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Acceleration - slope watch

1. take a look at the result attached.
Surely mg/cos(90-x) = ma?

Attached Images

2. Is it ?

Doesn't cos act downward?
3. (Original post by Diiiii)
Is it ?

Doesn't cos act downward?
Cos(90-x) is mathematically the same as Sin
4. "mg" is the hypotenuse. You're trying to find the adjacent i.e. ma

cos(90°-θ) = ma / mg
ma = mg*cos(90°-θ)

Here's a picture to help:
5. (Original post by raiden95)
Cos(90-x) is mathematically the same as Sin
Thanks, I completely forgot that
6. (Original post by raiden95)
Cos(90-x) is mathematically the same as Sin
So even if I use , I'll still get the answer won't I, and will I still get the marks for using this instead?
7. (Original post by crc290)
"mg" is the hypotenuse. You're trying to find the adjacent i.e. ma

cos(90°-θ) = ma / mg
ma = mg*cos(90°-θ)
why is it the hyp?
8. (Original post by jsmith6131)
why is it the hyp?
See the picture I've just uploaded
9. (Original post by crc290)
See the picture I've just uploaded
how was I to know that?
10. (Original post by jsmith6131)
how was I to know that?
Well you had the right idea, you just got the adjacent and the hypotenuse mixed up

You said: "Surely mg/cos(90-x) = ma?"

That rearranges to cos(90°-θ) = mg/ma

11. (Original post by Diiiii)
So even if I use , I'll still get the answer won't I, and will I still get the marks for using this instead?
Yes, but personally i like to have the equations in terms of sinx and cosx, I feel like I'm less likely to make a mistake
12. (Original post by raiden95)
Yes, but personally i like to have the equations in terms of sinx and cosx, I feel like I'm less likely to make a mistake
Thanks I find it easier the other way
13. (Original post by jsmith6131)
how was I to know that?
It's not a case of knowing that, it's a case of drawing that triangle. You're resolving the force (weight) into two perpendicular directions (parallel to slope, and normal to slope) and treating those directions as the axes of your problem.

Though, if you want to sanity check it - Cos is always less than 1, so mg/cos(...) would be larger than mg, while mgcos(...). Do balls roll down slopes faster or slower than they fall off cliffs?

The two resolved components of a force add vectorially to give the original force, so they must both be smaller in magnitude than the original force. The original force is always the hypotenuse, the components are the opposite and adjacent.

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