of F net = (mg sin θ)i a = (g sin θ)i free-body diagram for the car. The only forces acting on the car are the normal force e, which acts perpendicular to the plane, and the gravitational force F g 5 mg, which oblems involving inclined planes, it is convenient to choose the coordinate axes with x icular to it as in Figure 5.11b. With these axes, we represent the gravitational force by x mg cos u u Fg = m g x

37 Incline Example 24 above the horizontal. (a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the hor- A 65-kg skier izontal speeds is greater down that 24, a trail, and everything as shown. else remains The the same, surface is is the force exerted by the floor on his feet smooth and inclined at an angle of 22 greater than, less than, or the same as the value found in part (a)? with Explain. the horizontal. (a) Find the25. direction IP A65-kgand skier speeds magnitude down a trail, of as shown the inet Figure force acting on the The surface is smooth and inclined at an angle of 22 with the skier. horizontal. (a) Find the direction and magnitude of the net force (b) Does the net actingforce on the skier. exerted (b) Doeson the net the forceskier exertedincrease, on the skier increase, decrease, or stay the same as the slope becomes steeper? decrease, or stay the same as the slope becomes steeper? Explain. Explain. ck, as shown in Fige of 130 N at an angle f the raft. The second k, pulls at an angle of wman pull so that the ward direction? FIGURE 5 27 Problems 25 and 38 1 Walker, Ch 5, #32. 22

38 ome types of forces N OF NEWTON LAW Tension 6 2 trings and prings A common way to exert a force on an object is to pull on it with a string, a rope, a cable, or a wire. imilarly, you can push or pull on an object if you attach it to a The force exerted by a rope or chain to suspend or pull an object spring. In this section we discuss the basic features of strings and springs and how withthey mass. transmit forces. T T T T eavy rope FIGURE 6 5 Tension in a string A string, pulled from either end, has a tension, T. If the string were to be cut at any point, the force required to hold the ends together is T. Problems involving tensions often require solving systems of vector equations. trings and Tension 1 Imagine picking up a light string and holding it with one end in each hand. If you Figure from

39 T ome types of forces Example problems with gravity and tension. A common way to exert a force o cable, or a wire. imilarly, you c spring. In this section we discuss they transmit forces. T T 3 T 2 T 1 FIGURE 6 5 Tension in a st A string, pulled from either en at any point, the force required FIGURE 6 6 Tension in a heavy rope Because of the weight of the rope, the tension is noticeably different at points 1, 2, and 3. As the rope becomes lighter, however, the difference in tension decreases. In the limit of a rope of zero mass, the tension is the same throughout the rope. (ee also example 5-5 on pg 126 and 6-5 of the textbook.) 1 Figure from Walker, Physics. trings and Tension Imagine picking up a light string pull to the right with your right hand with a force T, the string be tension T in the string. To be mo some point, the tension T is the Figure 6 5 that is, T is the force to hold the cut ends together. N

40 Pulleys Pulleys turn tensions around a corner. y over a frictionless pulley of is called an Atwood machine. ermine the value of g. Deterbjects and the tension in the + T T m 1 m 1 gure 5.14a in action: as one m 2 m 1 g m 2 nward. Because the objects + lerations must be of equal For the moment, we are just considering massless, frictionless g pulleys. What does that mean? subject to the gravitational nnected to them. Therefore, particles under a net force. s are shown in Figure 5.14b. exerted by the string and a m 2 Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the b

41 Pulleys Pulleys turn tensions around a corner. y over a frictionless pulley of is called an Atwood machine. ermine the value of g. Deterbjects and the tension in the + T T m 1 m 1 gure 5.14a in action: as one m 2 m 1 g m 2 nward. Because the objects + lerations must be of equal For the moment, we are just considering massless, frictionless m 2 g pulleys. What does that mean? subject to the gravitational a b nnected to Massless: them. Therefore, we do not have to worry about force needed to particles under accelerate a net force. each atom Figure in the 5.14 pulley(example 5.9) The Atwood machine. (a) Two objects Frictionless: the axle of s are shown in Figure 5.14b. connected the pulley by a massless has no inextensible friction to resist the wheel turning string over a frictionless pulley. exerted by the string and (b) The free-body diagrams for the

42 Pulleys and Tension Because of the weight of the rope, the tension is noticeably different at points 1, 2, and 3. As the rope becomes lighter, however, the difference in tension decreases. In If the rope is light (massless) the limit of a rope and of zero themass, pulley the tension is massless and is the same throughout the rope. frictionless, the tension in the rope on both sides of the pulley is the same. FIGURE 6 7 A pulley changes the direction of a tension 1 Figure from Walker, Physics. T T hand with a force T, the string tension T in the string. To be m some point, the tension T is Figure 6 5 that is, T is the for to hold the cut ends togethe equally to the right and to the As an example, consider a to a box with a weight of 105 N tion, suppose the rope is unifo is the tension in the rope (i) w (iii) where it attaches to the ce First, the rope holds the bo to the box is simply the weig rope, the tension supports th rope. Thus, T 2 = 105 N sion supports the box plus all that the tension pulls down o From this discussion, we c from top to bottom because of difference in tension between rope s mass were to be vanish as well. In this text, we will a practically massless unless s tension is the same throughou Pulleys are often used to r Figure 6 7. In the ideal case, a Thus, an ideal pulley simply ch changing its magnitude. If a sys

43 ummary Newton s third law action-reaction pairs some types of forces Homework Ch 5 Ques: 9; Probs: 17, 29, 31, 33, 39, 45, 49, 53, 55, 87