Physics 201 Lecture 9

Transcription

1 Physcs 20 Lecture 9 l Goals: Lecture 8 ewton s Laws v Solve D & 2D probles ntroducng forces wth/wthout frcton v Utlze ewton s st & 2 nd Laws v Begn to use ewton s 3 rd Law n proble solvng Law : An obect subect to no external forces s at rest or oves wth a constant velocty f vewed fro an nertal reference frae. Law 2: or any obect, E = Σ = a Law 3: orces occur n pars: A, B = - B, A (or every acton there s an equal and opposte reacton.) Read: orce on A by B Physcs 20: Lecture 9, Pg Physcs 20: Lecture 9, Pg 2 Exercse ewton s 3rd Law l wo blocks are beng pushed by a assless fnger on a horzontal frctonless floor. l How any acton-reacton force pars are present n ths exercse ewton s Law he acceleraton of an obect s drectly proportonal to the net external force actng upon t. he constant of proportonalty s the ass. A. 2 B. 4 C. 6 D. Soethng else a b l hs s a vector expresson r = î + ĵ+ kˆ x y z Ex Ey Ez = a = a x = a y z Physcs 20: Lecture 9, Pg 3 Physcs 20: Lecture 9, Pg 4 Analyzng orces: ree Body Dagra A heavy sgn s hung between two poles by a rope at each corner extendng to the poles. ree Body Dagra Step one: Defne the syste 2 θ 2 A hangng sgn s an exaple of statc equlbru (depends on observer) What are the forces on the sgn and how are they related f the sgn s statonary (or ovng wth constant velocty) n an nertal reference frae Physcs 20: Lecture 9, Pg 5 Step two: Sketch n force vectors Step three: Apply ewton s st & 2 nd Laws (Resolve vectors nto approprate coponents) θ Physcs 20: Lecture 9, Pg 6 Page

2 Physcs 20 Lecture 9 ree Body Dagra 2 θ Pushng and Pullng orces l Strng or ropes are exaples of thngs that can pull l You ar s an exaple of an obect that can push or push Vertcal : y-drecton 0 = - + snθ + 2 sn Horzontal : x-drecton 0 = - cosθ + 2 cos Physcs 20: Lecture 9, Pg 7 Physcs 20: Lecture 9, Pg 8 Scale Proble Scale Proble l You are gven a ass and you hang t drectly on a fsh scale and t reads 50 (g s 0 /s 2 ). l ow you use ths ass n a second experent n whch the ass hangs fro a assless strng passng over a assless, frctonless pulley and s anchored to the floor. he pulley s attached to the fsh scale. l What force does the fsh scale now read 50 l Step : Identfy the syste(s). In ths case t s probably best to treat each obect as a dstnct eleent and draw three force body dagras. v One around the scale v One around the assless pulley (even though assless we can treat s as an obect ) v One around the hangng ass l Step 2: Draw the three BGs. (Because ths s a now a one-densonal proble we need only consder forces n the y-drecton.) Physcs 20: Lecture 9, Pg 9 Physcs 20: Lecture 9, Pg 0 3: 2: Scale Proble.0 kg W - - l Σ y = 0 n all cases : 0 = : 0 = = 3: 0 = W (not useful here) l Substtutng 2 nto yelds = 2 = 00 (We start wth 50 but end wth 00 ) : - - Physcs 20: Lecture 9, Pg ewton s 2 nd Law, orces are condtonal P s the upward force beng exerted on the crate by the person C s the contact or noral force on the crate by the floor, and Ws the weght (force of the earth on the crate). A. P + C < W B. P + C > W C. P = C D. P + C = W Physcs 20: Lecture 9, Pg 2 Page 2

