Cube in Water and Oil!

a) Use the formula for pressure. Consider the height in each case.

b) This is calculated directly with the previous result obtained in part (b).

c) Use Newton’s Second Law, and use the forces previously found.

d) From Newton’s Second Law, and solve for the mass and the density.

a) The hydrostatic pressure can be written as:

\begin{equation*}
P = \rho g H + P_0.
\end{equation*}

For the top, the height is \(0.2 \, \text{m}\), then:

\begin{equation*}
P_{top} = 103147.8 \ \text{Pa},
\end{equation*}

and for the bottom \(0.5 \, \text{m}\) is submerged in oil and \(0.2 \, \text{m}\) in water, then:

\begin{equation*}
P_{bot} = 107695 \ \text{Pa}.
\end{equation*}

b) By definition,

\begin{equation*}
F = P A.
\end{equation*}

The area is a square. Then, for the top:

\begin{equation*}
F_{top} = 25772.25 \ \text{N},
\end{equation*}

and for the bottom:

\begin{equation*}
F_{\text{bot}} = 26923.75 \ \text{N}.
\end{equation*}

c) Using Newton’s second law:

\begin{equation*}
F_{\text{bot}} – F_{\text{top}} – w = 0.
\end{equation*}

Then:

\begin{equation*}
w = 1151.5 \, \text{N}.
\end{equation*}

d) The density equation is:

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

where the volume is for a cube. Then:

\begin{equation*}
\rho = 940 \, \text{kg/m}^3.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) We can find the pressure at the top and bottom faces of the cube by using the equation for the hydrostatic pressure in terms of the height of the column of fluid that is exerting the pressure and the density of the fluids. In this case, we must be careful to include the contributions from both oil and water to the pressure when necessary.

The hydrostatic pressure P on a point in a fluid such that the height of the column of liquid above that point is H is given by

\begin{equation}
\label{EQ:hstat}
P = \rho g H + P_0,
\end{equation}

where \(\rho\) is the density of the fluid exerting the pressure and \(P_0\) is the atmospheric pressure. The upper face of the cube has a column of oil of height \(h\) above it (see the previous figure). Hence, we can write eq. \eqref{EQ:hstat} with \(\rho = \rho_{\text{oil}}\) and \(H = h\) i.e. the pressure \({P_\text{top}}\) at the top face is given by

\begin{equation}
\label{EQ:ptop}
P_{top} = \rho_{\text{oil}}gh + P_0.
\end{equation}

Assuming that the container is at sea level, and substituting numerical values, we get

\begin{equation}
P_{top} = \left(930 \frac{\text{Kg}}{\text{m}^3}\right)\left(9.8 \ \frac{\text{m}}{\text{s}^2}\right)(0.2 \ \text{m}) + 101325 \ \text{Pa},
\end{equation}

which is the same as

\begin{equation}
\label{EQ:nptop}
P_{top} = 1822.8 \ \text{Pa} + 101325 \ \text{Pa}
= 103147.8 \ \text{Pa}.
\end{equation}

Now, the bottom face has a column of oil and a column of water above it, as seen in figure 1. Hence, we need to modify eq. \eqref{EQ:hstat} to account for this. In this case, both oil and water contribute to the hydrostatic pressure according to the height of the column of liquid of each component above the bottom face. Following this logic, we can write the pressure \(P_{\text{bot}}\) at the bottom face of the block as

\begin{equation}
\label{EQ:pbot}
P_{\text{bot}} = \rho_{\text{oil}}gh_{\text{oil}} + \rho_wgh_w + P_0,
\end{equation}

where \(\rho_{\text{oil}}\) and \(\rho_w\) are the densities of oil and water, respectively, and \(h_{\text{oil}}\) and \(h_w\) are the heights of the columns of oil and water above the depth of the bottom face of the cube, as shown in figure 2. According to that figure, we can write

\begin{equation}
h_w = h,
\end{equation}

and

\begin{equation}
h_{\text{oil}} = (h + a) – h = a.
\end{equation}
Substituting these expressions in eq. \eqref{EQ:pbot} yields

\begin{equation}
\label{EQ:pbot2}
P_{\text{bot}} = \rho_{\text{oil}} g a + \rho_w g h + P_0
= (\rho_{\text{oil}} a + \rho_w h)g + P_0,
\end{equation}

and after inserting numerical values, we get

\begin{equation}
P_{\text{bot}} = \left(\left(900 \ \frac{\text{Kg}}{\text{m}^3}\right)(0.5\ \text{m}) + \left(1000 \ \frac{\text{Kg}}{\text{m}^3}\right)(0.2 \text{m})\right)\left(9.8 \ \frac{\text{m}}{\text{s}^2}\right) + 101325 \ \text{Pa},
\end{equation}

which is equivalent to

\begin{equation}
\label{EQ:npbot}
P_{\text{bot}} = 6370 \ \text{Pa} + 101325 \ \text{Pa}
= 107695 \ \text{Pa}.
\end{equation}

b) By definition, pressure is the force exerted per unit area. Hence, we can find the force on the top and bottom faces of the cube if we multiply the pressures found, in part, by the area of each face. The hydrostatic pressure will tend to compress the cube, so the direction of the forces produced by the fluid pressure will be towards the center of the cube.

