Physics 1501 Lecture 8

Transcription

1 Physcs 1501: Lecture 8 Announcements Homework #3 : due next Monday Topcs Revew of ewton s Laws. rcton Some applcatons of ewton s laws. Physcs 1501: Lecture 8, Pg 1 Revew ewton s Laws 1, 2, 3 Isaac ewton ( ) publshed Prncpa Mathematca n In ths work, he proposed three laws of moton: Law 1: An obect subect to no external forces s at rest or moves wth a constant velocty f vewed from an nertal reference frame. Law 2: or any obect, ET = Σ = ma Law 3: orces occur n pars: A,B = - B,A (or every acton there s an equal and opposte reacton.) Physcs 1501: Lecture 8, Pg 2 Page 1

2 Exercse: Inclned plane A block of mass m sldes down a frctonless ramp that makes angle θ wth respect to horzontal. What s ts acceleraton a? m a θ Physcs 1501: Lecture 8, Pg 3 Inclned plane... Defne convenent axes parallel and perpendcular to plane: Acceleraton a s n x drecton only. m a θ Physcs 1501: Lecture 8, Pg 4 Page 2

3 Inclned plane... Consder x and y components separately: : sn θ = ma a = g sn θ : - cos θ = 0. = cos θ ma sn θ θ cos θ Physcs 1501: Lecture 8, Pg 5 Angles of an Inclned plane ma = sn θ θ + φ = 90 θ θ φ θ Physcs 1501: Lecture 8, Pg 6 Page 3

4 Example Gravty, ormal orces etc. Consde a women on a swng: Actve gure When s the tenson on the rope largest. Is t : A) greater than B) the same as C) less than the force due to gravty actng on the woman Physcs 1501: Lecture 8, Pg 7 What does t do? It opposes moton! ew Topc: rcton How do we characterze ths n terms we have learned? rcton results n a force n a drecton opposte to the drecton of moton! APPLIED ma f RICTIO Physcs 1501: Lecture 8, Pg 8 Page 4

5 rcton... rcton s caused by the mcroscopc nteractons between the two surfaces: Physcs 1501: Lecture 8, Pg 9 rcton... orce of frcton acts to oppose moton: Parallel to surface. Perpendcular to ormal force. ma f Physcs 1501: Lecture 8, Pg 10 Page 5

6 Model for Sldng rcton The drecton of the frctonal force vector s perpendcular to the normal force vector. The magntude of the frctonal force vector f K s proportonal to the magntude of the normal force. f K = µ K ( = µ K n the prevous example) The heaver somethng s, the greater the frcton wll be...makes sense! The constant µ K s called the coeffcent of knetc frcton. Physcs 1501: Lecture 8, Pg 11 Model... Dynamcs: : : µ K = m a = so µ K = m a ma µ K Physcs 1501: Lecture 8, Pg 12 Page 6

7 Lecture 8, ACT 1 rcton and Moton A box of mass m 1 = 1 kg s beng pulled by a horzontal strng havng tenson T = 40. It sldes wth frcton (µ k =.5) on top of a second box havng mass m 2 = 2 kg, whch n turn sldes on an ce rnk (frctonless). What s the acceleraton of the second box? (a) a = 0 m/s 2 (b) a = 2.5 m/s 2 (c) a = 10 m/s 2 T m 1 sldes wth frcton (µ k =0.5 ) a =? m 2 sldes wthout frcton Physcs 1501: Lecture 8, Pg 13 Statc rcton... So far we have consdered frcton actng when somethng moves. We also know that t acts n un-movng statc systems: In these cases, the force provded by frcton wll depend on the forces appled on the system. f S Physcs 1501: Lecture 8, Pg 14 Page 7

8 Statc rcton... Just lke n the sldng case except a = 0. : f S = 0 : = Whle the block s statc: f S = (unlke knetc frcton) f S Physcs 1501: Lecture 8, Pg 15 Statc rcton... The maxmum possble force that the frcton between two obects can provde s f MAX = µ S, where µ s s the coeffcent of statc frcton. So f S µ S. As one ncreases, f S gets bgger untl f S = µ S and the obect breaks loose and starts to move. f S Physcs 1501: Lecture 8, Pg 16 Page 8

9 Statc rcton... µ S s dscovered by ncreasng untl the block starts to slde: : MAX µ S = 0 : = µ S = MAX / MAX µ S Physcs 1501: Lecture 8, Pg 17 Addtonal comments on rcton: Snce f = µ, the force of frcton does not depend on the area of the surfaces n contact. By defnton, t must be true that µ S > µ K for any system (thnk about t...). Graph of rctonal force vs Appled force: f = µ S f = µ K Actve gure f f = A A Physcs 1501: Lecture 8, Pg 18 Page 9

