Surface force due to pressure

${\displaystyle f_{s}=p\cdot A\ }$, where f = force, p = pressure, and A = area on which a uniform pressure acts

Pressure related surface force

Since pressure is ${\displaystyle {\frac {\mathit {force}}{\mathit {area}}}=\mathrm {\frac {N}{m^{2}}} }$,^[1] and area is a ${\displaystyle (length)\cdot (width)=\mathrm {m\cdot m} =\mathrm {m^{2}} }$,

a pressure of ${\displaystyle 5\ \mathrm {\frac {N}{m^{2}}} =5\ \mathrm {Pa} }$ over an area of ${\displaystyle 20\ \mathrm {m^{2}} }$ will produce a surface force of ${\displaystyle (5\ \mathrm {Pa} )\cdot (20\ \mathrm {m^{2}} )=100\ \mathrm {N} }$.