The pressure \(P\) exerted on a surface of area \(A\) by a force of magnitude \(F\) perpendicular to it is defined as

\begin{equation}
P = \frac{F}{A}
\end{equation}

Multiplying by \(A\) gives

\begin{equation}
F = PA.
\end{equation}

The area of a face of the cube is \(A = a^2\). Hence,

\begin{equation}
\label{EQ:f}
F = Pa^2.
\end{equation}

Therefore, the magnitude of the force \(F_{top}\) exerted on the top face is

\begin{equation}
\label{EQ:ftop_s}
F_{top} = P_{top}a^2,
\end{equation}

and substituting eq. \eqref{EQ:ptop} in the previous expression, we get

\begin{equation}
\label{EQ:ftop_l}
F_{top} = (\rho_{\text{oil}}gh + P_0)a^2.
\end{equation}

After inserting numerical values, we get

\begin{equation}
\label{EQ:ftop_n}
F_{top} = \left( \left(900 \ \frac{\text{Kg}}{\text{m}^3}\right)\left(9.8 \frac{\text{m}}{\text{s}^2}\right)(0.2\ \text{m}) + 101325\ \text{Pa}\right) (0.5\ \text{m})^2
= 25772.25 \ \text{N}
\end{equation}

Similarly, the magnitude of the force exerted on the bottom face is given by

\begin{equation}
\label{EQ:fbot_s}
F_{\text{bot}} = P_{\text{bot}}a^2,
\end{equation}

and substituting eq. \eqref{EQ:pbot} yields

\begin{equation}
\label{EQ:fbot_l}
F_{\text{bot}} = ( (\rho_{\text{oil}} a + \rho_w h)g + P_0 ) a^2.
\end{equation}

If we insert numerical values, we get

\begin{equation}
F_{\text{bot}} = \left( \left( \left(900 \ \frac{\text{Kg}}{\text{m}^3}\right) (0.5\ \text{m}) + \left(1000 \ \frac{\text{Kg}}{\text{m}^3}\right)(0.2\ \text{m})\right)\left(9.8 \frac{\text{m}}{\text{s}^2}\right) + 101325 \ \text{Pa} \right) (0.5\ \text{m})^2,
\end{equation}

namely

\begin{equation}
\label{EQ:fbot_n}
F_{\text{bot}} = 26923.75 \ \text{N}.
\end{equation}

Now we should determine the direction of these forces. The force produced by the hydrostatic pressure will tend to compress the block. In other words, it will be directed inwards towards the block in the direction that is perpendicular to its surface. Following this reasoning, if we establish a reference frame such that the \(y\) axis points in the upwards direction, \(F_{top}\) will point towards the \(-\hat{\textbf{j}}\) direction, and \(F_{\text{bot}}\) will point towards the \(+\hat{\textbf{j}}\) direction, as shown in figure 3. Therefore, if we include the directions in eqs. \eqref{EQ:ftop_l}, \eqref{EQ:ftop_n}, \eqref{EQ:fbot_l}, and \eqref{EQ:fbot_n}, we get

\begin{equation}
\label{EQ:ftop_vec}
\vec{F}_{top} = -(\rho_{\text{oil}}gh + P_0)a^2 \hat{\textbf{j}} = -25772.25 \ \text{N} \hat{\textbf{j}},
\end{equation}

and

\begin{equation}
\label{EQ:fbot_vec}
\vec{F}_{\text{bot}} = ( (\rho_{\text{oil}} a + \rho_w h)g + P_0 ) a^2 \hat{\textbf{j}} = 26923.75 \ \text{N} \hat{\textbf{j}}
\end{equation}

c) The only force acting on the cube that remains to be found in the weight, which we can find by using Newton’s second law and using the expressions for the forces found in part b.

