Problem in inertial reference frame

[balest]

Well, in principle, this problem seems very easy but I don't know the exact explanation.Here we go. We have a truck and a pack in its back.The bed of the truck is frictionless and the pack is stopped by a small piece of wood of something like that. The size of the pack is whatever(for instance,base=2m^2, height=1,75m)and its mass is 2kg so the question is what is the minimum acceleration of the truck to turn over the pack. In the frame of the truck is easy for me to solve the problem but I don't know how to solve it in an inertial frame. What I want to know is why the normal force, in the exact moment when the pack begins to turn over, has to pass through the centre of gravity.Why??? I'm mystified.

[Andrew Mason]

Well, in principle, this problem seems very easy but I don't know the exact explanation.Here we go. We have a truck and a pack in its back.The bed of the truck is frictionless and the pack is stopped by a small piece of wood of something like that. The size of the pack is whatever(for instance,base=2m^2, height=1,75m)and its mass is 2kg so the question is what is the minimum acceleration of the truck to turn over the pack. In the frame of the truck is easy for me to solve the problem but I don't know how to solve it in an inertial frame. What I want to know is why the normal force, in the exact moment when the pack begins to turn over, has to pass through the centre of gravity.Why??? I'm mystified.
Think of the pack as a point mass connected by a rigid rod to the fulcrum (the pack edge pressing against the block of wood). When the rod passes the vertical, the pack starts to turn over. At that point, where is the normal force? Where is the centre of gravity?

AM

[balest]

In the exact situation when the pack begins to turn over I can see those forces

gravity applied in the center of gravity
horizontal normal force(x) applied in the corner near the piece of wood
vertical normal force(y) applied in the corner near the piece of wood too.

but the point is why the net normal force , that's to say, normal force(x)+normal force(y)=black arrow must go through cg(center of gravity). What is the rule,law or whatever in order to infer that?

In the situation you describe it's quite difficult for me to see the forces.I suppose that the normal force is near the fulcrum and vertical as well as the centre of gravity but I'm not sure.

Think of the pack as a point mass connected by a rigid rod to the fulcrum (the pack edge pressing against the block of wood). When the rod passes the vertical, the pack starts to turn over. At that point, where is the normal force? Where is the centre of gravity?

AM

[balest]

Ok,now I know the answer.The point is incredibly the "equivalence principle" and the stability of an object. That seems crazy but just think of it. The horizontal aceleration of the truck(plus the gravity downward) is the same than a situation where the truck is inclined in such a way that the resulting force goes through the rear botton edge of the pack, that is to say the fulcrum(in this situation the pack is about to tip over) So that's why the resulting normal force is equivalent to a force going through the center of gravity, because of the equivalence principle,isn't it marvelous?