Engineering Mechanics: Force couple

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SUMMARY

The discussion focuses on replacing a 150-N force with an equivalent force-couple system at point A in engineering mechanics. The user initially calculated the force components as -150cos35j + 150sin35k and determined the position vector D as (-0.1m)i + (0.12m)j + (-0.2m)k. The moment was computed using the cross product, resulting in (14.25N m)i + (-8.6N m)j + (-12.29N m)k. The user received feedback to verify the orientation of the force components and coordinate axes, leading to a realization of an error in their calculations.

PREREQUISITES
  • Understanding of force and moment concepts in engineering mechanics
  • Proficiency in vector mathematics, specifically cross products
  • Familiarity with coordinate systems and their application in mechanics
  • Knowledge of force-couple systems and their equivalence in static equilibrium
NEXT STEPS
  • Study the principles of static equilibrium in engineering mechanics
  • Learn about the application of cross products in calculating moments
  • Explore the concept of force-couple systems and their significance in mechanics
  • Review coordinate systems and their impact on vector component orientation
USEFUL FOR

Students and professionals in engineering mechanics, particularly those studying static equilibrium and force systems, will benefit from this discussion.

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Homework Statement


Problem is:
Replace the 150-N force with an equivalent force-couple system at A
Since it is really hard to describe the scenario in words, attached is a link that shows the diagram of the question.
http://www.chegg.com/homework-help/questions-and-answers/replace-150-n-force-equivalent-force-couple-system--q3044348

Homework Equations


Cross Product and sum of forces

The Attempt at a Solution


So I have the feeling I screwed up somewhere. What I did was I first converted the force into its components, producing me with -150cos35j+150sin35k. Since one of the things that the question asked was for I just simply took -150cos35 and 150sin35. I then decided to find the D which yields me (-0.1m)i+(0.12m)j+(-0.2m)k. Since the moment is equal to FxD, I took the cross product of these two values to produce me a value of (14.25N m)i+(-8.6N m)j +(-12.29N m)k. I have the strange feeling it is wrong and I screwed up somewhere. Can anyone help me with the problem?
 
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Check the direction of the components of the force. Look at the coordinate axes to make sure the components are oriented in the correct sense.
 
Ah I see now! Thanks!
 

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