Forces on a ladder leaning against a box My question is based on the following problem:

A ladder of length $l=3\mathrm{m}$ is leaning against a box with height $h=2\mathrm{m}$ so that the angle between the ladder and the ground is 60 degrees. The ladder is held in place by a wire streched from the bottom of the box to the bottom of the ladder. What is the force exerted on the ladder by the wire? Ignore all friction forces.

My initial thought relating to the forces, was that there has to be four forces acting on the ladder:


*

*$m\vec{g}$ (The force of gravity)

*$\vec{S}$ (The force from the wire)

*$\vec{N_1}$ (The force from the ground at the base of the ladder)

*$\vec{N_2}$ (The force from the edge of the box at which the ladder is leaning)


I then wanted to use $\sum{\vec{F}}=0$ and $\sum{\vec{\tau}} = 0$ to find $\vec{S}$, but I discovered that I needed to know the direction of $\vec{N_2}$ for this method to work. When the ladder is leaning against a wall, I know that $\vec{N_2}$ is perpendicular to the wall, but in this case it was not clear to me what the direction of $\vec{N_2}$ should be. After some trial and error, I assumed that $\vec{N_2}$ would be perpendicular to the ladder, which gave me the correct value for $\vec{S}$. But this left me wondering: Is it always the case that $\vec{N_2}$ is perpendicular to the ladder, regardless of the angle between the ladder and the ground? Furthermore, is yes: Why is that? And if no: What would be the correct, non-trial&error approach to this problem?
 A: The question states that all friction forces are to be ignored. This implies that all contact forces are normal to the common surface of contact. Any component of force parallel to the contact surface would not be resisted and would result in motion. The question states that the ladder is held in place, which implies that all such motion has ceased and the parallel forces are now zero. 
It is not clear what direction the normal to the corner of the box is pointing in, because angular corners are discontinuous - the normal changes abruptly from horizontal to vertical. However, the normal to the ladder is the same at all points along its length, so we must assume that this is also the normal to the common surface.
More realistically we could imagine that the corner is rounded with a very small radius of curvature. The contact surface will be tangent to this arc and parallel to the ladder.
In reality an angular corner will exert a very high pressure on the wood of the ladder, biting into it so that any component of a force parallel to the ladder will be resisted by a friction force. This possibility has been excluded here by using an ideal model in which there is no friction - ie no resistance to forces which are parallel to the common surface. 
In some ideal models the direction of the surface of contact cannot be determined. For example, the collision between 2 point masses in 2D or 3D. The indeterminacy can be resolved by making the points into circles or spheres with a given radius, so that the collision forces can be resolved along a common normal at the point of contact.
A: Yes, the force of the box on the ladder will always be perpendicular to the ladder, because it is the Third Law complement of the force of the ladder on the box.  And in the absence of friction, a surface (the ladder's surface in this case) can exert a force only in the direction perpendicular to itself. 
