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Final Exam Review

Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!. Final Exam Wed. Wed. May 14 8 – 10 a.m. Review. Always work from first Principles!. Review. Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints. Review.

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Final Exam Review

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  1. Final Exam Review

  2. Please Return Loan Clickers to the MEG office after Class!Today!

  3. Final ExamWed. Wed. May 14 8 – 10 a.m.

  4. Review Always work from first Principles!

  5. Review Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints

  6. Review 1. Free-Body

  7. B_x mg B_y 1. Free-Body

  8. B_x mg B_y 2. Newton Moments about B: -mg*L/2 = IB*a with IB = m*L1/3

  9. B_x mg B_y 3. Constraint aG = a*L/2 = -3g/(2L)* L/2 = -3g/4

  10. mg A_x A_y N 1. Free-Body

  11. mg 2. Newton A_x A_y N Moments about Center of Cylinder:A_x From triangle at left: Ax*(R-h) –b*mg = 0 acart*(R-h) –b*g = 0

  12. mg 2. Newton A_x A_y N N = 0 at impending rolling, thus Ay = mg Ax = m*acart

  13. Kinematics (P. 16-126) 4r -2r*i + 2r*j CTR

  14. Point Mass Dynamics X-Y Coordinates

  15. Normal and Tangential Coordinates Velocity Page 53

  16. Normal and Tangential Coordinates

  17. Polar coordinates

  18. Polar coordinates

  19. Polar coordinates

  20. 12.10 Relative (Constrained) Motion We Solve Graphically (Vector Addition) vA vB vB/A

  21. Example : Sailboat tacking against Northern Wind 2. Vector equation (1 scalar eqn. each in i- and j-direction) 500 150 i

  22. Constrained Motion vA is given as shown. Find vB Approach: Use rel. Velocity: vB = vA +vB/A (transl. + rot.)

  23. NEWTON'S LAW OF INERTIA A body, not acted on by any force, remains in uniform motion. NEWTON'S LAW OF MOTION Moving an object with twice the mass will require twice the force. Force is proportional to the mass of an object and to the acceleration (the change in velocity). F=ma.

  24. Rules 1. Free-Body Analysis, one for each mass 2. Constraint equation(s): Define connections. You should have as many equations as Unknowns. COUNT! 3. Algebra: Solve system of equations for all unknowns

  25. Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown. M*g*sinq*i -M*g*cosq*j M*g

  26. Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Step 2: Apply Newton’s Law in each Direction: M*g*sinq*i N -M*g*cosq*j M*g

  27. Friction F = mk*N: Another horizontal reaction is added in negative x-direction. M*g*sinq*i mk*N N -M*g*cosq*j M*g

  28. Energy Methods

  29. Only Force components in direction of motion do WORK

  30. The work is defined as The potential energy V is defined as:

  31. The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.

  32. Work of Gravity

  33. Conservative Forces: Gravity is a conservativeforce: Gravity near the Earth’s surface: A spring produces a conservative force:

  34. Rot. about Fixed Axis Memorize!

  35. Page 336: an = w x ( w x r) at = a x r

  36. fig_05_012 Mathcad EXAMPLE

  37. fig_05_012 Mathcad Example part 2: Solving the vector equations

  38. fig_05_012 Mathcad Examples part 3 Graphical Solution

  39. y x • vB = 3 ft/s down, Q = 60o • and vA = vB/tanQ. The relative velocity vA/B is found from the vector eq. • vA = vB+ vA/B ,vA/B points • vA = vB+ vA/B ,vA/B points • (C)vB = vA+ vA/B ,vA/B points • (D) VB = vB+ vA/B ,vA/B points vA vB vA vB vA/B

  40. The instantaneous center of Arm BD is located at Point: • B • D • F • G • H

  41. Rigid Body Acceleration Stresses and Flow Patterns in a Steam Turbine FEA Visualization (U of Stuttgart)

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