Chapter 6. Force and motion II

Transcription

1 Chapter 6. Force and motion II Friction Static friction Sliding (Kinetic) friction Circular motion Physics, Page 1

2 Summary of last lecture Newton s First Law: The motion of an object does not change unless it is acted on by a net force Newton s Second Law: F net ma Newton s Third Law: F a,b - F b,a Types of Forces Graity Normal Force Friction Tension Physics, Page

3 Static Friction ( 정지쓸림힘 ) F N f s F W Static Friction: a force between two surfaces that preents motion f s <μ s F N F s,max (size is whateer is needed to preent motion) μ s coefficient of static friction a property of the two surfaces 1 μ s 0 direction is whateer direction needed to preent motion Physics, Page 3

4 Kinetic Friction ( 운동쓸림힘 ) F N f k direction of motion F W Kinetic (sliding) Friction: a force between two surfaces that opposes motion f k μ k F N μ k coefficient of sliding friction a property of the two surfaces 1 μ k 0 direction is opposite to motion Physics, Page 4

5 Static Kinetic f s μ s n f k μ k n, μ s : coefficient of static friction, μ k : coefficient of kinetic friction Physics, Page 5

6 Example 1 F N μ k m/s mg Find stopping distance Answer: 16.3 m F μ k F N μ k (mg), F ma, V V o -as Physics, Page 6

7 Example 1- Initial elocity 0 0 m/sec. Stop at l 115 m Find the friction coef. : μ k Example 5-1 F a t l μ μ k n μ k mg ma μ g k μ k g k a t g t μ k t + 0 gl 1 at 0 t 1 μ gt ( ) 0m / sec 9. 8m / sec 115 m k μ 0 k g Physics, Page 7

8 Example μ k 0. T50 N M 5 kg Find acceleration of block Answer: 8.04 m/s F T- μ k F N ma Physics, Page 8

9 Example 3 μ k 0. T50 N θ50 0 M 5 kg Find acceleration of block Answer: 6.0 m/s F T cosθ - μ k F N ma Physics, Page 9

10 Example 4 a b Which case requires the lesser force to oercome static friction? a) a b) b c) the same Physics, Page 10

11 Example 5 : Find a. Example 5- m 1 m 1 m μ k m 1 g m +) F 1 T - μ k m 1 g m 1 a F m g - T m a m g - μ k m 1 g m 1 a +m a a m m μ k m + m 1 1 g Physics, Page 11

12 Uniform Circular Motion (circular motion with constant speed) R a a R centripetal acceleration Instantaneous elocity is tangent to circle Instantaneous acceleration is radially inward There must be a force to proide the acceleration Physics, Page 1

13 Centripetal force ( 구심력 ) and Centrifugal force ( 원심력 ) y Angular elocity θ ω d θ dt T π ω x T F r mrω m r Centripetal force Centrifugal force? Physics, Page 13

14 Question Consider the following situation: You are driing a car with constant speed around a horizontal circular track. How many forces are acting on the car? correct F N f W Physics, Page 14

15 Question The net force on the car is F N 1. Zero. Pointing radially inward 3. Pointing radially outward correct W f a /R R ΣF ma m /R The car has a constant speed so acceleration is zero A car that is driing in a circle is accelerating towards the center of the circle. According to Newton's second law the net force must be pointing toward the center of the circle. when I sit in this car I can feel the FORCE...pulling me to the dark side...away from the center of the circle Physics, Page 15

16 Question Suppose you are driing through a alley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force F N exerted on you by the car seat as you drie past the bottom of the hill? 1. F N < mg. F N mg 3. F N > mg correct F N a /R R You feel like you are being pulled down in your seat so the mg must be greater then the normal force -- wrong! F N is always equal to mg --- wrong! The net force must point towards the center mg ΣF ma m /R F N -mg m /R F N mg + m /R Physics, Page 16

17 Question Going oer top of hill. F N 1. F N < mg. F N mg 3. F N > mg correct mg a /R R ΣF ma m /R mg - F N m /R F N mg - m /R Physics, Page 17

18 Example: Circular motion and graity Find the period for one orbit. M E R m Newton s Law of Graity: F GM E m/r Newton s nd law: Fma Centripetal acceleration: a /R Thus GM E m/r m /R GM E /R T time for one orbit πr/ T πr 3/ GM E Physics, Page 18

19 Example: Circular motion and graity M E R m T πr 3/ G 6.67 x N-m /kg GM E M E 5.58 x 10 4 kg 1. R moon-earth 3.85 x 10 8 m Plug in and calculate: T7.5 days. R shuttle R E 6.38 x 10 6 m Plug in and calculate: T 84.5 minutes 3. R synchronous 4.3 x 10 7 m Plug in and calculate: T1 day Physics, Page 19

20 Example: Carnial Ride R g Find minimum coefficient of static friction so that you don t fall. Answer: μ S > gr/ Physics, Page 0

21 Example : Find the angle θ. Tcosθ θ Tsinθ xˆ ŷ : : F tan F x y mg θ T cos θ mg mg T cos θ T tan sin gr θ θ ma m r gl sin 0 m θ r Example: Find the maximum elocity on the track. F f s ma m r f s, max μ s mg max f s, max m r μ s gr Physics, Page 1

22 Example : Find the normal forces at bottom, top, and at A-point. n A At Bottom: At Top: At A-point: m n bot mg r m n bot + r n top + mg m m n top r r mg n A m r mg Physics, Page

23 Terminal speed ( 종단속력 ) V terminal when a 0 R F b : 마찰력 (Assumption) m g b m a mg b ma Terminal speed a b g m t m b g 0 Physics, Page 3

24 Terminal speed ( 종단속력 ) in air Drag Coefficient R 1 Dρ A Air density Cross section 1 F mg R mg D ρ A t a g 0 D ρ A m r mg D ρ A 3 ma : For terminal elocity t r r Physics, Page 4