I am looking at this example from my engineering manual. Is there anyone that understands each step of the math so that the answer 0.866 can be reached. It would help me to understand if the steps could be shewn. The unit conversion is a bit confusing to me. There is talk of the inverse of gc but I'm not really sure what gc is a reference to in this case. My question is can someone who understands this post the steps of the formula to reach the answer.
Thanks!
Well, water is well-known to have density of very close to $1\,\frac{\text{kg}}{\text{l}}=1\,\frac{\text{kg}}{\text{dm}^3}=1000\,\frac{\text{kg}}{\text{m}^3}$, the gravitational acceleration is approximately $9.81\,\frac{\text{m}}{\text{s}^2}$ and the depth of two feet is something like about $61\,\text{cm}=0.61\,\text{m}$. So the pressure is seen to be (without magic constants) $$ p=\rho gh=1000\,\frac{\text{kg}}{\text{m}^3}\cdot 9.81\,\frac{\text{m}}{\text{s}^2}\cdot 0.61\,\text{m}\approx 5980\,\frac{\text{kg}}{\text{m}\,\text{s}^2}=5980\,\frac{\text{N}}{\text{m}^2}=5980\,\text{Pa}$$ I hope this can be converted to $0.866\,\text{psi}$.
Yes, the conversion factor $g_c$ is a little confusing. The reason we need to use it to calculate pressure in English units is due to the unusual definition that 1 pound of mass weighs 1 pound (exerts one pound of force). In contrast, for the SI system, the base unit of mass is the kilogram (kg), where 1 kg of mass weighs 9.8 Newtons (so that we do not need to divide again by 9.8).
