Juanito thinks that he is very strong because he can lift his dad up when he is totally submerged in a pool. Juanito’s father has a mass of 85 kg, and adults, on average, have a volume of 66 L. How much mass does Juanito’s father seem to have inside the pool? (Assume the pool has pure water without chlorine or other additives.)
Use Newton’s Second Law to find the net force, and then divide it by the acceleration due to gravity to solve for the apparent mass.
The buoyant force is:
\begin{equation*}
F_B = \rho V_{\text{sub}} g.
\end{equation*}
By Newton’s second law we get:
\begin{equation*}
F_{\text{net}} = F_B – mg = 0,
\end{equation*}
where the net force divided by \(g\) is:
\begin{equation*}
m_{\text{apparent}}= m-\rho V_{\text{sub}},
\end{equation*}
or with numerical values:
\begin{equation*}
m_{\text{apparent}}= 19 \, \text{kg}.
\end{equation*}
For a more detailed explanation of any of these steps, click on “Detailed Solution”.
To find the apparent mass of Juanito’s father, we must first draw his free-body diagram assuming he is totally submerged in water, as seen in figure 1.
Figure 1: Free-body diagram of Juanito’s father totally submerged in water. The coordinate system is chosen so that the positive Y-axis points upwards. The relevant forces are also shown: the weight \(\vec{W}\) and the buoyant force \(\vec{F}_B\).
From figure 1, we see that only two forces are exerted on Juanito’s father: the weight \(\vec{W}\) and the buoyant force \(\vec{F}_B\). The net force is equal to the force that Juanito will have to exert on his father. Thus, dividing this net force by the gravitational acceleration \(g\) gives us the apparent mass of Juanito’s father. Let’s start by calculating the net force \(\vec{F}_{\text{net}}\) as the sum of all the forces exerted on Juanito’s father, namely
\begin{equation}
\label{net}
\vec{F}_{\text{net}}=\vec{W}+\vec{F}_B.
\end{equation}
The weight has magnitude \(mg\) and is directed towards the negative Y axis, then
\begin{equation}
\label{weight}
\vec{W}=-mg\,\hat{\textbf{j}}.
\end{equation}
The buoyant force has a magnitude \(\rho V_{\text{sub}} g\), where \(\rho\) is the density of the fluid and \(V_{\text{sub}}\) is the volume of the body that is submerged. This buoyant force is directed upwards, towards the positive Y axis, then
\begin{equation}
\label{buoyant}
\vec{F}_B=\rho V_{\text{sub}}g\,\hat{\textbf{j}}.
\end{equation}
Using the explicit expressions for the weight and the buoyant force given in equations \eqref{weight} and \eqref{buoyant} into equation \eqref{net}, we obtain
\begin{equation}
\label{net2}
\vec{F}_{\text{net}}=-mg\,\hat{\textbf{j}}+\rho V_{\text{sub}}g\,\hat{\textbf{j}},
\end{equation}
which can be factorized as
\begin{equation}
\label{net3}
\vec{F}_{\text{net}}=-\left(m-\rho V_{\text{sub}}\right)g\,\hat{\textbf{j}}.
\end{equation}
The magnitude of the net force is then
\begin{equation}
\label{net4}
|\vec{F}_{\text{net}}|=\left(m-\rho V_{\text{sub}}\right)g,
\end{equation}
and dividing by \(g\) gives us the apparent mass \(m_{\text{apparent}}\).
\begin{equation}
m_{\text{apparent}}=\frac{|\vec{F}_{\text{net}}|}{g}=\frac{\left(m-\rho V_{\text{sub}}\right)g}{g}.
\end{equation}
Therefore, we can write
\begin{equation}
m_{\text{apparent}}=m-\rho V_{\text{sub}}.
\end{equation}
Using the numerical values in SI units (\(V_{\text{sub}}=0.066\,\text{m}^3\) and \(\rho=1000\,\text{kg/m}^3\)), we have
\begin{equation}
m_{\text{apparent}}=85\,\text{kg}-(1000\,\text{kg/m}^3)(0.066\,\text{m}^3),
\end{equation}
\begin{equation}
m_{\text{apparent}}=19\,\text{kg}.
\end{equation}
It turns out that Juanito is not as strong as he thought!
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