# Is it possible for density in pounds per cubic foot, times a volume, to equal a force in pounds?

In the Calculus book "Calculus: Early Transcendentals" by Edwards and Penney, 7th Edition, there is a chapter on Force and Work. In particular, there is a section explaining how to calculate the work done when filling a tank with fluid.

The assumptions of the problem are:

• the tank is filled in thin, horizontal layers of fluid, each lifted from its ground position to its final position in the tank
• the bottom of the tank is at height $$y=a$$ and its top at height $$y=b>a$$.
• $$A(y)$$ is the cross-sectional area of the tank at height $$y$$

The solution of the problem:

• partition $$[a,b]$$ into n subintervals of length $$\Delta y$$
• the volume of the i-th horizontal slice is thus $$\Delta V_i = \int_{y_{i-1}}^{y_i} A(y)dy = A(y_i^*) \Delta y$$ for some $$y_i^*$$ in $$[y_{i-1}, y_i]$$ (the last equality follows from the average value theorem for integrals, on the interval $$[y_{i-1}, y_i]$$

Now here is the part I don't understand, and I think it probably has to do with confusion regarding the system of measurement

suppose $$\rho$$ is the density of the fluid (in pounds per cubic foot, for example), then the force required to lift this slice from the ground to its final position in the tank is simply the (constant) weight of the slice: $$F_i= \rho \Delta V_i = \rho A(y_i^*) \Delta y$$

I see a density times a volume, giving a mass in pounds (which to me is the counterpart to the kilogram unit of mass if we are in a metric system). I googled and found out that indeed there is a unit called a pound of force, which is defined as the force that accelerates 1 pound of mass at 9.80665 meters per second square.

How is the expression above for force a force and not just a mass? • Abusive notation like that makes me shudder.
– noah
May 23, 2021 at 22:39

It's clearly just a mass, but the author seems to be abusing the fact that the proportionality constant between pounds - as in mass - and pounds of force is $$g = 1\,\frac{\mathrm{lbf}}{\mathrm{lb}}$$.
So the correct expression for a force would be $$F_i= g \, \rho \, \Delta V_i$$ which holds in any unit system, not just the $$\mathrm{lb/lbf}$$ version.