2. Physics 111: Lecture 8, Pg 1 Physics 111: Lecture 8 Today’s Agenda l Friction Recap l Drag Forces çTerminal speed l Dynamics of many-body systems çAtwood’s machine çGeneral case of two attached blocks on inclined planes çSome interesting problems

3. Physics 111: Lecture 8, Pg 2 Friction Review: l Friction is caused by the “microscopic” interactions between the two surfaces: çSee discussion in text

4. Physics 111: Lecture 8, Pg 3 Model for Friction f N l Arah vektor gaya gesek f F adalah tegak lurus dengan vektor gaya normal N, dan berlawanan arah dengan gaya yang bekerja pada sistem l Gesekan Kinetic (sliding): Besarnya vektor gaya gesek berbanding lurus dengan besarnya gaya normal l Gesekan Kinetic (sliding): Besarnya vektor gaya gesek berbanding lurus dengan besarnya gaya normal N. f F =  K N l Gesekan Static: Gaya gesekan menyeimbangkan total yang yang bekerja pada sistem sehingga sistem tidak bergerak. Besarnya gaya gesek statis maksimum berbanding lurus dengan l Gesekan Static: Gaya gesekan menyeimbangkan total yang yang bekerja pada sistem sehingga sistem tidak bergerak. Besarnya gaya gesek statis maksimum berbanding lurus dengan gaya normal N. f F   S N

5. Physics 111: Lecture 8, Pg 4 Kinetic Friction:  K koefisien gesek kinetik. i :F  K N = ma j :N = mg soF  K mg = ma amaama F gmggmg N i j KNKN

6. Physics 111: Lecture 8, Pg 5 Static Friction: Koefisien gesek statis  S, menentukan gaya gesek statis maksimum,  S N, yang timbul antara persentuhan dua buah benda.  S dapat ditentukan dengan cara memperbesar F sampai benda mulai bergeser: F MAX -  S N = 0 N = mg F MAX =  S mg  S  F MAX / mg F F MAX gmggmg N i j  S N

7. Physics 111: Lecture 8, Pg 6 Lecture 8, Act 1 Two-body dynamics Sebuah balok bermassa m, jika diletakkan pada permukaan bidang miring yang kasar (  > 0) dan diberikan gaya sesaat, akan bergerak turun pada bidang miring dengan kecepatan konstan. Jika balok yang sejenis(same  ) denganmassa 2m diletakkan pada bidang miring yang sama kemudian diberikan gaya sesaat, maka balok tersebut akan: (a) (a) berhenti (b) (b) dipercepat (c) (c) Bergerak dengan kecepatan tetap m

8. Physics 111: Lecture 8, Pg 7 Lecture 8, Act 1 Solution l Gambarkan diagram benda bebas dan tentukan gaya total dalam arah X  i j mg N  KNKN F NET,X = mg sin  K mg cos  = ma = 0 (kasus balok pertama) Jika massa digandakan maka semua komponen menjadi dua kali Lebih besar. Tetapi percepatan tetap nol Speed will still be constant!

9. Physics 111: Lecture 8, Pg 8 Drag Forces: l When an object moves through a viscous medium, like air or water, the medium exerts a “drag” or “retarding” force that opposes the motion of the object. F g = mg F DRAG j v

10. Physics 111: Lecture 8, Pg 9 Drag Forces: l This drag force is typically proportional to the speed v of the object raised to some power. This will result in a maximum (terminal) speed. F g = mg F D = bv n j v feels like n=2 Parachute

11. Physics 111: Lecture 8, Pg 10 Terminal Speed: l Suppose F D = bv 2. Sally jumps out of a plane and after falling for a while her downward speed is a constant v. çWhat is F D after she reaches this terminal speed? çWhat is the terminal speed v? l F TOT = F D - mg = ma = 0. çF D = mg l Since F D = bv 2 çbv 2 = mg F g = mg F F D = bv 2 j v

12. Physics 111: Lecture 8, Pg 11 Many-body Dynamics l Systems made up of more than one object l Objects are typically connected: çBy ropes & pulleystoday çBy rods, springs, etc.later on

13. Physics 111: Lecture 8, Pg 12 Atwood’s Machine: l Find the accelerations, a 1 and a 2, of the masses. l What is the tension in the string T ? Masses m 1 and m 2 are attached to an ideal massless string and hung as shown around an ideal massless pulley. Fixed Pulley m1m1 m2m2 j a1a1 a2a2 T1T1 T2T2

14. Physics 111: Lecture 8, Pg 13 Atwood’s Machine... l Draw free body diagrams for each object j l Applying Newton’s Second Law: ( j -components) ç T 1 - m 1 g = m 1 a 1 ç T 2 - m 2 g = m 2 a 2 But T 1 = T 2 = T since pulley is ideal and a 1 = -a 2 = -a. since the masses are connected by the string m2gm2g m1gm1g Free Body Diagrams T1T1 T2T2 j a1a1 a2a2

15. Physics 111: Lecture 8, Pg 14 T - m 1 g = -m 1 a(a) T - m 2 g = m 2 a(b) l Two equations & two unknowns çwe can solve for both unknowns (T and a). l subtract (b) - (a): çg(m 1 - m 2 ) = a(m 1 + m 2 ) ça = l add (b) + (a): ç2T - g(m 1 + m 2 ) = -a(m 1 - m 2 ) = çT = 2gm 1 m 2 / (m 1 + m 2 ) Atwood’s Machine... -

