33 Hydrostatic Weighing

The method of  allows us to determine the average  (\rho) of an object without the need for a  measurement. Instead, we measuring only the objects weight (F_g) and  apparent weight (F_a) when submerged and enter them into the equation below to calculate the density. To see how we arrive at this useful result, follow the steps in the at the end of this chapter.

The method of allows us to determine the average (\rho) of an object without the need for a measurement. Instead, we only need to know the object’s weight (F_g),  apparent weight (F_a) and the density of water (/rho_w) and then enter them into the equation below to calculate the density.

(1)   \begin{equation*} \rho = \frac{F_g}{F_g-F_a}\rho_w \end{equation*}

Keep in mind that if you look up these equations from other sources you might see different symbols used, but the equations are actually the same. For example, the image below shows how the body density, residual volume, and body fat equations are related, but the symbols used are: body density = D_b, water density = D_{H2O}, body weight = BW, and apparent weight = UWW (for under-water weight).

Reinforcement Exercises

Now that we understand mass, volume, density, weight, apparent weight, and static equilibrium, we are ready to see where the previous equation comes from. We start with the definition of the density of an object.

    \begin{equation*} \rho = \frac{m}{V} \end{equation*}

Using the standard model for weight near Earth’s surface we know that mass is can be calculated as weight divided by g.

Inserting that result for mass we have:

    \begin{equation*} \rho = \frac{F_g}{gV} \end{equation*}

When an object is completely submerged then it will displace a volume of water that is equal to its own total volume and we can replace V with the displaced water volume (V_d).

    \begin{equation*} \rho = \frac{F_g}{gV_d} \end{equation*}

Using the definition of density again, we can replace V_d with the displaced water mass (m_d) divided by water density (\rho_w) and then simplify a bit:

    \begin{equation*} \rho = \frac{F_g}{g(m_d/\rho_w)} = \frac{F_g}{m_dg}\rho_w \end{equation*}

Notice that we now have the mass of displaced water multiplied by g. That is exactly how we would calculate the weight of the displaced water (F_d), so we could make that substitution. However, we also know that Archimedes’ principle tells us that the weight of displaced water is equal to the buoyant force F_b, so we can actually replace the displaced water weight (m_dg) with F_b

    \begin{equation*} \rho = \frac{F_g}{F_b}\rho_w \end{equation*}

We also learned in the last chapter that the buoyant must be the difference in in the magnitudes of the weight and the apparent weight.

    \begin{equation*} F_b = F_g - F_a \end{equation*}

Making that replacement in our density equation we have:

    \begin{equation*} \rho = \frac{F_g}{F_g-F_a}\rho_w \end{equation*}

We now have an equation that allows us to calculate the density of an object by measuring only its weight and apparent weight, as long as we know the density of the fluid we are using for submersion.

Specific Gravity

The ratio of the of a substance to that of water is known as the . Specific gravity can be determined by . If we simply divide both sides of our density equation by the density of water we will have a formula for the specific gravity with weight and apparent weight as input:

(2)   \begin{equation*} SG = \frac{\rho}{\rho_W} = \frac{F_g}{F_g-F_a} \end{equation*}

Reinforcement Exercises


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