Forces on an oscillating skateboard A half pipe of a skateboard park consists of a concrete trough with a semicircular section of radius 5m, I hold a frictionless skateboard on the side of the trough, pointing down toward the bottom and release it; how long will it take to come to back to the point of release?
I got this from an example in the book Classical Mechanics by Taylor and I'm having trouble trying to understand the expression for the radial force. I've resolved $mg$ into $mg \cos \varphi$ and $mg \sin\varphi$ as seen below (sorry for the poor picture).

The book says $F_{r} = mg \cos \varphi - N$ and $F_{\varphi} = -mg \sin\varphi$.
My questions are:

*

*Why does $ F_{r} = mg \cos \varphi - N$ and not $N - mg \cos \varphi$?

*Why does $F_{\varphi}$  have a negative sign in front?

I believe all this depends on where we choose to place our reference frame but I'm not sure where that is. If I was to take an educated guess, it would be the skateboard itself. If that is the case, then I have my answer to question 2.
Could someone please explain this please and point out if I am incorrect? I'm trying to understand some of this stuff before I go into my mechanics module next semester which is why I'm slightly struggling.
 A: When you do this sort of analysis it's important you choose a clear sign convention then adhere to it strictly.
In this case the radial coordinate $r$ is the distance from the centre of the circle. $r=0$ at the centre and it increases as we move outwards away from the centre, so for the radial component outwards away from the centre is positive and inwards towards the centre is negative. If you look at your diagram you'll see that for $-\pi/2 < \varphi < \pi/2$ the gravitational force on the skateboard points outwards away from the centre so it is positive i.e. $+mg\cos\varphi$. Likewise the normal force on the skateboard points inwards towards the centre so it is negative i.e. $-N$.
If you manage to skate so fast you rise higher than the centre of the circle the gravitational force now points inwards and becomes negative, but you don't have to worry about this as the sign of $\cos\varphi$ becomes negative for $\varphi < -\pi/2$ and $\varphi > \pi/2$ so your equation automatically takes care of this for you.
For the angular coordinate the convention is that anticlockwise is positive (why this should be is lost in the mists of time). You are taking $\varphi=0$ for the vertically downward direction, and moving anticlockwise is the positive direction, so any tangential force is positive if it points anticlockwise and negative if it points clockwise. When you are to the right of the centre, i.e. $0 < \varphi < \pi/2$, the tangential force points left and that is the clockwise direction. Hence the tangential force is $-mg\sin\varphi$.
When you move the the left of the bottommost point the tangential force points right/anticlockwise so it is positive, but again this is taken care of automatically because the sign of $\sin\varphi$ changes to negative for $-\pi/2 < \varphi < 0$.
With experience most of us become casual about signs because our intuition tells us what is correct, but when you are first learning mechanics it makes life a lot easier if you are careful to define your sign convention and stick to it.
A: 
Put coordinate system at the center of the half circle
the position vector to the mass is
$$\vec R=r\,\left[ \begin {array}{c} \sin \left( \varphi  \right) 
\\ -\cos \left( \varphi  \right) \end {array}
 \right] 
$$
thus:
$$\vec e_r=\left[ \begin {array}{c} \sin \left( \varphi  \right) 
\\  -\cos \left( \varphi  \right) \end {array}
 \right] \\
\vec e_\varphi= \left[ \begin {array}{c} \cos \left( \varphi  \right) 
\\ \sin \left( \varphi  \right) \end {array}
 \right]\\
\vec e_r\cdot \vec e_\varphi=0\quad,\vec e_r\cdot \vec e_r=1\quad,\vec e_\varphi\cdot \vec e_\varphi=1
$$
you want to "write" the the force like this
$$ \vec F=F_r\,\vec e_r+F_\varphi\,\vec e_\varphi$$
with:
$$\vec F=N\,\begin{bmatrix}
   -\sin(\varphi) \\
  \cos(\varphi) \\
 \end{bmatrix}+
m\,g\begin{bmatrix}
   0 \\
  -1 \\
 \end{bmatrix}$$
from here
$$F_r=\vec F\cdot \vec e_r=m\,g\cos(\varphi)-N\\
F_\varphi=\vec F\cdot \vec e_\varphi=-m\,g\,\sin(\varphi)$$
