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I was watching a lecture discussing a mechanical problem. I understand the proposed solution so my question is in the setup. This is the problem:

A ladder with given length and mass is leaning on a wall without friction. The other end of the ladder is on the ground where there is friction with known coefficient. For what angles of tilt the ladder will be stable?

A better description of the problem can be found here

In the setup, the wall exerts force on the ladder. While this interpretation leads to solution that agrees with reality, I don't understand why there is such force (as shown at 4:18).

If I put two boxes on a table side to side, they won't exert forces on each other (ignoring the gravitational force between them). How is a ladder leaning on a wall different?

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    $\begingroup$ If you lean one box on the other, will one exert a force on the other? $\endgroup$ – Cort Ammon May 3 '17 at 22:42
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    $\begingroup$ Its a good question, I thought it might be explainable through torque but I tried drawing a diagram and that still doesn't help $\endgroup$ – Jaywalker May 3 '17 at 22:44
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If the wall wasn't there what would the ladder do?

The answer is obviously that it would tip over. Which means that without the wall the ladder would be subject to non-zero net force and torque.

With the wall there the ladder doesn't tip and indeed remains static. And that means some force exists in the system with the wall there that does not exist when the wall is not there.

The lack of friction means—by definition—that the contact force along the surface of contact is zero. So the only possibility is the normal force. A force that exists because the wall is a solid thing and the ladder can't occupy the same space as the wall.

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The reason there exists a force between the wall and the ladder is because the ladder is exerting a force against the wall due to the gravitational force. This is why your analogy of two boxes is not applicable, since there are no forces exerted by one box on the other.

Walter Lewin explans this at the 5:30 mark. Since the force exerted by the ladder on the wall is at an angle towards the wall, torque must be taken into consideration.

This also explains the $N_p = F_f$ condition he sets up. This means that the force exerted by the ladder on the wall, which by Newton's Third law is the normal force exerted on the ladder must be equal to the friction force exerted by the ladder at point $Q$ magnitude wise.

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