Solved Problem on Fluid Mechanics
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A right circular cylinder of height h = 30 cm and base area A = 10 cm2, floats in water in a vertical position with 2/3 of its height immersed, a force F is applied in the axial direction of the upper base passing through the cylinder to have 5/6 of its height immersed. Determine:
a) What is the density of the cylinder relative to water?
b) What is the magnitude of the force F?
Data: g = 9.8 m/s2, density of water = 1 g/cm3.


Problem data:
  • Cylinder height:    h = 30 cm;
  • Cylinder base area:    A = 10 cm2;
  • Fraction of cylinder height initially immersed:    \( h_{1}=\dfrac{2}{3}h \);
  • Fraction of the cylinder height immersed after the force is applied:    \( h_{2}=\dfrac{5}{6}h \);
  • Acceleration due to gravity:    g = 9.8 m/s2;
  • Density of water:    da = 1 g/cm3.
Solution

First, let us convert the acceleration due to gravity given in meters per second squared (m/s2) to centimeters per second squared (cm/s2).
\[ \begin{gather} g=9.8\;\frac{\cancel{\text{m}}}{\text{s}^{2}}\times\frac{100\;\text{cm}}{1\;\cancel{\text{m}}}=980\;\text{cm/s}^{2} \end{gather} \]
a) In the initial situation, the gravitational force \( {\vec F}_{g} \) and the buoyant force \( {\vec{B}}_{1} \) act on the cylinder due to the volume of water displaced by the immersed height h1 of the cylinder (Figure 1).
The gravitational force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{I} \end{gather} \]
The density is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {d=\frac{m}{V}} \end{gather} \]
Figure 1

\[ \begin{gather} m=dV \tag{II} \end{gather} \]
substituting expression (II) for the mass in equation (I)
\[ \begin{gather} F_{g}=dVg \tag{III} \end{gather} \]
The volume of a cylinder is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {V=Ah} \tag{IV} \end{gather} \]
substituting equation (IV) in equation (III)
\[ \begin{gather} F_{g}=dAhg \tag{V} \end{gather} \]
The buoyant force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {B=m_{\small L}g} \tag{VI} \end{gather} \]
Using expression (II), where mL = ma, the mass of water displaced will be
\[ \begin{gather} m_{a}=d_{a}V_{a} \tag{VII} \end{gather} \]
where da is the density of the water in which the body is immersed, and Va is the volume of water displaced.
Substituting expression (VII) in equation (VI)
\[ \begin{gather} B_{1}=d_{a}V_{a}g \tag{VIII} \end{gather} \]
using equation (IV) for the volume of water displaced
\[ \begin{gather} V_{a}=Ah_{1} \tag{IX} \end{gather} \]
substituting expression (IX) for the volume of water displaced in equation (VIII)
\[ \begin{gather} B_{1}=d_{a}Ah_{1}g \tag{X} \end{gather} \]
Since the body is in equilibrium, the gravitational force and the buoyant force cancel each other out
\[ \begin{gather} B_{1}=F_{g} \tag{XI} \end{gather} \]
substituting equations (V) and (X) in the condition (XI)
\[ \begin{gather} d_{a}\cancel{A}h_{1}\cancel{g}=d\cancel{A}h\cancel{g}\\[5pt] d_{a}h_{1}=dh\\[5pt] \frac{d}{d_{a}}=\frac{h_{1}}{h} \end{gather} \]
substituting the height of the cylinder immersed in water, given in the problem
\[ \begin{gather} \frac{d}{d_{a}}=\frac{\dfrac{2}{3}\cancel{h}}{\cancel{h}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\frac{d}{d_{a}}=\frac{2}{3}} \end{gather} \]

b) In the final situation, we have the external force \( \vec{F} \) acting on the cylinder. We choose a frame of reference pointing downward, in the same direction as the acceleration due to gravity, so that the cylinder remains in equilibrium and the sum of forces is equal to zero (Figure 2).
\[ \begin{gather} \bbox[#99CCFF,10px] {\sum F=0} \end{gather} \]
\[ \begin{gather} F_{g}+F-B_{2}=0 \tag{XII} \end{gather} \]
the buoyancy \( {\vec{B}}_{2} \) is due to the volume of water displaced by the immersed height h2 of the cylinder. Substituting equations (V) and (X), for height h2, in equation (XII)
\[ \begin{gather} dAhg+F=d_{a}Ah_{2}g\\[5pt] F=d_{a}Ah_{2}g-dAhg \end{gather} \]
Figure 2

substituting the density of the cylinder relative to water   \( \left(d=\frac{2}{3}d_{a}\right) \) found in the previous item
\[ \begin{gather} F=d_{a}A\frac{5}{6}hg-\frac{2}{3}d_{a}Ahg\\[5pt] F=d_{a}Ahg\left(\frac{5}{6}-\frac{2}{3}\right) \end{gather} \]
multiplying by 2, the numerator and denominator of the second term in parentheses
\[ \begin{gather} F=d_{a}Ahg\left(\frac{5}{6}-\frac{2}{3}.\frac{2}{2}\right)\\[5pt] F=d_{a}Ahg\left(\frac{5}{6}-\frac{4}{6}\right)\\[5pt] F=\left(\frac{1}{6}\right)d_{a}Ahg \end{gather} \]
substituting problem data
\[ \begin{gather} F=\left(\frac{1}{6}\right)\times 1\times 10\times 30\times 980\\[5pt] F=50\times 980 \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {F=49000\;\text{dinas}} \end{gather} \]

Note: dyne is the unit of force in the CGS system of units.
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