resolving forces

[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   

[RuiAce]

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

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