PHY183-Lecture14 - S t a t ic E q u ilib r iu m C o n d it...

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Unformatted text preview: S t a t ic E q u ilib r iu m C o n d it io n s Fnet , x = ! Fi , x = F1, x + F2 , x + ... + Fn , x = 0 n Physics for Scientists & Engineers 1 Fall Semester 2006 Lecture 14 i =1 n Fnet , y = ! Fi , y = F1, y + F2 , y + ... + Fn , y = 0 i =1 n Fnet , z = ! Fi , z = F1, z + F2 , z + ... + Fn , z = 0 i =1 ! net = " ! clockwise, i # " ! counter-clockwise, j = 0 i j K e y to s o lv in g a ll s ta ti c e q u ilib riu m p ro b le m s ! September 14, 2006 Physics for Scientists&Engineers 1 1 September 14, 2006 Physics for Scientists&Engineers 1 2 E x a m p le : T e e t e r -T o t t e r ( 1 ) ! Playground toy for which answer seems obvious • If you place a weight m1 on one end at a distance r1 from the center axle, then obviously that end goes down, simply because of the force and torque that the weight exerts on it. • Question: Where do you place m2 in order for the teeter- totter to balance? ! Question: ! Answer: E x a m p le : T e e t e r -T o t t e r ( 2 ) • Where do you place m2 in order for the teeter- totter to balance? • Force due to m1 on beam: F1=m1g. • Gravitational force on the beam (with mass M): F=Mg • Force due to m2 on beam: F2=m2g • All three forces act in negative y-direction, as shown • In addition, there is a yet unknown y normal force N acting on the beam, originating from the pivot x September 14, 2006 Physics for Scientists&Engineers 1 3 September 14, 2006 Physics for Scientists&Engineers 1 4 E x a m p le : T e e t e r -T o t t e r ( 3 ) ! Equilibrium conditions: • y direction: E x a m p le : T e e t e r -T o t t e r ( 4 ) ! Calculation of torques • Pick a pivot point • Natural choice: central axle of teeter-totter • Note: N and Mg attach exactly at this point and thus have moment arms of length 0 => no torque about this pivot from these two forces. • General strategy: pick pivot point so that an unknown force has 0 moment arm (here Mg is unknown). Fnet , y = ! Fi , y = " m1g " m2 g " Mg + N = 0 i # N = g( m1 + m2 + M ) • No force components in x direction => no additional information • Torque condition will provide y information on distance r2 that was asked for x • Only clockwise torque comes from m2g • Only counterclockwise torque from m1g 5 September 14, 2006 Physics for Scientists&Engineers 1 y x September 14, 2006 Physics for Scientists&Engineers 1 6 E x a m p le : T e e t e r -T o t t e r ( 5 ) ! Torque Equilibrium Condition: H a n g in g S ig n ( 1 ) ! Sign is hanging from a post that is attached to wall via a hinge and held by a diagonal cable. ! Question: What is the tension in the cable holding the post? ! net = " ! clockwise, i # " ! counter-clockwise, j i j = m2 gr2 sin 90° # m1gr1 sin 90° = 0 $ m2 r2 = m1r1 m $ r2 = r1 1 m2 ! This is our final answer! y x September 14, 2006 Physics for Scientists&Engineers 1 7 September 14, 2006 Physics for Scientists&Engineers 1 8 H a n g in g S ig n ( 2 ) ! First step: Draw in all force arrows acting on the post. • Sign has mass M. Its force on the post is Mg, straight down. • Post itself has a mass, m. Weight = mg, also straight down. • String tension due to the cable, T • Finally, the wall exerts a force the post, F. • For now, F and T are unknown. Also, the direction of F is not known. September 14, 2006 Physics for Scientists&Engineers 1 9 H a n g in g S ig n ( 3 ) ! Quick first accounting: • 3 unknowns: Fx , Fy , T • 3 equations: F net , x = 0, • There is hope! Fnet , y = 0, ! net = 0 ! We already can calculate the angle between the cable and the beam y x ! = arctan(d / l ) September 14, 2006 Physics for Scientists&Engineers 1 10 H a n g in g S ig n ( 4 ) ! Again, question: What is tension T? ! We can figure this out without knowing F by calculating torque about point where the posts attaches to wall. H a n g in g S ig n ( 5 ) ! Put in some numbers: • m = 19.7 kg, M = 33.1 