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# 6 newton law and gravitation

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### 6 newton law and gravitation

1. 1. Physics 111: Lecture 4, Pg 1 Physics 111: Lecture 4Physics 111: Lecture 4 Today’s AgendaToday’s Agenda Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? DynamicsDynamics
2. 2. Physics 111: Lecture 4, Pg 2 Review: Centripetal AccelerationReview: Centripetal Acceleration UCM results in acceleration: Magnitude: |a| = v2 / R = ω2 R Direction: - rr (toward center of circle) R aa ω ^^ Useful stuff:Useful stuff: f = rotations / sec T = 1 / f ω = 2π / T = 2π f = rad/sec v = ω R
3. 3. Physics 111: Lecture 4, Pg 3 Isaac Newton (1643 - 1727) published Principia Mathematica in 1687
4. 4. Physics 111: Lecture 4, Pg 4 DynamicsDynamics In this work, he proposed three “laws” of motion: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, FFNET = Σ FF = maa Law 3: Forces occur in pairs: FFA ,B = - FFB ,A (For every action there is an equal and opposite reaction.)
5. 5. Physics 111: Lecture 4, Pg 5 Newton’s First LawNewton’s First Law An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial referenceinertial reference frameframe. If no forces act, there is no acceleration. The following statements can be thought of as the definition of inertial reference frames. An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector! 1. Dishes 2. Monkey
6. 6. Physics 111: Lecture 4, Pg 6 Is Urbana a good IRF?Is Urbana a good IRF? Is Urbana accelerating? YES! Urbana is on the Earth. The Earth is rotating. What is the centripetal acceleration of Urbana? T = 1 day = 8.64 x 104 sec, R ~ RE = 6.4 x 106 meters . Plug this in: aU = .034 m/s2 ( ~ 1/300 g) Close enough to 0 that we will ignore it. Urbana is a pretty good IRF. a v R R T RU = = =       2 2 2 ω π2 Ice puck
7. 7. Physics 111: Lecture 4, Pg 7 Newton’s Second LawNewton’s Second Law For any object, FFNET = Σ FF = maa. The acceleration aa of an object is proportional to the net force FFNET acting on it. The constant of proportionality is called “mass”, denoted m. » This is the definition of mass. » The mass of an object is a constant property of that object, and is independent of external influences. Force has units of [M]x[L / T2 ] = kg m/s2 = N (Newton)
8. 8. Physics 111: Lecture 4, Pg 8 Newton’s Second Law...Newton’s Second Law... What is a force? A Force is a push or a pull. A Force has magnitude & direction (vector). Adding forces is like adding vectors. FF1 FF2 aa FF1 FF2 aa FFNET FFNET = maa
9. 9. Physics 111: Lecture 4, Pg 9 Newton’s Second Law...Newton’s Second Law... Components of FF = maa : FX = maX FY = maY FZ = maZ Suppose we know m and FX , we can solve for aX and apply the things we learned about kinematics over the last few weeks: tavv ta 2 1 tvxx xx0x 2 xx00 += ++=
10. 10. Physics 111: Lecture 4, Pg 10 Example: Pushing a Box on Ice.Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m? FF v = 0 m a ii
11. 11. Physics 111: Lecture 4, Pg 11 Example: Pushing a Box on Ice.Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ? d FF v m a ii
12. 12. Physics 111: Lecture 4, Pg 12 Example: Pushing a Box on Ice...Example: Pushing a Box on Ice... Start with F = ma. a = F / m. Recall that v2 - v0 2 = 2a(x - x0 ) (Lecture 1) So v2 = 2Fd / m v Fd m = 2 d FF v m a ii
13. 13. Physics 111: Lecture 4, Pg 13 Example: Pushing a Box on Ice...Example: Pushing a Box on Ice... Plug in F = 50 N, d = 10 m, m = 100 kg: Find v = 3.2 m/s. d FF v m a ii v Fd m = 2
14. 14. Physics 111: Lecture 4, Pg 14 Lecture 4,Lecture 4, Act 1Act 1 Force and accelerationForce and acceleration A force F acting on a mass m1 results in an acceleration a1. The same force acting on a different mass m2 results in an acceleration a2 = 2a1. If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? (a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1 F a1 m1 F a2 = 2a1 m2 F a = ? m1 m2
15. 15. Physics 111: Lecture 4, Pg 15 Lecture 4,Lecture 4, Act 1Act 1 Force and accelerationForce and acceleration Since a2 = 2a1 for the same applied force, m2 = (1/2)m1 ! m1+ m2 = 3m1 /2 (a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1 F a = F / (m1+ m2) m1 m2 So a = (2/3)F / m1 but F/m1 = a1 a = 2/3 a1
