Force vector in a new coordinate frame?

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SUMMARY

The discussion focuses on converting a force vector \(\mathbf{f}\) measured in the \(u, v\) coordinate basis, specifically \(\begin{bmatrix} 1 \\ -1\end{bmatrix}\text{N}\), into the \(x, y\) coordinate frame. The solution involves using trigonometric relationships to express the basis vectors \(\mathbf{e}_u\) and \(\mathbf{e}_v\) in terms of \(\mathbf{e}_x\) and \(\mathbf{e}_y\). A key hint provided is the formula \(\mathbf{A} \cdot \mathbf{B} = ||\mathbf{A}|| ||\mathbf{B}|| \cos \theta\), which is essential for understanding the angle between the vectors during the transformation.

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  • Understanding of vector representation in different coordinate systems
  • Knowledge of trigonometric functions and their application in vector analysis
  • Familiarity with the concept of basis vectors in physics
  • Basic proficiency in linear algebra
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  • Learn about basis vectors and their significance in physics
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Homework Statement


http://img811.imageshack.us/img811/9092/captureykj.png

The Attempt at a Solution


The answer is also in the image above. I have no clue how to start this question. Could anyone be so kind to give me a hint on how I should approach this question? Thanks!
 
Last edited by a moderator:
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theBEAST said:

Homework Statement


http://img811.imageshack.us/img811/9092/captureykj.png

The Attempt at a Solution


The answer is also in the image above. I have no clue how to start this question. Could anyone be so kind to give me a hint on how I should approach this question? Thanks!

The statement "[itex]\mathbf{f}[/itex] is measured as [itex]\begin{bmatrix} 1 \\ -1\end{bmatrix}\text{N}[/itex] in the [itex]u[/itex], [itex]v[/itex] coordinate basis" means that [itex]\mathbf{f} = (1 \text{N})\mathbf{e}_u +(-1\text{N}) \mathbf{e}_v[/itex]. So, if you can do a little trig to express [itex]\mathbf{e}_u[/itex] & [itex]\mathbf{e}_v[/itex] in terms of [itex]\mathbf{e}_x[/itex] & [itex]\mathbf{e}_y[/itex], you can express [itex]\mathbf{f}[/itex] in the [itex]x[/itex], [itex]y[/itex] coordinate basis.

Hint: [itex]\mathbf{A} \cdot \mathbf{B} = ||\mathbf{A}||||\mathbf{B}|| \cos \theta[/itex], where [itex]\theta[/itex] is the angle between the two vectors.
 
Last edited by a moderator:

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