Direction of the normal force when acting on a tilted object

[Like Tony Stark]

Homework Statement

See below

Relevant Equations

$\Sigma F=0$
$\Sigma M=0$

My question is: given a rigid body which interacts with a surface, what's the direction of the normal force? Because, as the word says, it has to be normal to the surface. But when treating problems of a vertical rod which is slightly pushed and forms an angle $\theta$ with the surface, some people draw it as parallel to the rod, not normal to the surface.
Also, in this example, why does friction act on the body? I mean, there's no force that has to be compensated (in the x axis), normal force and weight are the only forces acting on it (in the y direction).
This is not like the case of the leaning ladder, which has a horizontal normal force and so the friction must act for the body to be static. If there was no friction, would the ladder be "pushed" by the wall?

 

[robphy]

I think of it this way.
Assume the reaction force is approximately uniform over a small region
$\begin{eqnarray*} \vec R &= \vec R_{\|} + \vec R_{\bot} \\ &= \vec f + \vec N\\ \end{eqnarray*}$
The "reaction force" [by the surface on the object] has two components,
one parallel to the surface ("the friction force")
and
one perpendicular to the surface ("the normal force").

Furthermore,
the "reaction force" [on the object] is whatever it needs to be [together with the other forces [and torques] on the object] to satisfy Newton's Law of Motion.

 

[Lnewqban]

Could you draw the vertical rod and the forces acting on it?
Sorry, I can't see it clearly based on your description.

 

[kuruman]

This is not like the case of the leaning ladder, which has a horizontal normal force and so the friction must act for the body to be static. If there was no friction, would the ladder be "pushed" by the wall?

The wall always pushes on the ladder in a direction perpendicular to the wall and away from it. Walls don't pull on ladders. This happens whether there is friction or not. Imagine what would happen to the ladder if the wall were suddenly snatched away.

 

[Like Tony Stark]

Could you draw the vertical rod and the forces acting on it?
Sorry, I can't see it clearly based on your description.

The picture shows the two cases of the normal force. I would say the the first one is the correct one, since the Normal force is normal to the surface. But the second one makes me think that, as the normal force has a component on x, this would make the friction force come into play.

 

[jbriggs444]

The wall always pushes on the ladder in a direction perpendicular to the wall and away from it. Walls don't pull on ladders.

If one imagines a ladder fairly near vertical leaning against a vertical wall...

And if one further imagines that the top of this ladder has nice fresh sticky rubber compound on it with a reasonably high coefficient of friction...

And if one proceeds to examine the bottom of this ladder to find that it rides on an ice skating rink and that the risers are sitting in a pair of ice skates pointed away from the wall...

Then the frictional force of wall on ladder can have the effect of putting the top of the ladder in tension. So walls can indeed pull on ladders.

 

[haruspex]

what's the direction of the normal force?

It is not always clear what surface it is normal to. The principle is that the normal force is the force of minimum magnitude that prevents interpenetration of the bodies.
See section 1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/.

Likewise, the static frictional force is the force of minimum magnitude which prevents relative tangential motion of the surfaces in contact,

 

[Lnewqban]

The picture shows the two cases of the normal force. I would say the the first one is the correct one, since the Normal force is normal to the surface. But the second one makes me think that, as the normal force has a component on x, this would make the friction force come into play.

Thank for the schematic.
That body is not in equilibrium, there is a moment created by the normal force and the weight.

The resistive force of friction can't exist without the tangential component of the reactive force of contact between both surfaces.
One is the cause, the other is the effect.

Simplifying a very complex phenomena to the extreme, I would suggest trying to understand the basic principle behind normal force related to static friction in the following crude and very far from academic way:

In orther to start relative movement betwen the surfaces, the tangential force must do the required work to "lift" the body from the surface.
The magnitude of the required force to do that work is what we call static friction.
The direction of that minimum required force is more or less perpendicular to the plane of contact of the two surfaces.

Static friction happens only because there is a "locking" mechanism between the body and the surface, and because there is a "clamping" (normal) force that keeps them "locked".

 

[haruspex]

The picture shows the two cases of the normal force. I would say the the first one is the correct one, since the Normal force is normal to the surface.

Quite so, assuming it is a flat level floor. Any force parallel to the floor would come from friction.

The second picture is wrong even if we change the force designated N to mean the total reaction from the floor. It will not line up with the ladder.
If we count just three forces, gravity, reaction from floor, reaction from wall, then torque balance requires that the three lines of action pass through a single point. For the diagram to be correct, the reaction from the wall must also be in the line of the ladder. But the horizontal components of those reactions must be equal and opposite, making their vertical components also equal and opposite. So now we cannot balance the vertical forces.

 

[Lnewqban]

I have just found this nice summary with easy to understand illustrations for you:

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/6-2-friction/