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

mg/ma is hypotenuse/adjacent, not adjacent/hypotenuse

Reply 10

.raiden.

Original post by Diiiii

So even if I use

cos(90-x)

, I'll still get the answer

g sin x = a

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

Reply 11

Diiiii

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 :smile:

Reply 12

The Mr Z

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.