Dynamics Laws of Motion Elevators, Pulleys, and Friction

Transcription

1 Dynamics Laws of Motion Elevators, Pulleys, and riction Lana heridan De Anza College Oct 12, 2017

2 Last time equilibrium nonequilibrium Problem solving with tensions inclines

3 Overview Problem solving with more complex scenarios elevators pulleys friction Extending these ideas

4 5 mg cos That is consistent with eparate Objects Pushed Together ntact force between the car and the plane. Consider a force that acts on two objects, masses m 1 and, free to slide on a frictionless surface: m 1 a y n1 n2

5 eparate Objects Pushed Together Question. What is the acceleration of object? m 1 y (A) (B) m 1 (C) (D) m 1 + x P 21 m a n1 n2 P 12

6 eparate Objects Pushed Together Question. What is the acceleration of object? m 1 y (A) (B) m 1 (C) (D) m 1 + x P 21 m a n1 n2 P 12

7 eparate Objects Pushed Together Question. What is the net force on object? m 1 a y (A) 0 (B) (C) (D) m 1 m 1 + m 1 + x P 21 n1 n2 P 12

8 eparate Objects Pushed Together Question. What is the net force on object? m 1 a y (A) 0 (B) (C) (D) m 1 m 1 + m 1 + x P 21 n1 n2 P 12

9 sin 90 5 g eparate Objects Pushed Together er than a x 5 2g because we have chosen positive x to be cos u Main gives Idea: us n 5 if objects mg cos are 90 pushed 5 0. That or pulled is consistent together, with then they must se there all is accelerate no contact at the force same between rate. the car and the plane. t in contact ure 5.12a. A d e m 1 That means that the individual net forces on each must be different: na 1 n2 n y 1 n2 y P 21 P a and on because x x a P 21 m 1 m m g 2 1 g b m 1 c P 12 g g

10 eparate Objects Pulled Along m T m 2 1 The accelerations of the two blocks are the same.

11 eparate Objects Pulled Along m m 2 1 m T m 2 1 The accelerations of the two blocks are the same. Imagining them as a single block gives the acceleration straight away: a = m 1 +

12 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.

13 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.

14 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.

15 Elevator Problems 5.7 Analysis Models Using Newton s econd Law 127 An example with the a value greater normal than theforce replaced reads a value less than bythe a tension instead: reading on the scale is related to the ng in the scale, which is related to the the spring as in igure 5.2. Imagine ing on a string attached to the end of se, the magnitude of the force exerted l to the tension T in the string. Therefor T. The force T pulls down on the n the fish. categorize this problem by identifyicle in equilibrium if the elevator is not particle under a net force if the elevator When the elevator accelerates upward, the spring scale reads weight of the fish. a When the elevator accelerates downward, the spring scale weight of the fish. T T a diagrams of the forces acting on the mg mg d notice that the external forces acting ownward gravitational force g 5 mg a b erted by the string. If the elevator is igure 5.13 (Example 5.8) A fish is weighed on a spring scale in ng at constant velocity, the fish is a parso o y 5 T 2 g 5 0 or T 5 g 5 mg. an accelerating elevator car. scalar mg is the weight of the fish.) elevator is moving with an acceleration a relative to an observer standing outside the elevator in e fish is now a particle under a net force. nd law to the fish: o y 5 T 2 mg 5 ma y Whether the reaction of the floor of the tension in the cable cause the upward force, the net force is still calculated in the same way. (1) T 5 ma y 1 mg 5 mg a a y g 1 1b 5 g a a y g 1 1b

16 Elevator Problems Question. You are hired to design an elevator that can lift people up to the top of a 70 story building in a short amount of time. The weight of the elevator car is 3500 N and the max load the elevator should be rated to carry is 2000 N and the weight of the entire cable used to lift the elevator car is 3000 N. You add these numbers together and decide that you can choose a low-cost cable rated to carry a load of 9000 N without breaking as the elevator cable. Will you end up fired or with a commendation? (A) Commendation! (B) ired! (C) I m not sure.

