Hey tomato89.

This is going to be a little messy if you did a real world example due to the distribution of mass of the see-saw and how this affects the rest of it, but if you treat it has some standard object, you consider the net force of that object moving and how to decompose it.

So you will have a gravitational force of the see-saw acting in the direction it acts (which is approximated to be negative y axis) and you decompose that force into the x and y components by doing a projection onto the <1,0> and <0,1> vectors. Same for friction (find out the friction magnitude in the direction of the incline opposing the object and decompose it into x and y parts).

Now you've got the forces provided by the thing the see-saw is sitting on which is that plane. You've got the gravitational forces of the see-saw that will give it a net force down the plane. If the frictional and object forces don't completely cancel this out, then the object will accelerate.

Now what can happen in these toy examples is that we only consider the net-force in the direction opposing the frictional force and disgregard the rest.

So if you have a gravitational force acting in the negative y direction with some magnitude, you need to find the component acting in the direction that is parallel to the inclined plane as it heads down.

When you add the two opposing forces together (they will be two vectors but they will point in the opposite directions) you will get a net force. If that force is non-zero, the object will accelerate.

So if you have a gravitational force down, it has a magnitude of mg where g = 9.8 m/s^2. and m is 50. So now you have to find the component of that force in the direction going down the plane. This is a projection.

You said the plane is inclined at 30 degrees to the horizontal which means that you have to project your vector that is <0,-50*9.8> onto the vector that is counterclockwise of 30+180 degrees from the x-axis. This vector is basically <cos(210),sin(210)> which has a length 1.

If you find the length of the gravitational component in that direction, you get the magnitude to be 0*cos(210) + -50*9.8*sin(210) = -50*9.8*-sin(30) = 50*9.8*sin(30) which is usually a trick you use in these inclined problems.

That's the real derivation of how you get this result: it is simply a projection operation of projecting one vector onto another.

So now you need to get the magnitude of the friction but you make this negative since it is going in the opposite direction (but it will always be between 0 and the magnitude you just calculated above).

If it is non-zero you now have to project this vector back onto the x and y vectors. The x vector is <1,0> and the y vector is <0,1>. The original vector though is <cos(210),sin(210)> but it has a magnitude of whatever you got (call it m) so your new forces in x and y co-ordinates will be m*cos(210) Newtons in the x direction and m*sin(210) in the y direction, and since cos(210) = -cos(30) and sin(210) = -sin(30) you have the x component being -m*sin(30) for x and -m*cos(30) for y which is intuitive since it means it is accelerating both left and downwards which is what you would expect for an inclined plane heading in that direction.