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Force vector in a new coordinate frame?

  • Thread starter theBEAST
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Homework Statement


http://img811.imageshack.us/img811/9092/captureykj.png [Broken]

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!
 
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Answers and Replies

  • #2
gabbagabbahey
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Homework Statement


http://img811.imageshack.us/img811/9092/captureykj.png [Broken]

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.
 
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