Encyclopedia>Numbers and Quantities>Quantities>Measuring>Imperial>How to Find Volumetric Mass Density (Imperial)

This entry uses the imperial system of measurement. It is more common to use the metric system when you measure density. Click here to go to the metric version of this page.

I choose to use ${\text{lb/ft}}^{3}$ for this page.

Mass density is a topic you may recognize from science class. The density of a substance says something about how compact a substance is. This means that you can have two objects that are the same size (volume), but due to different densities, one object will weigh more (have larger mass) than the other. Mass density is given by mass ($m$) divided by volume ($V$). The symbol for mass density is the Greek letter $\rho $ (rho).

Rule

$$\rho =\frac{m}{V}$$ |

Example 1

Iron has density of $\rho =\text{}7.88\text{}\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}$. An iron cannon ball has a volume $\text{}1800\text{}\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}$. What does the cannon ball weigh?

Insert into the formula and solve for $m$: $$\begin{array}{llll}\hfill \rho =\frac{m}{V}\iff m& =\rho \cdot V\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7.88\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}\cdot 1800\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =14\phantom{\rule{0.17em}{0ex}}184\phantom{\rule{0.17em}{0ex}}\text{lb}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =7.092\text{shortton}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Water has a mass density equal to $62.4\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}$. If an object has a lower density than water, it will float on the water. However, if an object has a higher mass density than water, it will sink.

Example 2

A plank weighs $\text{}1\text{}\phantom{\rule{0.17em}{0ex}}\text{lb}$, and has volume $\text{}600\text{}\phantom{\rule{0.17em}{0ex}}{\text{in}}^{3}$. Will the plank float on the water?

First, make sure the units are the same. Since one foot is 16 inches, $1\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}=16\cdot 16\cdot 16=1728\phantom{\rule{0.17em}{0ex}}{\text{in}}^{3}$. This means you need to divide the ${\text{in}}^{3}$ by $1728$

$$600\phantom{\rule{0.17em}{0ex}}{\text{in}}^{3}=\frac{600}{1728}\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}\approx 0.347\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}$$ |

The plank has density

$$\rho =\frac{m}{V}=\frac{1\phantom{\rule{0.17em}{0ex}}\text{lb}}{0.347\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}}\approx 2.857\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}.$$ |

The plank has a much lower density than water, and will thus float.

Example 3

Annie has found two bars in a hidden cabin. They are gold colored, and Annie wonders if she has found real gold bars, or if they are only painted on the outside. She has weighed and measured the volume of one of the bars. It weighed $26.5\phantom{\rule{0.17em}{0ex}}\text{lb}$, and had a volume of about $0.022\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}$. Annie has found that the mass density of gold is around $1206.11\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}$. Is it a real gold bar? $$\begin{array}{llll}\hfill \rho & =\frac{m}{V}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{26.54\phantom{\rule{0.17em}{0ex}}\text{lb}}{0.022\phantom{\rule{0.17em}{0ex}}{\text{ft}}^{3}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 1206.37\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 1206.11\phantom{\rule{0.17em}{0ex}}{\text{lb/ft}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Annie’s bar definitely seems to be made of gold.

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