Why do we need extra force to pull an upside down mug out of a bucket? If I try to pull an upside down mug (full of water) out of a bucket of water, extra force is required  to break free from the water surface.
In the case of an inverted  mug (rotated by $180^0$ from the previous case), pulling out of water is straight forward i.e. you apply the force equal to the weight of the mug + water.
 A: Atmospheric pressure is at play here. If the upside-down mug had any amount of water in it, and you start to pull it out of the water, the water has to leave the mug. But since the water leaving the mug needs to be replaced with air, this creates a sort of vacuum. This vacuum is behind the additional force required to break free from the surface (i.e. the force required to create this vacuum). 
Edit: 
One way to think about this is in terms of energy. Creating a vacuum requires energy (think of the created vacuum as a store for this potential energy). Given the relation between potential energy and force, you can see how this would require additional force. 
The extra force is mostly still just the weight of the water. As the bottom (closed) of the mug leaves the surface of the water, it pulls the water up with it above the surface of the water because of said vacuum. 
A: I was asking  the same  question to myself the other day and today i finally figured it out.Actually it's not due to vacuum , it is due to the fact that the mug has tiny amount of air trapped inside it.
I am using an approximation that the mug is pulled out slowly i.e without accelerating.

Let the :

*

*Density of water be $\rho_w$

*The external force that you apply be F

*Atmospheric pressure be $p_o$
The Key thing to notice is that air inside the mug is not exerting the same pressure.


*Pressure that trapped air is exerting be $p$


*Area of cross section of the mug be $A$
Now since the water is at same horizontal level (at my dashed line ) and the water is at rest . Pressure at the horizontal line is $p_o$.
Now : $p_o=p+\rho_w g h'  ====> (p<p_o) $
Let's us say difference in pressure is  $\Delta p$
Therefore according my free body diagram we get :
$$F +pA = mg + p_oA $$
$$F  = mg + \Delta pA $$
$$F>mg$$
After that, the air enters into the mug and the water spills out.
