PHY183-Lecture16 - N e w t o n’ s T h r e e L a w s !...

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Unformatted text preview: N e w t o n’ s T h r e e L a w s ! Newton’s First Law: • In the absence of an external force on an object, this object will remain at rest, if it was at rest, or, if it was moving, it will remain in motion with the same velocity. Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 16 ! Newton’s Second Law: • If there is a net external force Fnet acting on an object with ! mass m, then the force will! cause an acceleration, a: ! ! Newton’s Third Law: ! Fnet = ma • The forces that two interacting objects exert on each other are always exactly equal in magnitude and opposite in ! ! direction to each other. F1, 2 = ! F2 ,1 September 26, 2006 Physics for Scientists&Engineers 1 1 September 26, 2006 Physics for Scientists&Engineers 1 2 R e v ie w : In c lin e d P la n e • Step 1: pick coordinate system such that x-axis is along plane • Step 2: y-component has no acceleration => determines normal force In c lin e d P la n e + F r ic t io n ! Now add friction force (note blue arrow) • Direction: opposite to motion, i.e. up the hill • Magnitude (use kinetic friction, the boarder is moving) fk = µ k N = µ k mg cos! • Newton’s Second Law in x-direction: Fy = 0 ! N = mg cos" • Step 3: Use Newton’s Second Law to determine acceleration in x-direction Fx = max ! ax = g sin " • Please note that the mass of the object has canceled out of this equation => All objects accelerate at the same rate on this plane, independent of their mass. • We can write down the equation for the acceleration vector: ! N mg sin ! y mg cos! ! ! ! a = ( g sin ! )ex ! Fg x ! !F i x,i = mg sin " # µ k mg cos" = max ax = g(sin ! " µ k cos! ) Physics for Scientists&Engineers 1 4 • Limiting case: acceleration approaches 0 as angle approaches 0. Expected! September 26, 2006 Physics for Scientists&Engineers 1 3 • Final results: September 26, 2006 D e t e r m in in g K in e t ic F r ic t io n C o e ffic ie n t ! (Review): Slide object across horizontal surface, with constant velocity, and measure force F ! Also, measure weight of object = normal force D e t e r m in in g S t a t ic F r ic t io n C o e ffic ie n t s ! We can determine coefficient of static friction by using an inclined plane with variable angle • S lo w ly r a is e p la n e fo r m h o r iz o n ta l to w a r d v e r tic a l • R e c o r d a n g le a t w h ic h o b je c t ju s t s ta r ts s lip p in g • R ig h t b e fo r e s lip p in g , a c c e le r a tio n w a s 0 , a n d s ta tic fr ic tio n fo r c e w a s a t m a x im u m , ju s t b a la n c in g c o m p o n e n t o f w e ig h t a lo n g p la n e : N=m g ! F Top view !F i x,i = 0 = mg sin " # µ s mg cos" • S o lv e fo r c o e ffic ie n t o f s ta tic fr ic tio n : mg sin ! = µ s mg cos! " sin ! = µ s cos! " ! Coefficient of kinetic friction: September 26, 2006 Physics for Scientists&Engineers 1 µk = F mg 5 µ s = tan ! September 26, 2006 Physics for Scientists&Engineers 1 6 T w o B lo c k s - R e v ie w ! F r e e -b o d y d ia g r a m s T w o B lo c k s + F r ic t io n ! F r e e -b o d y d ia g r a m f o r a ls o c o n t a i n s f r i c t i o n m 2 st ays t he sam e ! N e w f r e e -b o d y d ia g r a m f o r m 1 m1a = T a=g September 26, 2006 T ! m2 g = ! m2 a m2 m1 + m2 Physics for Scientists&Engineers 1 7 m1a = T ! f T ! m2 g = ! m2 a a= September 26, 2006 m2 g ! f m1 + m2 Physics for Scientists&Engineers 1 8 T w o B lo c k s + F r ic t io n ( 2 ) ! W h at is f i n t h is e q u at io n : T r ib o lo g y - S c ie n c e o f F r ic t io n ! Microscopic origin of friction is still under active discussion and intense research ! Atomic force microscopes • (AFM) measure the forces between individual atoms on the surface of samples • Very sharp tip is dragged across surface and measures mechanical resistance • Measure forces as as small as 10-11 N = 10 pN ! ! m g! f a= 2 m1 + m2 Answer: fk = µ k m1g; T w o cases: 1. fs , max = µ s m1g; The coefficient of friction is small enough µs < m2 ! m2 g > fs , max m1 a= 2 k 1 In this case, the force of static friction is overcome, and the system of blocks