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183_notes:examples:vectordecomposition [2014/07/10 19:02] – caballero | 183_notes:examples:vectordecomposition [2014/07/10 19:08] (current) – [Solution] caballero | ||
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Determining the [[183_notes: | Determining the [[183_notes: | ||
- | In the figure below, a position vector has been drawn. It has a magnitude of $5 m$ and makes an angle of 35$^{\circ}$ with the negative y-axis. Determine the components of this vector in the coordinate system that is drawn. | + | In the figure below, a position vector has been drawn. It has a magnitude of $5\:m$ and makes an angle of 35$^{\circ}$ with the negative y-axis. Determine the components of this vector in the coordinate system that is drawn. |
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You might want to immediately apply the formulae: $ r_x = |\vec{r}| \cos \theta$ and $ r_y = |\vec{r}| \sin \theta $. But you should first check that they apply, and if not, you need to re-derive the formulae. If we draw the triangle where the base and height are the $x$ and $y$ components respectively, | You might want to immediately apply the formulae: $ r_x = |\vec{r}| \cos \theta$ and $ r_y = |\vec{r}| \sin \theta $. But you should first check that they apply, and if not, you need to re-derive the formulae. If we draw the triangle where the base and height are the $x$ and $y$ components respectively, | ||
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+ | The angle that you know ($\theta = 35^{\circ}$) is the one that the vector makes with the negative y-axis. So the opposite side is the $x$-component. Hence the above formula do not apply and, | ||
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+ | $$ r_x = |\vec{r}| \sin(\theta) = (5\:m) \sin(35^{\circ}) = 2.87\:m$$ | ||
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+ | The $y$-component is the adjacent side and is negative. Hence, | ||
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+ | $$ r_y = -|\vec{r}| \cos(\theta) = -(5\:m) \cos(35^{\circ}) = -4.10\: | ||
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+ | where the minus sign was introduced because the measure of the angle was less than 90$^{\circ}$. |