Forces on a simple flat slope 
This diagram is of a block (b) moving down a ramp unaided.
Taking gravity (g) as $9.8ms^{-2}$, and the mass of the block (b) as $0.2kg$, how can I find the force acting along the slope (f).
BTW: the ramp is inclined 1.8 degrees, not radians.
I thought I could use basic trigonometry, and $F=ma$:
$$ Sin(1.8) = \frac{0.2*9.8}{f} $$
$$ f = \frac{0.2*9.8}{sin(1.8)} $$
However this results in: $f = 62.40$
Seeing as force is a vector I thought I could use trig to split into it's components but obviously I'm doing something wrong.
(the answer should be ~0.06N)
Sorry for the noob question, but it's got me clueless.
 A: Here's something that might be worth remembering: when you split a vector into components (as you were trying to do), the original vector is always the hypotenuse. See if you can use that fact to find the error you made in writing the formula 
$$Sin(1.8) = \frac{0.2*9.8}{f}$$
A: When in doubt, always draw a force diagram.  Remember, $g$ will always point straight down, $f$ will always run parallel to the plan upon which you're moving, and your third vector will always be perpendicular to the plane.  With that in mind, your problem will have the following force diagram:

In this case, $a$ is the force component of gravity perpendicular to the plane.  If you were working with friction, it would be important ... but you're not, so feel free to ignore it.  The angle directly opposite $f$ is $1.8^{\circ}$.
You can use the law of sines (basic trig) to calculate f at this point.
$$\frac{f}{sin 1.8} = \frac{g}{sin 90}$$
Or ...
$$f = g * sin 1.8$$
I'll let you plug in the numbers (gravity and the mass of the block) to solve for $f$.
