Moment about a point in an incline plane

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SUMMARY

The discussion centers on calculating the moment produced by a force of 460 N acting perpendicular to an inclined plane about point A. The participants provided vectors for the position and unit vectors, ultimately computing the moment as r(CA) x F, resulting in the vector 1382i - 1767j - 2356k. However, there is confusion regarding the correctness of the calculations, as the online tool masteringengineering.com indicates the answer is incorrect. Clarification is sought on whether the magnitude of the force is indeed 460 N.

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  • Understanding of vector cross products in physics
  • Knowledge of force and moment calculations
  • Familiarity with inclined plane mechanics
  • Proficiency in using vector notation and components
NEXT STEPS
  • Review vector cross product calculations in physics
  • Study the principles of moments in mechanics
  • Explore inclined plane dynamics and forces
  • Practice problems involving force vectors and moments
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Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators seeking to clarify concepts related to moments and forces on inclined planes.

Robb
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Homework Statement



Probs._448_49.jpg


Force F = 460 N acts perpendicular to the inclined plane

Determine the moment produced by F about point A.
Enter the x, y, and z components of the moment separated by commas.

Homework Equations

The Attempt at a Solution



r(CA)=4j-3k
u(CA)=4/5j-3/5k

r(CB)=-3i+4j
u(CB)=-3/5i+4/5j

u(CA) x u(CB)=1.28i+.36j+.48k
F=460(1.28i+.36j+.48k)=589i+166j+221k

r(CA) x F=1382i-1767j-2356k

masteringengineering.com says the answer is wrong. Please help!
 
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Robb said:
F=460(1.28i+.36j+.48k)=589i+166j+221k
Is the magnitude of this force 460 N?
 

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