10 Lecture 8, ACT 2 Two-body dynamcs A block of mass m, when placed on a rough nclned plane (µ > 0) and gven a bref push, keeps movng down the plane wth constant speed. If a smlar block (same µ) of mass 2m were placed on the same nclne and gven a bref push, t would: m (a) stop (b) accelerate (c) move wth constant speed Physcs 1501: Lecture 8, Pg 19 Example wth pulley A mass M s held n place by a force. nd the tenson n each segment of the rope and the magntude of. Assume the pulleys massless and frctonless. Assume the rope massless. T 1 T 2 T 4 T 3 We use the 5 step method. Draw a pcture: what are we lookng for? What physcs dea are applcable? Draw a dagram and lst known and unknown varables. ewton s 2 nd law : =ma < M T 5 ree-body dagram for each obect Physcs 1501: Lecture 8, Pg 20 Page 10

11 BD for all obects Pulleys: contnued T 4 T 2 T 3 T 4 T 1 T 2 T 3 T 5 < M T 5 M T 5 Mg =T 1 T 2 T 3 Physcs 1501: Lecture 8, Pg 21 Pulleys: fnally Step 3: Plan the soluton (what are the relevant equatons) =ma, statc (no acceleraton: mass s held n place) T 5 M T 5 =Mg Mg T 4 T 1 +T 2 +T 3 =T 4 =T 1 T 2 T 3 T 2 +T 3 =T 5 =T 1 T 3 T 5 T 2 Physcs 1501: Lecture 8, Pg 22 Page 11

12 Pulleys: really fnally! Step 4: execute the plan (solve n terms of varables) We have (from BD): =T 1 T 5 =Mg T 2 +T 3 =T 5 T 1 +T 2 +T 3 =T 4 Pulleys are massless and frctonless T 1 =T 3 T 2 =T 3 T 4 T 2 +T 3 =T 5 gves T 5 =2T 2 =Mg T 2 =Mg/2 T 1 T 2 T 3 T 1 =T 2 =T 3 =Mg/2 and T 4 =3Mg/2 T 5 =Mg and =T 1 =Mg/2 Step 5: evaluate the answer (here, dmensons are OK and no numercal values) < M T 5 Physcs 1501: Lecture 8, Pg 23 Lecture 8, ACT 3 ewton s Second Law I push wth a force of 2 ewtons on a cart that s ntally at rest on an ar table wth no ar. I push for a second. Because there s no ar, the cart stops after I fnsh pushng. It has traveled a certan dstance. = 2 t o, v o = 0 Cart Δx 1 t 1 =1s, v 1 t 2 =2s, v 2 Cart Ar Track Δx 2 Cart or a second shot, I push ust as hard but keep pushng for 2 seconds. The dstance the cart moves the second tme versus the frst s, A) 4 x as long B) 2 x as long C) Same D) 1/2 as long E) 1/4 x as long Physcs 1501: Lecture 8, Pg 24 Page 12

13 Lecture 8, Act 4 You are gong to pull two blocks (m A =4 kg and m B =6 kg) at constant acceleraton (a= 2.5 m/s 2 ) on a horzontal frctonless floor, as shown below. The rope connectng the two blocks can stand tenson of only 9.0. Would the rope break? (A) YES (B) CA T TELL (C) O A rope B a= 2.5 m/s 2 Physcs 1501: Lecture 8, Pg 25 Example Three blocks are connected on the table as shown. The table has a coeffcent of knetc frcton of 0.350, the masses are m 1 = 4.00 kg, m 2 = 1.00kg and m 3 = 2.00kg. m 2 T 1 m 1 m 3 a) What s the magntude and drecton of acceleraton on the three blocks? b) What s the tenson on the two cords? Physcs 1501: Lecture 8, Pg 26 Page 13

14 T 12 T 23 T 12 m 2 T 1 T 23 m 1 m 3 a =-m 2 g T 12 T 12 T 23 m m 1 2 a µ k m 2 g m 1 g m 2 g a T 23 m 3 m 3 g T 12 - m 1 g = - m 1 a T 23 - m 3 g = m 3 a -T 12 + T 23 + µ k m 2 g = - m 2 a SOLUTIO: T 12 = = 30.0, T 23 = 24.2, a = 2.31 m/s2 left for m 2 Physcs 1501: Lecture 8, Pg 27 Page 14