Stupid question of forces from one who skipped Physics B

[CoreanJesus]

Sooooo yeah...
My question is something very basic: Internal Forces vs External Forces.
From my knowledge, Internal forces does not change the total mechanical energy while external does.
But than why is gravity an internal force while normal force is external?
Is there any other ways of figuring out external and internal forces?

This question popped up when I was solving a center of mass problem where two skaters pull on a pole of negligible mass on a frictionless surface. The assumption here is that the forces exerted by the two people are internal and therefore does not affect the system. I didn't understand this concept and would appreciate it if someone clarified this for me.
Thanks~~~

Edit: Second question :P
if a dog walks on a boat one way and if friction of the boat and water can be ignored, why are there no external forces? Isn't the dog doing work?

 

[Nugatory]

Newton's third law says that if A exerts a force on B, then B exerts an equal and opposite force on A.

If you're considering A and B to both be part of the same system, then both forces are internal ("inside the system").

If you're considering A and B to be part of separate systems, then the force that A exerts on B is an external ("from outside") force on the system that contains B; and the force that B exerts on A is an external force on the system containing A.

In your problem, we can think of the two skaters and the pole as a single system with no outside forces acting on it. The forces exerted by the two people are internal because they're acting only on other things inside the system. Although the two skaters move towards one another, the center of mass of the combined system is unchanged.

Or we could choose to analyze just a single skater being pulled on by the pole, and think of the other skater tugging on the other end as the source of a force applied to our one-skater system. In this case the center of mass of our one-skater system moves along with the skater, and the force is considered external.

 

[CoreanJesus]

Newton's third law says that if A exerts a force on B, then B exerts an equal and opposite force on A.

If you're considering A and B to both be part of the same system, then both forces are internal ("inside the system").

If you're considering A and B to be part of separate systems, then the force that A exerts on B is an external ("from outside") force on the system that contains B; and the force that B exerts on A is an external force on the system containing A.

In your problem, we can think of the two skaters and the pole as a single system with no outside forces acting on it. The forces exerted by the two people are internal because they're acting only on other things inside the system. Although the two skaters move towards one another, the center of mass of the combined system is unchanged.

Or we could choose to analyze just a single skater being pulled on by the pole, and think of the other skater tugging on the other end as the source of a force applied to our one-skater system. In this case the center of mass of our one-skater system moves along with the skater, and the force is considered external.

What you said makes perfect sense thank you!
But to extend on what you said about center of mass being constant, why is it that when no external force is applied the center of mass doesn't change?

 

[Nugatory]

why is it that when no external force is applied the center of mass doesn't change?

Because if a force is internal, then its equal and opposite third law partner will also be internal, so all internal forces must add to zero. Zero net force means zero movement of the center of mass.

Another way of seeing it: if an internal force pushes one component of the system around in a way that might move the center of mass, an equal and opposite internal force will move some other component of the system around in a way that moves the center of gravity in the opposite direction to cancel the first effect.

For example in your ice skater problem, suppose that the left-hand skater is pulled one meter to the right (which should shift the center of mass of the two-skater system 1/2 meter to the right). What happens to the right-hand skater, and what effect does that have on the position of the center of mass of the two-skater system?

 

[CoreanJesus]

Because if a force is internal, then its equal and opposite third law partner will also be internal, so all internal forces must add to zero. Zero net force means zero movement of the center of mass.

Another way of seeing it: if an internal force pushes one component of the system around in a way that might move the center of mass, an equal and opposite internal force will move some other component of the system around in a way that moves the center of gravity in the opposite direction to cancel the first effect.

For example in your ice skater problem, suppose that the left-hand skater is pulled one meter to the right (which should shift the center of mass of the two-skater system 1/2 meter to the right). What happens to the right-hand skater, and what effect does that have on the position of the center of mass of the two-skater system?

So if it is frictionless, the right-handed skater should travel one meter to the right as well so the center of mass would not change?

 

[PeroK]

So if it is frictionless, the right-handed skater should travel one meter to the right as well so the center of mass would not change?

Yes. It's the same with momentum. A system's momentum can only change through an external force. Internal forces, as @Nugatory explained above, come in equal and opposite pairs. All internal changes in momentum, therefore, cancel out, leaving the momentum and the motion of the centre of mass unchanged.

In the two skater problem (assuming the skaters are of equal mass), the motion of one skater will be the equal and opposite of the other. Not only will they both move the same distance, but always with equal and opposite velocity and equal and opposite acceleration.

They can only change this through an external force - in this case pushing on the ice.