17 Elevator Problems Question. You are hired to design an elevator that can lift people up to the top of a 70 story building in a short amount of time. The weight of the elevator car is 3500 N and the max load the elevator should be rated to carry is 2000 N and the weight of the entire cable used to lift the elevator car is 3000 N. You add these numbers together and decide that you can choose a low-cost cable rated to carry a load of 9000 N without breaking as the elevator cable. Will you end up fired or with a commendation? (A) Commendation! (B) ired! (C) I m not sure.

18 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 1 g nward. Because the objects + lerations must be of equal or 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 igure 5.14b. exerted by the string and a igure 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

19 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 1 g nward. Because the objects + lerations must be of equal or the moment, we are just considering massless, frictionless 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 igure in the 5.14 pulley(example 5.9) The Atwood machine. (a) Two objects rictionless: the axle of s are shown in igure 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

20 Pulleys ree-body diagrams: over a frictionless pulley of s called an Atwood machine. rmine the value of g. Deterjects and the tension in the + T T ure 5.14a in action: as one ward. Because the objects lerations must be of equal m 1 subject There to the is gravitational another way ofarepresenting this, b that for these massless, nected to them. Therefore, frictionless rope & pulley articles under a net force. igure systems 5.14 (Example that I find 5.9) The helpful. Remove the pulley and lay everything Atwood outmachine. flat on(a) atwo frictionless objects surface: are shown in igure 5.14b. connected by a massless inextensible string over a frictionless pulley. exerted by the string and m g 1 (b) The free-body T m g diagrams for the 2 as this one in which the two objects. nsion in the string on both or is subject to friction, the tensions m 1 on either side are not the + m 1 m 1 g g

21 Pulleys and the Atwood Machine The Atwood Machine can be used to make careful determinations of g, as well as explore the behavior of forces and accelerations. nless pulley of twood machine. lue of g. Detertension in the + T T action: as one se the objects st be of equal e gravitational em. Therefore, a m 1 + m 1 m 1 g 1 machine b g

22 Pulleys and the Atwood Machine Notice that again, like the pushed blocks, the two objects must accelerate together at the same rate, because they are connected through an inextensible rope.

23 Pulleys and the Atwood Machine Notice that again, like the pushed blocks, the two objects must accelerate together at the same rate, because they are connected through an inextensible rope. (Is that realistic?)

24 Pulleys and the Atwood Machine Notice that again, like the pushed blocks, the two objects must accelerate together at the same rate, because they are connected through an inextensible rope. (Is that realistic?) The tension in all parts of the rope will be the same.

25 Pulleys and the Atwood Machine We can consider the motion for each mass separately: net,1 = (T m 1 g)j = m 1 a j (1) net,2 = (T g)j = a j (2)

26 Pulleys and the Atwood Machine We can consider the motion for each mass separately: net,1 = (T m 1 g)j = m 1 a j (1) net,2 = (T g)j = a j (2) Be careful about the signs! Both masses must accelerate together - one up, one down. Two equations, two unknowns. olve as you like!

27 Pulleys and the Atwood Machine Take eq (1) eq (2): net,1 = (T m 1 g)j = m 1 a j (1) net,2 = (T g)j = a j (2) g m 1 g = m 1 a + a a = ( m 1 )g m 1 +

28 Pulleys and the Atwood Machine Take eq (1) eq (2): net,1 = (T m 1 g)j = m 1 a j (1) net,2 = (T g)j = a j (2) g m 1 g = m 1 a + a a = ( m 1 )g m 1 + T = 2m 1 g m 1 + We get the same result if we use the horizontal layout picture.

29 ummary more problem solving pulleys irst Test this riday, Oct 13. (Uncollected) Homework erway & Jewett, Ch 5, onward from page 136. Obj.Q 1; Problems: 21