is moving with acceleration ( m ! µ m )g m1 + m2 2. The coefficient of static friction is bigger than this limit. Then the friction force will be just big enough to compensate the force from the hanging mass, and the two-block system will stay at rest September 26, 2006 Physics for Scientists&Engineers 1 9 September 26, 2006 Physics for Scientists&Engineers 1 10 A F M Im a g e s N a n o -w ire P e r s o n o n a la d d e r ( 1 ) ! Q: What friction force is needed to keep ladder from slipping? ! A: first, draw in all forces on ladder • N = normal force from floor • R = normal force from wall (assume no friction between wall and ladder) S in g le - a to m s te p s o n S a p p h ir e A lu m in u m M e m b ra n e C o m p u te r H a r d D r iv e • Wl = weight of ladder • Wm = weight of man • fs = static friction force between ladder and floor September 26, 2006 Physics for Scientists&Engineers 1 11 September 26, 2006 Physics for Scientists&Engineers 1 12 P e r s o n o n a la d d e r ( 2 ) ! Free-body diagram ! 3 equilibrium conditions (x,y,torque) P e r s o n o n a la d d e r ( 3 ) ! Solve torque equation for R: 1 ( m g )! sin ! + ( mm g )r sin ! R= 2 l ! cos! r% " = $ 1 ml g + mm g ' tan ! 2 # !& ! Since fs=R,our final answer is: r$ ! fs = # 1 ml g + mm g & tan ' "2 !% !F i x,i =0 R " fs = 0 # R = fs ! Fy,i = 0 i N " mm g " ml g = 0 # N = g( mm + ml ) "! i i =0 Piv o t h e r e ! For the ladder to not slip, we need: fs ! fs , max = µ s N = µ s g( ml + mm ) 13 September 26, 2006 Physics for Scientists&Engineers 1 14 ( ml g )(! / 2 ) sin # + ( mm g )r sin # $ R! cos# = 0 September 26, 2006 Physics for Scientists&Engineers 1 P e r s o n o n a la d d e r ( 4 ) ! Combine last two equations: r$ !1 # 2 ml + mm & tan ' ( µ s ( ml + mm ) " !% • T h e h ig h e r th e m a n c lim b s , th e m o r e lik e ly th e la d d e r w ill s lip • R e d u c in g th e a n g le o f th e la d d e r w ith th e w a ll w ill p r e v e n t th e la d d e r fr o m s lip p in g • In c r e a s in g th e fr ic tio n c o e ffic ie n t w ill h e lp s o th a t th e la d d e r w ill n o t s lip F r e e -F a ll w it h A ir R e s is t a n c e ! Cannot ignore friction with air when moving fast ! General form of this force should depend on velocity relative to air 2 with constants to be determined ! Macroscopic objects moving at relatively high speeds through air: can neglect the linear term ! Drag force Fdrag = Kv 2 ! Direction of force: opposite to direction of motion Ffrict = K 0 + K1v + K 2 v + ... September 26, 2006 Physics for Scientists&Engineers 1 15 September 26, 2006 Physics for Scientists&Engineers 1 16 T e r m in a l V e lo c it y ! An object in free-fall gets accelerated by gravity and falls faster and faster ! As its velocity increases, so does the drag force ! Once the drag force is equal to gravity, there is no more acceleration, and the velocity stays the same: Terminal velocity T e r m in a l V e lo c it y ( 2 ) ! Need to know the value of the constant K: ! Empirically found: 1 K = 2 cd A! Fg = Fdrag ! mg = Kv 2 ! Solve this for the velocity (speed, really): • A = area exposed to the air stream (in m2) • ! = density of air (approximately 1 kg/m3) • cd = drag coefficient, number between 0 and 1 ! Insert this into expression for terminal velocity v= September 26, 2006 mg K 17 September 26, 2006 v= 2 mg cd A! Note: this velocity depends on the mass and area of object! 18 Physics for Scientists&Engineers 1 Physics for Scientists&Engineers 1 E x a m p le S k y D iv in g ! Area of parachute = 42.7 m2, drag coefficient 0.63, density of air = 1.15 kg/m3, mass of the man + equipment = 76.4 kg. What is the terminal velocity? 2 " 76.4 " 9.81 m ! Answer: v = 2 mg = = 6.96 m/s cd A! 0.63 " 42.7 " 1.15 s ! What is the magnitude of the drag force at terminal velocity? ! Answer: F = mg = 76.4 kg · 9.81 m/s2 = 750 N (Please note: this makes use of Newton’s 1st Law) September 26, 2006 Physics for Scientists&Engineers 1 ˆ Kv 2 y ˆ ! mgy 19 ...
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This note was uploaded on 03/01/2010 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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