16. Physics 111: Lecture 8, Pg 15 Atwood’s Machine... m1m1 m2m2 j a a T T l So we find: Atwood’s Machine

17. Physics 111: Lecture 8, Pg 16 Is the result reasonable? Check limiting cases! l Special cases: i.) m 1 = m 2 = ma = 0 and T = mg. OK! ii.) m 2 or m 1 = 0 |a| = g and T= 0. OK! l Atwood’s machine can be used to determine g (by measuring the acceleration a for given masses). -

18. Physics 111: Lecture 8, Pg 17 Attached bodies on two inclined planes All surfaces frictionless m1m1 m2m2 smooth peg 11 22

19. Physics 111: Lecture 8, Pg 18 How will the bodies move? From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law: Taking “x” components: 1) T 1 - m 1 g sin  1 = m 1 a 1X 2) T 2 - m 2 g sin  2 = m 2 a 2X But T 1 = T 2 = T and a 1X = -a 2X = a (constraints) m1m1 y x T1T1 N m1gm1g m2m2 m2gm2g T2T2 N x y 11 22

20. Physics 111: Lecture 8, Pg 19 Solving the equations Using the constraints, solve the equations. T - m 1 gsin  1 = -m 1 a(a) T - m 2 gsin  2 = m 2 a(b) Subtracting (a) from (b) gives: m 1 gsin  1 - m 2 gsin  2 = (m 1 +m 2 )a So:  a mm mm g   1122 12 sin 

21. Physics 111: Lecture 8, Pg 20 Special Case 1: m1m1 m2m2 11 22 m1m1 m2m2 If  1 = 0 and  2 = 0, a = 0. Boring  a mm mm g   1122 12 sin 

22. Physics 111: Lecture 8, Pg 21 Special Case 2: If  1 = 90 and  2 = 90, m2m2 T T m1m1 Atwood’s Machine m1m1 m2m2 11 22  a mm mm g   1122 12 sin 

23. Physics 111: Lecture 8, Pg 22 Special Case 3: If  1 = 0 and  2 = 90, m1m1 m2m2 Lab configuration m1m1 m2m2 11 22  a mm mm g   1122 12 sin   Air-track

24. Physics 111: Lecture 8, Pg 23 Lecture 8, Act 2 Two-body dynamics l In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless. (a) (b) (c) (a) Case (1) (b) Case (2) (c) same m 10kg a m a F = 98.1 N Case (1)Case (2)

25. Physics 111: Lecture 8, Pg 24 Lecture 8, Act 2 Solution m 10kg a l Add (a) and (b): 98.1 N = (m + 10kg)a l Note: (a) (b) T = ma(a) (10kg)g -T = (10kg)a (b) l For case (1) draw FBD and write F NET = ma for each block:

26. Physics 111: Lecture 8, Pg 25 Lecture 8, Act 2 Solution l The answer is (b) Case (2) T = 98.1 N = ma l For case (2) m 10kg a m a F = 98.1 N Case (1)Case (2)

27. Physics 111: Lecture 8, Pg 26 Problem: Two strings & Two Masses on horizontal frictionless floor: m2m2 m1m1 T2T2 T1T1 l Given T 1, m 1 and m 2, what are a and T 2 ? T 1 - T 2 = m 1 a (a) T 2 = m 2 a (b) çAdd (a) + (b): T 1 = (m 1 + m 2 )a a a i çPlugging solution into (b):

28. Physics 111: Lecture 8, Pg 27 Lecture 8, Act 3 Two-body dynamics l Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? (a) T 1 > T 2 > T 3 (b) T 3 > T 2 > T 1 (c) T 1 = T 2 = T 3 T3T3 T2T2 T1T1 3m 2m m a

29. Physics 111: Lecture 8, Pg 28 Lecture 8, Act 3 Solution l Draw free body diagrams!! T3T3 3m T 3 = 3ma T3T3 T2T2 2m T 2 - T 3 = 2ma T 2 = 2ma +T 3 > T 3 T2T2 T1T1 m T 1 - T 2 = ma T 1 = ma + T 2 > T 2 T 1 > T 2 > T 3

30. Physics 111: Lecture 8, Pg 29 Lecture 8, Act 3 Solution l Alternative solution: T3T3 T2T2 T1T1 3m 2m m a Consider T 1 to be pulling all the boxes T3T3 T2T2 T1T1 3m 2m m a T 2 is pulling only the boxes of mass 3m and 2m T3T3 T2T2 T1T1 3m 2m m a T 3 is pulling only the box of mass 3m T 1 > T 2 > T 3

31. Physics 111: Lecture 8, Pg 30 Problem: Rotating puck & weight. l A mass m 1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m 2. çWhat is the tension (T) in the string? çWhat is the speed (v) of the sliding mass? m1m1 m2m2 v R

32. Physics 111: Lecture 8, Pg 31 Problem: Rotating puck & weight... l Draw FBD of hanging mass: çSince R is constant, a = 0. so T = m 2 g m2m2 m2gm2g T m1m1 m2m2 v R T

33. Physics 111: Lecture 8, Pg 32 Problem: Rotating puck & weight... l Draw FBD of sliding mass: m1m1 T = m 2 g m1gm1g N m1m1 m2m2 v R T Use F = T = m 1 a where a = v 2 / R m 2 g = m 1 v 2 / R T = m 2 g Puck

34. Physics 111: Lecture 8, Pg 33 Recap of today’s lecture l Friction Recap.(Text: 5-1) l Drag Forces. (Text: 5-3) çTerminal speed. l Dynamics of many-body systems.(Text: 4-7) çAtwood’s machine. çGeneral case of two attached blocks on inclined planes. çSome interesting special cases. l Look at Textbook problems l Look at Textbook problems Chapter 6: # 3, 7, 21, 75