Should deformation be considered when considering motion of particle along a curve? 
So, in the image I've shown above the external force on the particle moving along in a fixed curve in the direction perpendicular to the curve is given as :
$$ F   = -mg \sin \theta + N$$
At first I naively thought the force should be zero, because if it is non zero then that'd mean the particle is deforming the curve. However, the expression for the net force in the direction normal to curve is given as: $ \frac{mv^2}{r}$ where $r$ is the radius of the osculating circle. So the net force is definitely not zero; which means that the curve should deform (third law).
But if the curve deforms that contradicts the premise of the situation. What exactly am I doing wrong?
On further analysis, it was found that a rigid wire would support the forces completely which in turn could be supported by whatever the rigid wire is attached to. In the answers, I seek discussion on non-rigid wires.

Reference:
Page-19,20 of Fundamental Laws of Mechanics by I.e irodov
I found the book here
 A: If you parametrized your path with the line element s you obtain the position vector for a arbitrary path $$\vec R=X(s)\,\vec {\hat{e}}_x+Y(s)\,\vec{\hat{e}}_y$$
with $$\left( {\frac {d}{ds}}X \left( s \right)  \right) ^{2}+ \left( {
\frac {d}{ds}}Y \left( s \right)  \right) ^{2}
=1$$
the magnitude of the normal force (the contact force between the mass and the path $\vec N=N\,\vec{\hat{n}}~$)  is:
$$N=m \underbrace{\left(  \left( {\frac {d^{2}}{d{s}^{2}}}Y \left( s \right)  \right) 
{\frac {d}{ds}}X \left( s \right) - \left( {\frac {d^{2}}{d{s}^{2}}}X
 \left( s \right)  \right) {\frac {d}{ds}}Y \left( s \right)  \right) }_{1/r}
{{\dot{s}}}^{2}+mg{\frac {d}{ds}}X \left( s \right) 
$$
with  $\dot{s}=v~$ and $~r$ the radius of curvature
Example: circle path
$$X(s)=\rho\,\cos(\frac s\rho)$$
$$Y(s)=\rho\,\sin(\frac s\rho)$$
you obtain for N:
$$N={\frac {m{{\dot s}}^{2}}{\rho}}-mg\sin \left( {\frac {s}{\rho}}
 \right) 
$$
A: Your equation is correct. The net force perpendicular to the surface and acting on the mass is : N - mg sin(θ) which equals ma = m $v^2$/R. There is no need for deformation. The normal force from the surface will adjust itself to produce the needed centripetal acceleration (perpendicular to the surface).  There is another component of gravity acting parallel to the surface, mg cos(θ), which causes a tangential acceleration, dv/dt.
