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## Homework Statement

A cart of mass 357g is sent travelling down a frictionless sloped track.

It's acceleration is 1.65ms

^{-2}.

1. Draw a labelled free body diagram of the cart while it is accelerating down the track.

2. Using Newton's second law, calculate the slope of the track.

## Homework Equations

SOH CAH TOA

F=ma

## The Attempt at a Solution

I've got the model answers and I got the correct free body diagram (see attached).

What I don't understand is that when they change the x,y axis, where does θ end up?

Same diagram as attached, but I tried calculating like this without changing the axis...

F due to gravity = mg

F = 3.57kg x 9.8

F = 34.986N

Which would be the vertical component.

Hypotenuse would just be the acceleration at and angle down the slope.

F=ma

F = 3.57kg x 1.65

F = 5.8905N

Then by looking at SOH CAH TOA (I do need these to see why sin or cos is used!), we have the components opposite to the angle and the hypotenuse. Therefore we use sin.

sinθ = [itex]\frac{Opposite}{Hypotenuse}[/itex]

θ = sin

^{-1}[itex]\frac{Opposite}{Hypotenuse}[/itex]

θ = sin

^{-1}[itex]\frac{34.986}{5.8905}[/itex]

θ = MATH ERROR [itex]\leftarrow[/itex] That's what appears on my calculator when I plug it all in.

Can someone please tell me what I'm doing wrong? Can I calculate the answer without altering the axis? And if altering the axis would make it easier, where does θ end up?

I don't understand how the model answer has been worked.