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resolving forces

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mikamika:
Hello everyone!

For this question, I am particularly confused with the answers. Why is it for the horizontal force, the mrw^2 is also multiplied by sin(alpha) ? (this is from 2006HSC q5)

Thanks in advanced   :D

RuiAce:
Note that they used \( \ell\), not \(r\).

Simple trigonometry shows that \(\sin \alpha = \frac{r}{\ell} \), and thus \( r = \ell \sin \alpha\).

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