16. 16. Physics 111: Lecture 4, Pg 16 ForcesForces We will consider two kinds of forces: Contact force: » This is the most familiar kind. I push on the desk. The ground pushes on the chair... Action at a distance: » Gravity » Electricity
17. 17. Physics 111: Lecture 4, Pg 17 Contact forces:Contact forces: Objects in contact exert forces. Convention: FFa,b means “the force acting on a due to b”. So FFhead,thumb means “the force on the head due to the thumb”. FFhead,thumb
18. 18. Physics 111: Lecture 4, Pg 18 Action at a distanceAction at a distance Gravity:
19. 19. Physics 111: Lecture 4, Pg 19 GravitationGravitation (Courtesy of Newton)(Courtesy of Newton) Newton found that amoon / g = 0.000278 and noticed that RE 2 / R2 = 0.000273 This inspired him to propose the Universal Law of Gravitation:Universal Law of Gravitation: |FMm |= GMm / R2 R RE amoon g where G = 6.67 x 10 -11 m3 kg-1 s-2
20. 20. Physics 111: Lecture 4, Pg 20 Gravity...Gravity... The magnitude of the gravitational force FF12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is: The direction of FF12 is attractive, and lies along the line connecting the centers of the masses. F G m m R 12 1 2 12 2 = R12 m1 m2 FF12 FF21
21. 21. Physics 111: Lecture 4, Pg 21 Gravity...Gravity... Near the Earth’s surface: R12 = RE » Won’t change much if we stay near the Earth's surface. » i.e. since RE >> h, RE + h ~ RE. RE m M h 2 E E g R mM GF =FFg
22. 22. Physics 111: Lecture 4, Pg 22 Gravity...Gravity... Near the Earth’s surface...       == 2 E E 2 E E g R M Gm R mM GF So |Fg| = mg = ma a = g All objects accelerate with acceleration g, regardless of their mass! 2 2 E E s/m81.9 R M Gg ==Where:  =g Leaky Cup
23. 23. Physics 111: Lecture 4, Pg 23 Example gravity problem:Example gravity problem: What is the force of gravity exerted by the earth on a typical physics student? Typical student mass m = 55kg g = 9.8 m/s2 . Fg = mg = (55 kg)x(9.8 m/s2 ) Fg = 539 N FFg The force that gravity exerts on any object is called its Weight W = 539 N
24. 24. Physics 111: Lecture 4, Pg 24 Lecture 4,Lecture 4, Act 2Act 2 Force and accelerationForce and acceleration Suppose you are standing on a bathroom scale in 141 Loomis and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? You are told that RX ~ 20 REarth and MX ~ 300 MEarth. (a)(a) 00.75.75 W (b)(b) 1.5 W (c)(c) 2.25 W E X
25. 25. Physics 111: Lecture 4, Pg 25 Lecture 4,Lecture 4, Act 2Act 2 SolutionSolution The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by: F G Mm R = 2 W W F F X E X E =Ratio of weights = ratio of forces: = G M m R G M m R X X E E 2 2 = ⋅       M M R R X E E X 2 W W X E = ⋅      =300 1 20 75 2 .
26. 26. Physics 111: Lecture 4, Pg 26 Newton’s Third Law:Newton’s Third Law: Forces occur in pairs: FFA ,B = - FFB ,A. For every “action” there is an equal and opposite “reaction”. We have already seen this in the case of gravity: F F12 1 2 12 2 21= =G m m R R12 m1 m2 FF12 FF21 Newton’s Sailboard
27. 27. Physics 111: Lecture 4, Pg 27 Newton's Third Law...Newton's Third Law... FFA ,B = - FFB ,A. is true for contact forces as well: FFm,w FFw,m FFm,f FFf,m 2 Skateboards
28. 28. Physics 111: Lecture 4, Pg 28 Example of Bad ThinkingExample of Bad Thinking Since FFm,b = -FFb,m, why isn’t FFnet = 0 and aa = 0 ? a ??a ?? FFm,b FFb,m ice
29. 29. Physics 111: Lecture 4, Pg 29 Example of Good ThinkingExample of Good Thinking Consider only the boxonly the box as the system! FFon box = maabox = FFb,m Free Body Diagram (next time). aaboxbox FFm,b FFb,m ice
30. 30. Physics 111: Lecture 4, Pg 30 Lecture 4,Lecture 4, Act 3Act 3 Newton’s 3rd LawNewton’s 3rd Law Two blocks are stacked on the ground. How many action-reaction pairs of forces are present in this system? (a) 2 (b) 3 (c) 4 a b
31. 31. Physics 111: Lecture 4, Pg 31 Lecture 4,Lecture 4, Act 3Act 3 Solution:Solution: (c) 4 FFE,aE,a FFa,Ea,E a b FFE,bE,b a b FFb,Eb,E FFb,ab,a FFa,ba,b a b FFg,bg,b FFb,gb,g a b
32. 32. Physics 111: Lecture 4, Pg 32 Recap of today’s lectureRecap of today’s lecture Newton’s 3 Laws: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, FFNET = Σ FF = maa Law 3: Forces occur in pairs: FFA ,B = - FFB ,A. Extinguisher Cart