kg • l = 2.40 m, r = 1.95 m T= ( ml + 2 Mr )g 2l sin ! ((19.7 kg) " (2.40 m)+2 " (33.1 kg) " (1.95 m)) " (9.81 m/s 2 ) = 2 " (2.40 m) " sin25.4 ° = 840 N l mg sin 90° + Mgr sin 90° ! Tl sin " = 0 2 ( ml + 2 Mr )g T= 2l sin ! September 14, 2006 Physics for Scientists&Engineers 1 11 September 14, 2006 Physics for Scientists&Engineers 1 12 H a n g in g S ig n ( 6 ) ! Question 2: What is magnitude and direction of F? ! Answer: Since we now know all other forces, we can use our two force equilibrium conditions to determine the components of F. • x-direction: H a n g in g S ig n ( 7 ) ! Question 2: What is magnitude and direction of F? ! Since we now know the two components, we can calculate: • Magnitude Fx ! T cos" = 0 # Fx = T cos" = (840 N)cos25.4 °=759 N Fy + T sin ! " mg " Mg = 0 # Fy = ( m + M )g " T sin ! F = Fx2 + Fy2 = 775 N • Direction • y-direction: arctan( Fy / Fx ) = 11.8° = (19.7 kg+33.1 kg)(9.81 m/s 2 ) " (840 N)sin25.4 ° =158 N September 14, 2006 Physics for Scientists&Engineers 1 13 September 14, 2006 Physics for Scientists&Engineers 1 14 S ir Is a a c N e w t o n ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! Born Jan 4, 1643, Woolsthorpe, Lincolnshire, England June 5, 1661: Entered Trinity College, Cambridge April 1665: Bachelor’s Degree Summer 1665 to 1667: University is closed due to plague; Newton returns home and makes his major advances in math, physics, astronomy 1666: Universal law of gravitaton 1669: Newton is appointed Lucasian Chair at Cambridge 1670: Particle theory of light 1671: Publication of Calculus (later independently invented by German Leibniz) 1687: Publication of Principia (Newton’s 3 Laws of Motion) 1689: Election to Parliament 1696: Warden of the Royal Mint 1699: Master of the Royal Mint (becomes very rich!) 1703: Elected President of Royal Society 1705: Knighted by Queen Anne March 31, 1727: Died in London September 14, 2006 Physics for Scientists&Engineers 1 15 September 14, 2006 Physics for Scientists&Engineers 1 16 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. M ass ! Gravitational Mass • Already covered: Fg = mg = weight = gravitational force • Mass m entering this equation is source of interaction ! Newton’s Second Law: • ! Newton’s Third Law: ! If there is a net external force Fnet acting on an object with ! mass , then the force will cause an acceleration, a : ! ! Fnet = ma ! Inertial mass • = Resistance to changes in motion, to acceleration ! Newton’s Insight: These two masses are identical ! Where does mass come from? • Current thinking: Higgs particle • Still unknown, search underway at largest particle accelerators in the World • `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 17 September 14, 2006 Physics for Scientists&Engineers 1 18 1st 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. • First part now obvious, was basis of static equilibrium • Second part was far less obvious and represented a giant intellectual leap during Newton’s times • Prevailing view was Aristotelian: Need to keep pushing to keep an object moving • Example: Sliding a refrigerator across the kitchen floor. If you stop pushing, the fridge will stop moving. We will return to this later this week, when we cover friction forces! September 14, 2006 Physics for Scientists&Engineers 1 19 A c c e le r a t io n ! A lr e a d y i n t r o d u c e d : g r a v i t a t i o n a l a c c e le r a t i o n • Can be experienced by jumping off a cliff (not recommended!) => falling faster and faster ! E x p e r i e n c e a c c e le r a t i o n i n a c a r • Step on the gas and get accelerated forward • Step on the brake and slow down (negative acceleration) • Drive through a curve and feel sideward pull, another form of acceleration (will be studied in chapter on circular motion) ! A c c e le r a t i o n i s a v e c t o r • Has magnitude and direction ! P h y s i c a l d i m e n s i o n o f a c c e le r a t i o n : m/s2 • Sometimes acceleration is expressed in multiples of g (as in: “pulling 3 g’s”) September 14, 2006 Physics for Scientists&Engineers 1 20 A c c e le r a t io n a n d 2 n d L a w ! Fnet acting on an ! If there is a net external force object with mass , then the force will cause an ! acceleration, a : ! ! 