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Forces and Moments MET 2214

Forces and Moments MET 2214. Moments and Forces. Part 2. Calculating the moment using rectangular components. The moment of a force F about the axis passing through point O and perpendicular to the plane containing O and F can be expressed using the cross product: M o = r x F

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Forces and Moments MET 2214

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  1. Forces and Moments MET 2214 Statics (MET 2214) Prof. S. Nasseri

  2. Moments and Forces Part 2 Statics (MET 2214) Prof. S. Nasseri

  3. Calculating the moment using rectangular components • The moment of a force F about the axis passing through point O and perpendicular to the plane containing O and F can be expressed using the cross product: • Mo = r x F • The magnitude of the moment is the area shown below: Statics (MET 2214) Prof. S. Nasseri

  4. Calculating the moment using rectangular components • Mo = | r x F | = r F sinθ • r = x i + y j + z k • F = Fx i + Fy j + Fz k Statics (MET 2214) Prof. S. Nasseri

  5. Resultant Moment Statics (MET 2214) Prof. S. Nasseri

  6. Moments and Forces Part 3 Statics (MET 2214) Prof. S. Nasseri

  7. Moment about an axis • Sometimes the moment about a pointis known and you are supposed to calculate its component about an axis. To find the moment, consider the dot product of Mo and unit vector along axis a: O: any point on a-a Statics (MET 2214) Prof. S. Nasseri

  8. Moment about an axis You can also find the tangent force Fθ and then r x Fθ is the moment about aa: Statics (MET 2214) Prof. S. Nasseri

  9. Example 1 • Force F causes a moment MO about point O. What is the component of MO along axis oy (My)? Statics (MET 2214) Prof. S. Nasseri

  10. Scalar Analysis • There are 2 methods to find My • Scalar Analysis • Vector Analysis • (1) Scalar Analysis (first way): • MO = (20)(0.5) = 10 N.m • Imagine that we have found the direction of this moment (shown in figure) • MO tends to turn the pipe around axis ob. The component of MO along the y-axis, My, tends to unscrew the pipe from the flange at O. • Thus it is important to know its value. • My = (3/5)(10) = 6 N.m Statics (MET 2214) Prof. S. Nasseri

  11. Scalar Analysis • Scalar Analysis (second way): • To find My directly (not form MO) it is necessary to determine the moment-arm, knowing that the distance from F to the y-axis is 0.3m: • My = (20)(0.3) = 6 N.m • In general, If the line of action of a force F is perpendicular to any specific axis aa thus: • Ma = F .da Statics (MET 2214) Prof. S. Nasseri

  12. Vector Analysis • First, use the cross product formula to calculate the moment about O: • MO = rA × F • MO = (0.3i + 0.4j) × (-20k) • MO = {-8i + 6j} N.m • Then use the dot product of MO and the unit vector along y-axis to get My: • My = MO . ua • My = (-8i + 6j). (j) • My = 6 N.m Statics (MET 2214) Prof. S. Nasseri

  13. My = (rA × F) . ua OrMy = ua. (rA × F) Which is called triple scalar product Vector Analysis • We can always combine the two previous equations that you saw into one: • MO = rA × F yielding a scalar • My = MO . Ua Statics (MET 2214) Prof. S. Nasseri

  14. Magnitude of moment M about axis aa’ Vector of moment M about axis aa’ Remember the triple scalar product from Math section?!! Statics (MET 2214) Prof. S. Nasseri

  15. Example 2 Statics (MET 2214) Prof. S. Nasseri

  16. Example 2 Statics (MET 2214) Prof. S. Nasseri

  17. Example 2 Statics (MET 2214) Prof. S. Nasseri

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