How to Find the Projection of a Force Along the Y-Axis in 3D?

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SUMMARY

The discussion focuses on calculating the projection of a force vector along the Y-axis in a three-dimensional space. The force vector is defined as F = 600N with angles α = 120°, β = 60°, and ω = 45°. The correct force vector is derived as F = -300i + 300j + 424.2k. The projection along the Y-axis is accurately determined using the dot product with the unit vector of the Y-axis, yielding a result of 300N.

PREREQUISITES
  • Understanding of vector representation in three dimensions
  • Knowledge of trigonometric functions for calculating vector components
  • Familiarity with dot product operations
  • Concept of unit vectors in vector analysis
NEXT STEPS
  • Study vector decomposition techniques in three-dimensional space
  • Learn about the application of dot products in physics
  • Explore the concept of projections in vector calculus
  • Review trigonometric identities and their use in force analysis
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Students in physics or engineering courses, particularly those focusing on mechanics and vector analysis, will benefit from this discussion.

pearss
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Homework Statement


I have a force in three dimensions such that:
F=600N
Alpha(angle with x axis) = 120
Beta(angle with y axis) = 60
Omega(angle with z axis) = 45

I'm to find the projection of F along the y axis.

The Attempt at a Solution



I found the force vector for this force by

F=F(cos120i+cos60j+cos45k)
F=600N(-.5i+.5j+.707k)
F= -300i+300j+424.2k

now i know to use a dot product between the force vector and the unit vector of the vector that I'm trying to find a projection onto...but since the vector is the y-axis I'm unsure about what to do from here.

I thought that u_OY = oy/magnitude oy

0i+j+0k / 1 = j

and a dot product between this and my force vector would just give me the y component of that force vector...300.

but I'm pretty sure this is wrong. does anybody have any hints? Thanks much
 
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pearss: Your answer is correct. Nice work.
 
Thanks for the comfirmation nvn
 

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