Solved Problem on Fluid Mechanics

A hydrogen balloon weighing 600 N is attached to a string in vertical static equilibrium. Its volume is equal to 80 m³, and the air density is equal to 1.25 kg/m³. Determine:
a) The buoyant force due to the air in the balloon;
b) The tension force on the string holding the balloon;
c) The acceleration acquired by the balloon when the string breaks.
Assume the air buoyant force is constant.

Problem data:

Balloon weight: W = 600 N;
Balloon volume: V_b = 80 m³;
Air density: ρ_ar = 1.25 kg/m³;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

In the balloon acts the weight, \( \vec W \), the buoyant force due to the air, \( \vec B \), and the tension force of the string, \( \vec T \), which holds the balloon (Figure 1).

Figure 1

Solution

a) The buoyant force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {B=m_{\small L} g} \tag{I} \end{gather} \]

where m_L is the mass of liquid displaced by the body, in this case, the air, m_a.
The density is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\rho=\frac{m}{V}} \end{gather} \]

with the volume, V, of the body equal to the volume of air displaced by the balloon, V_b, and ρ is the density of the air, ρ_a

\[ \begin{gather} m_a=\rho_a V_b \tag{II} \end{gather} \]

substituting the equation (II) into equation (I)

\[ \begin{gather} B=\rho_a V_b g\\[5pt] B=\left(1.25\;\mathrm{\frac{kg}{\cancel{m^3}}}\right)\times\left(80\;\mathrm{\cancel{m^3}}\right)\times\left(9.8\;\mathrm{\frac{m}{s²}}\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {B=980\;\mathrm N} \end{gather} \]

b) Since the balloon is in static equilibrium, the resultant of the forces acting on the balloon is equal to zero

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum F=0} \end{gather} \]

We choose a frame of reference pointing upwards (Figure 2), and the resultant force will be

\[ \begin{gather} B-W-T=0\\[5pt] T=B-W\\[5pt] T=980\;\mathrm N-600\;\mathrm N \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {T=380\;\mathrm N} \end{gather} \]

Figure 2

c) When the string breaks, the tension force no longer acts, and the balloon moves under the action of the resultant force between the weight and the buoyant force (Figure 3). We choose a frame of reference pointing upwards and applying Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \end{gather} \]

\[ \begin{gather} B-W=ma \tag{III} \end{gather} \]

The weight of the balloon is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {W=mg} \end{gather} \]

Figure 3

the mass of the balloon will be

\[ \begin{gather} m=\frac{W}{g} \tag{IV} \end{gather} \]

substituing the equation (IV) into equation (III)

\[ \begin{gather} B-W=\frac{W}{g}a\\[5pt] a=(B-W)\frac{g}{W}\\[5pt] a=(980\;\mathrm N-600\;\mathrm N)\times\frac{9.8\;\mathrm{\frac{m}{s^2}}}{600\;\mathrm N}\\[5pt] a=\frac{380\;\mathrm{\cancel N}\times 9.8\;\mathrm{\frac{m}{s^2}}}{600\;\mathrm{\cancel N}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {a=6,21\;\mathrm{m/s^2}} \end{gather} \]

Note: the problem tells us to consider the air buoyancy constant, this means we can apply Newton's Second Law with a constant acceleration. In a real situation, the air density decreases with height, so the buoyant force also decreases, and it is not constant. When we use Newton's Second Law with a variable force the acceleration is no longer constant.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .