PHY183-Lecture15 - 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 15 ! 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 14, 2006 Physics for Scientists&Engineers 1 1 September 14, 2006 Physics for Scientists&Engineers 1 2 Y e s t e r d a y : 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 T w o B lo c k s ( 1 ) ! Setup: • One object supported by horizontal surface • Second object connect to it by a string, redirected over pulley Fy = 0 ! N = mg cos" • Step 3: Use Newton’s Second Law to determine accelerationin 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! ! Question: • What is the acceleration? ! ! First observation: • Both objects will have same acceleration, because they are connected by the string, making them move at the same rate ! ! a = ( g sin ! )ex ! Fg x ! • Limiting case: acceleration approaches 0 as angle approaches 0. Expected! September 14, 2006 Physics for Scientists&Engineers 1 3 September 14, 2006 Physics for Scientists&Engineers 1 4 T w o B lo c k s ( 2 ) ! A g a i n , w e w i ll u s e f r e e -b o d y d i a g r a m s • First, for mass 1: T w o B lo c k s ( 3 ) ! F r e e -b o d y d ia g r a m f o r m a s s 2 • All motion in vertical direction • F1 = m1g = gravity • Balanced by normal force N • Only remaining force is that of the string tension, T • Use Newton’s Second Law, F=ma, and get in this case in x direction: • F2 = m2g = gravity • Use Newton’s Second Law, F=ma, and get in this case in y direction: T ! m2 g = ! m2 a • Combine with previous equation to eliminate string tension m1a = T • a is the acceleration we are looking for, but what is T? For the answer, we turn to the other mass… m1a = T = m2 g ! m2 a " a=g m2 m1 + m2 • This is the result for the acceleration that we had been looking for. • Please note that we can insert a into either equation above to get T. September 14, 2006 Physics for Scientists&Engineers 1 5 September 14, 2006 Physics for Scientists&Engineers 1 6 F r ic t io n ! Introduction of a new force: friction ! Friction forces are present in practically all kinematical processes ! Concerning friction we observe the following • If a n o b je c t is a t r e s t, th e n it ta k e s a th r e s h o ld o f a n e x te r n a l fo r c e to make it move (experiment 1) • If a n o b je c t is a t r e s t, th e n th e fo r c e o n e m u s t e x e r t to s ta r t i t m o v in g is larger than the force required to keep it moving (experiment 2) • T h e fr ic tio n fo r c e is p r o p o r tio n a l to th e n o r m a l fo r c e (e x p e r im e n t 3 ) • T h e fr ic tio n fo r c e is in d e p e n d e n t o f th e s iz e o f th e c o n ta c t a r e a (experiment 4) • T h e fr ic tio n fo r c e d e p e n d s o n th e r o u g h n e s s o f th e c o n ta c t s u r fa c e (experiment 5) T w o T y p e s o f F r ic t io n ! There are two types of friction • Kinetic friction • O b je c t in m o tio n • Static friction • O b je c t a t r e s t; s ta tic fr ic tio n fo r c e h a s a m a x im u m v a lu e ! Both types of friction are proportional to the normal force f = µN ! The coefficient µ is always greater than zero and usually smaller than 1 ! The coefficient is different for kinetic friction and for static friction 7 September 14, 2006 Physics for Scientists&Engineers 1 8 September 14, 2006 Physics for Scientists&Engineers 1 K in e t ic F r ic t io n ! Kinetic friction deals with objects in motion ! The magnitude of kinetic friction is given by fk = µ k N S t a t ic F r ic t io n ! If an object is at rest, it takes a threshold amount of force to make it move ! If you push on a stationary object with a force below the threshold, it will not move ! If you push on the stationary object hard enough, it will begin to move ! The static friction force is always equal to and opposite the force exerted on the stationary object ! We can write the static friction force as fs ! µ s N = fs , max 9 September 14, 2006 Physics for Scientists&Engineers 1 10 ! N is the magnitude of the normal force ! µk is the coefficient of kinetic friction ! The direction of the kinetic friction force on an object is always opposite to the direction of the motion of the object ! If we push on an object to keep it sliding at a constant speed, the magnitude of the friction force is equal to the magnitude of the force with which we are pushing. Why? • O n ly tw o fo r c e s a r e a c tin g , fr ic tio n fo r c e a n d p u s h in g fo r c e • N e w to n ’ s F ir s t L a w : n e t fo r c e m u s t b e z e r o , b e c a u s e o b je c t m o v e s w ith c o n s ta n t v e lo c ity • = > fr ic tio n fo r c e in th is c a s e is e x a c tly o p p o s ite o f p u s h in g fo r c e September 14, 2006 Physics for Scientists&Engineers 1 Comparison of Kinetic and Static Friction ! How does the friction coefficient depend on an external force? Red path: increase external force from 0 Blue path: Decrease external force back down to 0 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 skier is moving) fk = µ k N = µ k mg cos! • Newton’s Second Law in x-direction: !F i x,i = mg sin " # µ k mg cos" = max ax = g(sin ! " µ k cos! ) Physics for Scientists&Engineers 1 12 ! Note that µ s > µ k September 14, 2006 Physics for Scientists&Engineers 1 11 • Final results: September 14, 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 14, 2006 Physics for Scientists&Engineers 1 µk = F mg 13 µ s = tan ! September 14, 2006 Physics for Scientists&Engineers 1 14 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 14, 2006 T ! m2 g = ! m2 a m2 m1 + m2 Physics for Scientists&Engineers 1 15 m1a = T ! f T ! m2 g = ! m2 a a= September 14, 2006 m2 g ! f m1 + m2 Physics for Scientists&Engineers 1 16 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 ! ! A n sw er : T w o cases: 1. m g! f a= 2 m1 + m2 fk = µ k m1g; fs , max = µ s m1g; The coefficient of friction is small enough µs < m2 ! m2 g > fs , max m1 In this case, the force of static friction is overcome, and the system of blocks is moving with acceleration a= 2. ( m2 ! µ k m1 )g m1 + m2 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 tow-block system will stay at rest Physics for Scientists&Engineers 1 17 • (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 September 14, 2006 September 14, 2006 Physics for Scientists&Engineers 1 18 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 14, 2006 Physics for Scientists&Engineers 1 19 September 14, 2006 Physics for Scientists&Engineers 1 20 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 ! For the ladder to not slip, we need: fs ! fs , max = µ s N = µ s g( ml + mm ) 21 September 14, 2006 Physics for Scientists&Engineers 1 22 ( ml g )(! / 2 ) sin # + ( mm g )r sin # $ R! cos# = 0 September 14, 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 Ffrict = K 0 + K1v + K 2 v 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 September 14, 2006 Physics for Scientists&Engineers 1 23 September 14, 2006 Physics for Scientists&Engineers 1 24 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: K = 1 cd A! 2 • 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 Fg = Fdrag ! mg = Kv 2 ! Solve this for the velocity (speed, really): v= mg K 25 September 14, 2006 v= 2 mg cd A! Note: this velocity depends on the mass and area of object! 26 September 14, 2006 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 14, 2006 Physics for Scientists&Engineers 1 ˆ Kv 2 y ˆ ! mgy 27 ...
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