3 Physcs 20 Lecture 9 Another experent A block s connected by a horzontal assless strng. here s a known ass 2.0 kg and you apply a constant force of 0. What s the acceleraton of the block f the table top s frctonless BD Σ x = a x = - Σ y = 0 = - a x = -/ = 5 /s 2 Physcs 20: Lecture 9, Pg 3 Verson 2 A block s connected by a horzontal assless strng. here s a known ass 2.0 kg and you apply a constant horzontal force of 0. he surface s frctonless and t s nclned 30 fro horzontal. What s the acceleraton of the block down the slde Σ x = a x = x +W x Σ y = 0 = + y +W y a x = - cos 30 +W sn 30 a x = -/ 3 ½ /2 - g / 2 a x = -5(.732)/2-5 = -7.2 /s 2 What s the apparent weght of the block on the slde = y +W y = sn 30 -W cos 30 = -0(0.5) + 20(0.866) = 2.3 BD = W Physcs 20: Lecture 9, Pg 4 Verson 3 A block s connected by a horzontal assless strng. here s a known ass 2.0 kg and you apply a constant horzontal force of he surface s frctonless and t s nclned 30 fro horzontal. What s the acceleraton of the block down the slde Σ x = a x = x +W x Σ y = 0 = + y +W y a x = - cos 30 +W sn 30 a x = -/ 3 ½ /2 - g / 2 a x = -7.3(.732)/ 2-5 = -35 /s 2 What s the apparent weght of the block on the slde = y +W y = sn 30 -W cos 30 = -34.6(0.5) + 20(0.866) = 0.0!!! BD = W Physcs 20: Lecture 9, Pg 5 Another experent: A odfed Atwood s achne wo blocks, & 2, are connected by a assless frctonless strng/pulley on the table as shown. he table surface s frctonless and lttle g acts downward. What s the acceleraton of the horzontal block Requres two BDs and 2 ewton s 3 rd Law. Mass Σ y = a y = g Σ x = 2 a 2x = - Σ y = 0 = 2 g g Physcs 20: Lecture 9, Pg 6 2 g Correlated oton: a y = a 2x a If oves up oves 2 rght Another experent: A odfed Atwood s achne wo blocks, & 2, are connected by a assless frctonless strng/pulley on the table as shown. he table surface s frctonless and lttle g acts downward. What s the acceleraton of the horzontal block. Requres two BDs and 2 ewton s 3 rd Law. Mass A. Σ y = a= g B. Σ x = 2 a = - Σ y = 0 = 2 g g Subbng n fro B nto A a= - 2 a g a + 2 a = g a = g / ( + 2 ) Physcs 20: Lecture 9, Pg 7 2 g A specal contact force: rcton l What does t do v It opposes oton (velocty, actual or that whch would occur f frcton were absent!) l How do we characterze ths n ters we have learned v rcton results n a force n a drecton opposte to the drecton of oton (actual or, f statc, then nferred )! APPLIED f RICIO a Physcs 20: Lecture 9, Pg 8 Page 3

4 Physcs 20 Lecture 9 l o net force If no acceleraton l So frctonal force ust equals appled force l Key pont: It s condtonal! rcton... l rcton s caused by the croscopc nteractons between the two surfaces: APPLIED f RICIO Physcs 20: Lecture 9, Pg 9 Physcs 20: Lecture 9, Pg 20 rcton: Statc frcton Statc equlbru: A block wth a horzontal force appled, Σ x = 0 = - + f s f s = BD Σ y = 0 = - + = As ncreases so does f s f s Statc frcton, at axu (ust before slppng) Equlbru: A block, ass, wth a horzontal force appled, Drecton: A force vector to the noral force vector and the vector s opposte to the drecton of acceleraton f µ were 0. Magntude: f S s proportonal to the agntude of f s = µ s f s Physcs 20: Lecture 9, Pg 2 Physcs 20: Lecture 9, Pg 22 Knetc or Sldng frcton (f k < f s ) Dynac equlbru, ovng but acceleraton s stll zero BD Σ x = 0 = - + f k f k = v Σ y = 0 = - + = As ncreases f k reans nearly constant (but now there acceleraton s acceleraton) f k = µ k f k Model of Statc rcton (sple case) Magntude: f s proportonal to the appled forces such that f s µ s µ s called the coeffcent of statc frcton Drecton: Opposte to the drecton of syste acceleraton f µ were 0 Physcs 20: Lecture 9, Pg 23 Physcs 20: Lecture 9, Pg 24 Page 4

5 Physcs 20 Lecture 9 Sldng rcton Coeffcents of rcton l Drecton: A force vector to the noral force vector and the vector s opposte to the velocty. Materal on Materal steel / steel µ s = statc frcton 0.6 µ k = knetc frcton 0.4 l Magntude: f k s proportonal to the agntude of add grease to steel v f k = µ k ( = µ K n the prevous exaple) etal / ce l he constant µ k s called the coeffcent of knetc frcton brake lnng / ron tre / dry paveent l Logc dctates that µ S > µ K for any syste tre / wet paveent Physcs 20: Lecture 9, Pg 25 Physcs 20: Lecture 9, Pg 26 Other orces are Condtonal l otce what happens f we change the drecton of the appled force l he noral force can ncrease or decrease f Let a=0 Physcs 20: Lecture 9, Pg 27 An experent wo blocks are connected on the table as shown. he table has unknown statc and knetc frcton coeffcents. Desgn an experent to fnd µ S Statc equlbru: Set 2 and add ass to f 2 S to reach the breakng pont. Requres two BDs g 2 g Mass Σ y = 0 = g Σ x = 0 = - + f s = - + µ S Σ y = 0 = 2 g = g = µ S 2 g µ S = / 2 Physcs 20: Lecture 9, Pg 28 Statc rcton wth a bcycle wheel (not sple) l You are pedalng hard and the bcycle s speedng up. What s the drecton of the frctonal force l Read Chapter 6 or hursday l You are breakng and the bcycle s slowng down What s the drecton of the frctonal force Physcs 20: Lecture 9, Pg 29 Physcs 20: Lecture 9, Pg 30 Page 5