According to the force diagram shown in figure 3, we can write Newton’s second law for the cube in vector form as

\begin{equation}
\vec{F}_{top} + \vec{F}_{\text{bot}} + \vec{w} =\ \text{m}\vec{a},
\end{equation}
where \(\vec{w}\) is the weight of the cube, and \(\vec{a}\) is its acceleration. Since the block is stationary, \(\vec{a}=0\) yielding

\begin{equation}
\vec{F}_{top} + \vec{F}_{\text{bot}} + \vec{w} = 0.
\end{equation}

Solving for \(\vec{w}\), we get

\begin{equation}
\label{EQ:w}
\vec{w} = – (\vec{F}_{top} + \vec{F}_{\text{bot}}),
\end{equation}

and substituting eqs. \eqref{EQ:ftop_vec} and \eqref{EQ:fbot_vec}, we get

\begin{equation}
\vec{w} = – ( -(\rho_{\text{oil}}gh + P_0)a^2 \hat{\textbf{j}} + ( (\rho_{\text{oil}} a + \rho_w h)g + P_0 ) a^2 \hat{\textbf{j}}),
\end{equation}

factoring out \(a^2 \hat{\textbf{j}}\), we get,

\begin{equation}
\vec{w} = – ( -(\rho_{\text{oil}}gh + P_0) + ( (\rho_{\text{oil}} a + \rho_w h)g + P_0 )) a^2 \hat{\textbf{j}}
= – ( -\rho_{\text{oil}}gh – P_0 + \rho_{\text{oil}} a g + \rho_w h g + P_0) a^2 \hat{\textbf{j}}.
\end{equation}

Now, the two terms containing \(P_0\) cancel out, yielding

\begin{equation}
\vec{w} = – ( -\rho_{\text{oil}}gh + \rho_{\text{oil}} a g + \rho_w h g ) a^2 \hat{\textbf{j}}.
\end{equation}

We can factor out g to obtain

\begin{equation}
\vec{w} = – ( -\rho_{\text{oil}}h + \rho_{\text{oil}} a + \rho_w h ) a^2 g \hat{\textbf{j}}.
\end{equation}

Finally, factoring \(\rho_oil\), we get

\begin{equation}
\label{EQ:w1.5}
\vec{w} = – ( \rho_{\text{oil}}(a – h) + \rho_w h) a^2 g \hat{\textbf{j}}.
\end{equation}

Notice that the first term corresponds to the weight of the oil that was displaced by the cube, and the second term corresponds to the weight of water displaced by it. Hence, the previous equation is equal to the weight of the liquid displaced by the cube, as predicted by Archimedes’ principle!

Finally, inserting numerical values gives

\begin{equation}
\vec{w} = – \left(\left(900 \ \frac{\text{Kg}}{\text{m}^3}\right)(0.5\ \text{m} – 0.2\ \text{m}) + \left(1000 \ \frac{\text{Kg}}{\text{m}^3}\right)(0.2\ \text{m})\right) (0.5\ \text{m})^2 \left(9.8 \frac{\text{m}}{\text{s}^2}\right) \hat{\textbf{j}},
\end{equation}

which gives us

\begin{equation}
\vec{w} = -1151.5 \ \text{N} \hat{\textbf{j}}.
\end{equation}

The minus sign refers to the direction of \(\vec{w}\), which is opposed to \(\hat{\textbf{j}}\).

d) Density is defined as mass per unit volume. In this case, we can find the density of the cube by using its weight to calculate its mass and using its dimensions to find its volume.

By definition, the density of the cube is given by

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

where \(m\) is the mass of the cube and \(V\) is its volume. The volume of a cube of side a is \(V = a^3\). Hence,

\begin{equation}
\label{EQ:rho2}
\rho = \frac{m}{a^3}.
\end{equation}

Now, the mass of the cube is related to its weight by

\begin{equation}
\vec{w} = m\vec{g},
\end{equation}

or in terms of the vector magnitudes:

\begin{equation}
w = mg.
\end{equation}

Dividing by g, we get

\begin{equation}
m =\frac{w}{g}.
\end{equation}

Therefore, substituting eq. \eqref{EQ:w1.5}, we get

\begin{equation}
m = ( \rho_{\text{oil}}(a – h) + \rho_w h) a^2 \frac{g}{g}
= ( \rho_{\text{oil}}(a – h) + \rho_w h) a^2,
\end{equation}

and inserting this in eq. \eqref{EQ:rho2}, we get

\begin{equation}
\rho = ( \rho_{\text{oil}}(a – h) + rho_w h) \frac{a^2}{a^3}
= \frac{\rho_{\text{oil}}(a – h) + rho_w h}{a}
\end{equation}

Finally, substituting numerical values, we get

\begin{equation}
\rho = \frac{\left(900 \ \frac{\text{Kg}}{\text{m}^3}\right) (0.5\ \text{m}- 0.2\ \text{m}) + \left(1000 \ \frac{\text{Kg}}{\text{m}^3}\right) (0.2\ \text{m})}{0.5\ \text{m}}
= 940 \ \frac{\text{Kg}}{\text{m}^3}.
\end{equation}

The value of the density of the cube is smaller than the density of the water but larger than the density of the oil. This is expected since the block does not float in the oil layer, but it does not sink in water.

You need to be registered and logged in to take this quiz. Log in or Register