Why normal force is greater than weight? The explanation of the banking angle of the road is said to be that one part of the normal force exerted by the road on a moving object neutralizes the object's weight and another part is providing the centripetal force necessary to turn the object in a circular path.
So the normal force is greater than the weight of the object. What is the source of this force? Why is this force greater than the weight?
 A: If the car is on the flat the normal force and the weight of the car are equal in magnitude and opposite in direction.
When going around a corner more force is needed so that there is a contribution from the force to provide the centripetal acceleration.
More force manifests itself as an increase in the normal force so that the vertical component balances the weight (as on the flat) and the horizontal component produces the centripetal acceleration.
A: If the bank is on a bend that is an arc of a circle then there must be a centripetal force directed horizontally towards the centre of circle to make the car follow the bend instead of travelling in a straight line. The car's weight cannot provide this centripetal force since the weight acts vertically and has no horizontal component.
If there is no friction between the car and the roadway then the only other force acting on the car is the normal force from the roadway. So the horizontal component of the normal force must equal the centripetal force required to make the car follow the bend. This is what determines the size of the normal force. So the size of the normal force depends on the angle of the bank, as well as the car's mass and speed and the radius of the bend. The relevant equation is
$\displaystyle N_{h} = \frac {mv^2} r$
where $N_h$ is the horizontal component of the normal force, $m$ is the car's mass, $v$ is the car's speed and $r$ is the radius of the bend. If the bank is at an angle $\theta$ to the horizontal then we know that $N_h = N \sin \theta$ and so
$\displaystyle N = \frac {mv^2} {r \sin \theta}$
The vertical component of the normal force may be less than or greater than the car's weight, depending on the angle of the bank. If the vertical component of the normal force is less than the car's weight then the car will slide down the bank. If the vertical component of the normal force is greater than the car's weight then the car will slide up the bank. If the angle of the bank (together with the car's speed and the radius of the bend) is such that the vertical component of the normal force is exactly equal to the car's weight, then the car will stay at the same height on the bend.
The vertical component of the normal force is
$\displaystyle N_v = N \cos \theta = \frac {mv^2} {r \tan \theta}$
In a typical question you are given the car's speed and the radius of the bend, and asked to find the bank angle for which the vertical component of the normal force equals the car's weight. To find the correct angle you must solve
$\displaystyle N_v = mg
\\ \displaystyle \Rightarrow mg = \frac {mv^2} {r \tan \theta}
\\ \displaystyle \Rightarrow \tan \theta = \frac {v^2} {gr}$
Note that $m$ cancels out in the final equation, so the bank angle for which the car stays at the same height does not depend on the car's mass.
A: Wikipedia

from the free body diagram you obtain two equations
$$N\cos(\theta)=m\,g\tag 1$$
$$N\sin(\theta)=\frac{m\,v^2}{r}\tag 2$$
those equations described a steady  state circular motion (velocity v is constant) where  the road has a
banking angle $~\theta~$
according to equation (1) the source of the normal force $~N~$ is the weight.
and with $~N=\frac{m\,g}{\cos(\theta)}~$ , N is grater then $~m\,g~$ because the banking angle $~\theta~$
