Unable to understand how these two forces are equal

[tbn032]

In the solution given in the above image, I am unable to understand and prove why N₁=f and N₂=W. I have tried balancing the torque on different point but still unable to prove. Explain how N₁=f and N₂=W can be proved.
The justification for N₁=f and N₂=W which I have so far read is that it is just balancing the vertical force with vertical force and the horizontal force with horizontal force which are being applied on the object since the object is at equilibrium. My confusion with that is that the vertical and horizontal forces are being applied at different position of the object, how can they be directly compared so that the ladder is in equilibrium.

 

[Lnewqban]

The object is solid and strong enough to transfer these forces from one end to the other.
The object is in static equilibrium; therefore, all the forces and moments created by them must be cancelling each other.
All the reactive forces and moments counteract the force of the weight and any moment that it induces.

Without those being present, the weight force will acelerate the object downwards, without inducing any rotation.
Imagining that it could be possible, without the weight force, those reactive forces will move and rotate the object in diferent directions.

 

[tbn032]

The object is solid and strong enough to transfer these forces from one end to the other.
The object is in static equilibrium; therefore, all the forces and moments created by them must be cancelling each other.
All the reactive forces and moments counteract the force of the weight and any moment that it induces.

Since the object is at equilibrium, I understand that the vector sum N1+N2+W+f=0 and the vector sum of torque generated by these forces =0.but I do not understand how can the vertical forces be compared directly with vertical forces and horizontal force directly compared with horizontal force resulting in N1=f and N2=W even after they are being applied at different position.

 

[Lnewqban]

Since the balance of moments is not confusing to you, just relocate all those forces to the center of mass of the object.
The actual location of each of those forces is only important for the moment it induces.

 

[tbn032]

jus relocate all those forces to the center of mass of the object.

Is relocation of forces which being applied at different position of the object to the center of mass of thr object which is at equilibrium allowed?if it is allowed, can you explain the concept behind it.

 

[Lnewqban]

Is relocation of forces which being applied at different position of the object to the center of mass of thr object which is at equilibrium allowed?if it is allowed, can you explain the concept behind it.

Yes, you may do that when you are analyzing translational equilibrium only.
For this first equilibrium condition for the static equilibrium (no translational acceleration) of a rigid body, the distances of the external forces to the center of mass are irrelevant.

Please, see:
https://courses.lumenlearning.com/s...apter/12-1-conditions-for-static-equilibrium/

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/9-6-center-of-mass/