Scenario 1:
Here, $O$ is the center of mass of a uniform, rigid and solid block. $F_g$ is the force of gravity acting on the body vertically downward, and $N$ is the normal force acting on the body vertically upward. The coefficient of friction of the floor is $\mu$ (distinguishing between $\mu_s$ and $\mu_k$ is probably unnecessary here). So, if the body is pushed horizontally, then the frictional force will be $F_f=\mu N$.
Scenario 2:
Here, $O$ is the center of mass, and we are in space, so no force of gravity exists. The coefficient of friction of the floor is $\mu$. As no normal force exists in this scenario, the frictional force is zero.
My comments:
We can evidently see the two scenarios are different. One has friction, the other doesn't, but why is this the case. The net force on the block in both cases is zero, so why does friction exist in scenario 1 and doesn't exist in scenario 2? Aren't scenarios 1 and 2 essentially the same as the net force on the block is zero?
My question:
- Why does friction exist in scenario 1, but doesn't exist in scenario 2?
Best Answer
At the microscopic level, friction is usually modeled as due to irregularities (hills and valleys) of the contacting surfaces such that when you put two surfaces together they actually make contact at very few places.
In scenario 1, when trying to move the block horizontally to move it over the surface below, the high part parts on each surface get "stuck" on one another due to the continued downward force of gravity and upward normal reaction force that keeps the surfaces in contact with one another. The applied horizontal force keeps unsticking the high points of the surfaces to maintain relative horizontal motion.
In scenario 2, however, there is no force to maintain contact between the surfaces. This allows the block, initially in contact with the surface below, to ride up the high points and separate completely from the surface below losing contact.
In short, pressure needs to be maintained between the surfaces for there to be friction. That pressure does not exist in scenario 2.
Hope this helps.