2 n d L a w in C o m p o n e n t s ! ! ! Fnet = ma is a vector equation. Fnet = ma ! We can write down the equations for the Cartesian components of the acceleration: • (Most famous formula in all of science) • Magnitude and direction of acceleration are proportional to those of the net force • A p p ly in g m o r e fo r c e r e s u lts in h ig h e r a c c e le r a tio n Fx = max Fy = may Fz = maz ! The Second Law holds independently for each Cartesian component of the acceleration September 14, 2006 Physics for Scientists&Engineers 1 22 • For a given external force, the magnitude of the acceleration is inversely proportional to the mass • H e a v ie r o b je c ts a r e h a r d e r to a c c e le r a te th a n lig h te r o n e s September 14, 2006 Physics for Scientists&Engineers 1 21 In c lin e d P la n e ( 1 ) ! Typical situation: an object slides down a plane that is at some angle relative to the horizontal ! Example: Snowboarding • S te p 1 : D r a w th e p la n e a n d th e a n g le • D r a w a ll fo r c e s th a t a c t o n th e s lid in g o b je c t • H e r e : g r a v ity , n o r m a l fo r c e • In g e n e r a l a ls o fr ic tio n ( n e g le c t fo r n o w ) • F ir s t im p o r ta n t o b s e r v a tio n : T h e fo r c e v e c to r s d o n o t a d d to 0 ! In c lin e d P la n e ( 2 ) ! N ! Fg September 14, 2006 Physics for Scientists&Engineers 1 23 September 14, 2006 Physics for Scientists&Engineers 1 ! 24 • S te p 2 : P ic k a c o n v e n ie n t c o o r d in a te s y s te m • F o r in c lin e d - p la n e p r o b le m s , it is a lm o s t a lw a y s a d v is a b le to s e le c t th e x a x is a lo n g th e p la n e ( p o s itiv e to th e r ig h t, y - a x is p e r p e n d ic u la r to it, o f c o u r s e ) • N o te : n o r m a l fo r c e h a s o n ly a c o m p o n e n t in y d ir e c tio n • D e c o m p o s e w e ig h t fo r c e in to c o m p o n e n ts In c lin e d P la n e ( 3 ) In c lin e d P la n e ( 3 ) • S te p 3 : N o te s im ila r tr ia n g le s ! W h e r e d o e s th e a n g le ! a p p e a r a g a in ? • Triangle abc (gravity force components) is similar to triangle ABC ( g e o m e tr y ) . • B e c a u s e a is p e r p e n d ic u la r to C , a n d c i s p e r p e n d ic u la r to A , it fo llo w s th a t th e a n g le b e tw e e n a a n d c is th e s a m e a s th e a n g le b e tw e e n A a n d C. • R e s u lt: T h e a n g le ! a p p e a r s in b o th tr ia n g le s • W e ig h t v e c to r c o m p o n e n ts in x a n d y d ir e c tio n : ! N mg sin ! y mg cos! September 14, 2006 ! ! Fg x Fg , x = mg sin ! Fg , y = " mg cos! B c C a! b ! 25 September 14, 2006 A Physics for Scientists&Engineers 1 ! 26 Physics for Scientists&Engineers 1 In c lin e d P la n e ( 4 ) • No motion in y direction => no net force in y direction • All force components have to add to 0 in this direction In c lin e d P la n e ( 5 ) ! E x a m p le : s n o w b o a r d i n g ( c o n t i n u e d ) - x c o m p o n e n t • Force component in x direction was already given. • Now we can use Newton’s Second Law and determine the acceleration Fy = Fg , y + N = 0 ! " mg cos# + N = 0 ! N = mg cos# ! N Fx = Fx , g = mg sin ! = max mg sin ! y " 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 cos! (very typical result) Fg ! • This equation determines the normal force • Normal force balances the y component of the skiers weight ! ! x mg sin ! y mg cos! ! ! Fg x ! ˆ a = ( g sin ! ) x ! • Please note: acceleration approaches 0 as angle approaches 0. Expected! September 14, 2006 Physics for Scientists&Engineers 1 27 September 14, 2006 Physics for Scientists&Engineers 1 28 ...
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