1. The problem statement, all variables and given/known data P=hρg is the pressure difference but where does this pressure act upon? In a standard barometer there is a vacuum on top so 0 Pa, and at the bottom the normal force acts on the glass base. So hρg=Pweight. But theoretically, where would the hρg calculated be acted upon? Or will that pressure just be exerted at all directions? If so in this example http://postimage.org/image/53k5i2cqz/full/ [Broken] will the hρg be exerted onto the air as well? 2. Relevant equations P=hρg 3. The attempt at a solution I think so because under forces, we learned that by newton's third law besides the normal force balancing out weight, there is also a reaction force acting on the other body that applied the normal force. So now in terms of pressure in that scenario the liquid would exert the pressure on the gas below. However, the main part that confuses me is that the normal force balancing the weight will be countered by weight itself. So there is no net force acting on the liquid. So why should there be a pressure being exerted on it? Thanks for the help!