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Physics 111: Lecture 8 Today’s Agenda

Physics 111: Lecture 8 Today’s Agenda. Friction Recap Drag Forces Terminal speed Dynamics of many-body systems Atwood’s machine General case of two attached blocks on inclined planes Some interesting problems. Friction Review:.

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Physics 111: Lecture 8 Today’s Agenda

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  1. Physics 111: Lecture 8Today’s Agenda • Friction Recap • Drag Forces • Terminal speed • Dynamics of many-body systems • Atwood’s machine • General case of two attached blocks on inclined planes • Some interesting problems

  2. Friction Review: • Friction is caused by the “microscopic” interactions between the two surfaces: • See discussion in text

  3. Model for Friction • Arah vektor gaya gesek fFadalah tegak lurus dengan vektor gaya normalN, dan berlawanan arah dengan gaya yang bekerja pada sistem • Gesekan Kinetic (sliding): Besarnya vektor gaya gesek berbanding lurus dengan besarnya gaya normalN. fF= KN • Gesekan Static: Gaya gesekan menyeimbangkan total yang yang bekerja pada sistem sehingga sistem tidak bergerak. Besarnya gaya gesek statis maksimum berbanding lurus dengan gaya normalN. fFSN

  4. Kinetic Friction: • Kkoefisien gesek kinetik. i :F KN = ma j :N = mg so F Kmg = ma j N F i ma KN mg

  5. Static Friction: • Koefisien gesek statis S, menentukan gaya gesek statis maksimum, SN, yang timbul antara persentuhan dua buah benda. • Sdapat ditentukan dengan cara memperbesar Fsampai benda mulai bergeser: FMAX -SN = 0 N = mg FMAX = Smg SFMAX / mg j N FMAX i S N mg

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

  7. Lecture 8, Act 1Solution • Gambarkan diagram benda bebas dan tentukan gaya total dalam arah X FNET,X = mg sin q - mKmgcos q = ma = 0 (kasus balok pertama) mKN Jika massa digandakanmaka semua komponen menjadi dua kali Lebih besar. Tetapi percepatan tetap nol Speed will still be constant! j N q q mg i

  8. Drag Forces: • 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. FDRAG j v Fg = mg

  9. Drag Forces: Parachute • 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. FD = bvn j feels like n=2 v Fg = mg

  10. Terminal Speed: • Suppose FD = bv2. Sally jumps out of a plane and after falling for a while her downward speed is a constant v. • What is FDafter she reaches this terminal speed? • What is the terminal speed v? • FTOT = FD- mg = ma = 0. • FD= mg • Since FD = bv2 • bv2 = mg FD = bv2 j v Fg = mg

  11. Many-body Dynamics • Systems made up of more than one object • Objects are typically connected: • By ropes & pulleys today • By rods, springs, etc. later on

  12. Atwood’s Machine: Masses m1 and m2are attached to an ideal massless string and hung as shown around an ideal massless pulley. • Find the accelerations, a1and a2, of the masses. • What is the tension in the string T ? Fixed Pulley j T1 T2 m1 a1 m2 a2

  13. Atwood’s Machine... • Draw free body diagrams for each object • Applying Newton’s Second Law: ( j-components) • T1 - m1g = m1a1 • T2 - m2g = m2a2 • But T1 = T2 = T since pulley is ideal • and a1 = -a2= -a.since the masses are connected by the string Free Body Diagrams T1 T2 j a1 a2 m2g m1g

  14. Atwood’s Machine... T - m1g = -m1 a(a) T - m2g = m2 a(b) • Two equations & two unknowns • we can solve for both unknowns (T and a). • subtract (b) - (a): • g(m1- m2 ) = a(m1+m2 ) • a= • add (b) + (a): • 2T - g(m1+ m2 ) = -a(m1 -m2 ) = • T = 2gm1m2 / (m1+ m2 ) -

  15. j T T m1 a m2 a Atwood’s Machine... Atwood’s Machine • So we find:

  16. Is the result reasonable?Check limiting cases! • Special cases: i.) m1 = m2 = m a = 0 and T = mg. OK! ii.) m2 or m1 = 0 |a| = g and T= 0. OK! • Atwood’s machine can be used to determine g (by measuring the acceleration a for given masses). -

  17. Attached bodies on two inclined planes smooth peg m2 m1 1 2 All surfaces frictionless

  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: x y Taking “x” components: 1) T1 - m1g sin1= m1a1X 2) T2- m2g sin 2 = m2a2X ButT1 = T2 = T anda1X= -a2X= a (constraints) x y N T1 T2 m2 N m1 2 1 m2g m1g

  19. q - q m sin m sin 1 1 2 2 = a g + m m 1 2 Solving the equations Using the constraints, solve the equations. T - m1gsin 1 = -m1 a(a) T - m2gsin 2= m2 a (b) Subtracting (a) from (b) gives: m1gsin 1-m2gsin 2 = (m1+m2 )a So:

  20. m2 m1 1 2 q - q m sin m sin 1 1 2 2 = a g + m m 1 2 Special Case 1: Boring m2 m1 If1 = 0 and 2 = 0,a = 0.

  21. m2 m1 1 2 q - q m sin m sin 1 1 2 2 = a g + m m 1 2 Special Case 2: T Atwood’s Machine T m1 m2 If1 = 90 and 2 = 90,

  22. m2 m1 1 2 q - q m sin m sin 1 1 2 2 = a g + m m 1 2 Special Case 3: Air-track m1 Lab configuration m2 - If1 = 0 and 2 = 90,

  23. m a 10kg Lecture 8, Act 2Two-body dynamics • 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. m a F = 98.1 N Case (1) Case (2) (a)Case (1) (b)Case (2)(c)same

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

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

  26. Problem: Two strings & Two Masses onhorizontal frictionless floor: • Given T1, m1 and m2, what are a and T2? T1 - T2 = m1a (a) T2 = m2a (b) • Add (a) + (b): T1 = (m1 + m2)a a • Plugging solution into (b): a i m1 m2 T2 T1

  27. a T3 T2 T1 3m 2m m Lecture 8, Act 3Two-body dynamics • 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) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3

  28. T3 3m T3= 3ma T2- T3=2ma T2=2ma+T3 > T3 T3 T2 2m T1- T2=ma T1= ma +T2 > T2 T2 T1 m Lecture 8, Act 3Solution • Draw free body diagrams!! T1 > T2 > T3

  29. Consider T1 to be pulling all the boxes a a a T3 T3 T3 T2 T2 T2 T1 T1 T1 T2is pulling only the boxes of mass 3m and 2m 3m 3m 3m 2m 2m 2m m m m T3is pulling only the box of mass 3m Lecture 8, Act 3Solution • Alternative solution: T1 > T2 > T3

  30. v m1 R m2 Problem: Rotating puck & weight. • A mass m1 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 m2. • What is the tension (T) in the string? • What is the speed (v) of the sliding mass?

  31. v m1 R T m2 Problem: Rotating puck & weight... T • Draw FBD of hanging mass: • Since R is constant, a = 0. so T = m2g m2 m2g

  32. v m1 R T m2 Problem: Rotating puck & weight... T = m2g Puck N T = m2g • Draw FBD of sliding mass: m1 Use F = T = m1a wherea= v2 /R m1g m2g = m1v2 / R

  33. Recap of today’s lecture • Friction Recap. (Text: 5-1) • Drag Forces. (Text: 5-3) • Terminal speed. • Dynamics of many-body systems. (Text: 4-7) • Atwood’s machine. • General case of two attached blocks on inclined planes. • Some interesting special cases. • Look at Textbook problems Chapter 6: # 3, 7, 21, 75

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