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ME 221 Statics Lecture #34 Sections 8.1 – 8.2

ME 221 Statics Lecture #34 Sections 8.1 – 8.2. Homework #11. Chapter 6 problems: 2, 3, 6 & 7 – Method of Joints 32, 36, 47 & 53 – Method of Sections 68 & 75 Due Today. Homework #12. Chapter 8 problems: 1, 2, 4, 8 & 15 Due Monday, December 1. Quiz #7. Today Analysis of Structures

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ME 221 Statics Lecture #34 Sections 8.1 – 8.2

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  1. ME 221 StaticsLecture #34Sections 8.1 – 8.2 Lecture 34

  2. Homework #11 Chapter 6 problems: • 2, 3, 6 & 7 – Method of Joints • 32, 36, 47 & 53 – Method of Sections • 68 & 75 Due Today Lecture 34

  3. Homework #12 Chapter 8 problems: • 1, 2, 4, 8 & 15 Due Monday, December 1 Lecture 34

  4. Quiz #7 Today Analysis of Structures Method of Joints or Method of Sections Lecture 34

  5. No ClassWednesday, November 26 Lecture 34

  6. No ClassFriday, December 5Mechanical Engineering Design ConferenceMSU Union 9am to noon Demonstrations1pm to 2pm Awards in BallroomME371 Mechanical Design IME412 Heat Transfer LabME456 Mechatronics Systems LabME471 Mechanical Design IIME481 Mech Engineering Design Projects Lecture 34

  7. Chapter 8 - Friction • Overview • Types • Coulomb friction (dry friction) • Static and kinematic • Fluid friction • Wedges Lecture 34

  8. Friction Angle • A body on an incline, acted upon by gravity alone, will slip at an angle related to the coefficient of friction • From the FBD: • N – W cos α = 0 & • ƒ – W sin α = 0 • Such that: • ƒ = N tan α (friction force) W f N  Lecture 34

  9. Friction Angle • The Coefficient of Static Friction (ms) is defined as the tangent of the maximum angle a body may be inclined before slip occurs ms = tan max =ƒmax / N W • The angle max is called the friction angle f N  Lecture 34

  10. Coulomb Friction (dry friction) • Friction force (ƒ) is proportional to the normal force (N) and opposes motion • Coefficient of friction, m, is the proportionality constant f • Static coefficient, ms msN • Kinematic coefficient, mk mkN W f=P P f P N Lecture 34

  11. Common Static Friction Coefficients Steel on steel 0.75 Rubber on concrete 0.5-0.9 Rubber on ice 0.05-0.30 Metal on wood 0.2-0.6 Teflon on Teflon 0.04 Tires on gravel 0.5 Lecture 34

  12. Example Problem Lecture 34

  13. Quiz #7 Lecture 34

  14. Friction: Wedges • Wedges are a simple means for lifting • Again, friction force opposes the motion • Disassembled FBDs are essential for solving wedge problems • Apply equilibrium equations to disassembled FBDs • Examples: 8.71, 8.72, 8.80, 8.45 Lecture 34

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