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# resolving components watch

1. Can someone point to me which of these two statements I've written is correct and the reasoning why. I thought that R =mgcos(#) would be correct but it does not work when using centripetal motion, does anyone know why?

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2. (Original post by BDunlop)
Can someone point to me which of these two statements I've written is correct and the reasoning why. I thought that R =mgcos(#) would be correct but it does not work when using centripetal motion, does anyone know why?

Attachment 699358
Attachment isn't showing up for me.
3. (Original post by Pangol)
Attachment isn't showing up for me.
now?
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4. (Original post by BDunlop)
now?
Attachment is sideways, so not that easy to read, but...

R = mg cos(theta) is correct. The entire weight is acting downwards, and resolving perpendicular to the slope, you get mg cos(theta) as the perpendicular component, which is therefore R (assuming there is no acceleration in this direction).

As to why it doesn't work for your circular motion question - you'll have to show us the question...
5. (Original post by BDunlop)
Can someone point to me which of these two statements I've written is correct and the reasoning why. I thought that R =mgcos(#) would be correct but it does not work when using centripetal motion, does anyone know why?

Attachment 699358
Did you do anything differently when posting the attachment the second time compared to the first?

I can view the attachment in your first post when looking in the app but not in a web browser. This has been a problem for a while and I'm just wondering if there's something specific that causes the bug.
6. (Original post by Pangol)
Attachment is sideways, so not that easy to read, but...

R = mg cos(theta) is correct. The entire weight is acting downwards, and resolving perpendicular to the slope, you get mg cos(theta) as the perpendicular component, which is therefore R (assuming there is no acceleration in this direction).

As to why it doesn't work for your circular motion question - you'll have to show us the question...
It is accelerating towards the right, which has the component RSin(theta).
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7. If you use mgcos(theta) you end out having to shift sine number > 1, but using rcos(theta) will leave you with tan so you can inverse that.

I'm wondering if there's some sort of rule or reason why mgcos(theta) shouldn't be used
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