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ME221 Statics LECTURE # 10 Sections 3.4 & 3.6

ME221 Statics LECTURE # 10 Sections 3.4 & 3.6. Homework #4. Chapter 3 problems: 1, 4, 8, 11, 17, 25, 26, 28, 35 & 40 To be solved using hand calculations May check work using MathCAD, etc. Due Friday, September 26. Quiz #3. Wednesday, September 24. Last Lecture. Cross Product.

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ME221 Statics LECTURE # 10 Sections 3.4 & 3.6

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  1. ME221 StaticsLECTURE # 10Sections 3.4 & 3.6 Lecture 10

  2. Homework #4 • Chapter 3 problems: • 1, 4, 8, 11, 17, 25, 26, 28, 35 & 40 • To be solved using hand calculations • May check work using MathCAD, etc. • Due Friday, September 26 Quiz #3 • Wednesday, September 24 Lecture 10

  3. Last Lecture • Cross Product • Rigid Bodies • Moment of a Force about a Point Today • Moment about an Axis • Moment of Couple • Equivalent Force Couple System Lecture 10

  4. F O rA/O A Vector Moment Definition The moment about point O of a force acting at point A is: MO = rA/O x F Compute the cross product with whichever method you prefer. Lecture 10

  5. 200 N 60 o O.4 0.2 60 o 0.285 x d MA =200N *0.247m= 49.4 Nm Example Method # 1 tan 60°=0.2m/x A x=0.115m sin 60°=d/0.285m d = 0.247 m Lecture 10

  6. 200 sin 60 200 N 60 o O.4 0.2 A + M Method # 2 200 cos60 =200N (sin 60)(0.4m)- 200N (cos 60)(0.2m) = 49.4 Nm Lecture 10

  7. 200 N 60 o O.4 0.2 r A F=200N cos 60 i + 200N sin 60 j r =0.4 i + 0.2 j ^ ^ ^ i j k MA= Method # 3 0.4 0.2 0 =200 (sin 60)(0.4) - 200 (cos 60)(0.2) 200cos60 200sin60 0 = 49.4 Nm Lecture 10

  8. 200 N 60 o O.4 0.2 A F=200N cos 60 i + 200N sin 60 j r =0.285 i MA= Method # 4 r =0.285 i i j k 0.285 0 0 = 49.4 Nm 200cos60 200sin60 0 Lecture 10

  9. y ^ |Mn| =MA·n ^ n ^ =n·(rB/A x F ) rAB=rB/A O x B A z Moment of a Force about an Axis F Same as the projection of MA along n nx ny nz rB/Ax r B/A y r B/Az |Mn|= F x F y F z Lecture 10

  10. ^ n y F A ^ Mn=|Mn|n rAB=rB/A B O Mp = MA - Mn x z ^ ^ =n x [(r B/Ax F) x n] Resolve the vector MA into two vectors one parallel and one perpendicular to n. MA Mp Mn Lecture 10

  11. B Let F1 = -F2 y A Mo=rA x F2+ rB x F1 =(rB - rA ) x F1 =rAB x F1= C O x z Moment of a Couple F1 d rAB=rB/A F2 rB rA The Moment of two equal and opposite forces is called a couple |C|=|F1| d Lecture 10

  12. Moment of a Couple (continued) • The two equal and opposite forces form a couple (no net force, pure moment) • The moment depends only on the relative positions of the two forces and not on their position with respect to the origin of coordinates Lecture 10

  13. Moment of a Couple (continued) • Since the moment is independent of the origin, it can be treated as a free vector, meaning that it is the same at any point in space • The two parallel forces define a plane, and the moment of the couple is perpendicular to that plane Lecture 10

  14. Example Lecture 10

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