The vacuum between cup and ceilling exerts no force on either. The atmospheric pressure of the air below the cup pushes up on it with a force (Patm)(A). If the cup barely supports the student's weight, then the normal force of the ceiling is approximately zero, and we have the following. E = 0 + (Patm)(A) – mg = 0 Solving for the area of the suction cup, we have mg A = atm 830.907 = 830.907
Fluid Pressure
The term fluid pressure is coined as, the measurement of the force per unit area of a given surface of a closed container. It is a branch of physics that helps to study the properties of fluid under various conditions of force.
Gauge Pressure
Pressure is the physical force acting per unit area on a body; the applied force is perpendicular to the surface of the object per unit area. The air around us at sea level exerts a pressure (atmospheric pressure) of about 14.7 psi but this doesn’t seem to bother anyone as the bodily fluids are constantly pushing outwards with the same force but if one swims down into the ocean a few feet below the surface one can notice the difference, there is increased pressure on the eardrum, this is due to an increase in hydrostatic pressure.


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