Question

A consumer upset with the latest trend of postal rate increases has decided to try to send letters by balloon even though they may not reach their intended destinations. A 66400 cm366400 cm3 gas-filled balloon will provide enough lift for a 43.4 g43.4 g package to be accelerated upward at a rate of 3.10 m/s23.10 m/s2 . For these circumstances, calculate the density of the gas the consumer fills the balloon with. The acceleration due to gravity is ????=9.81 m/s2g=9.81 m/s2 and the density of air is ????air=1.16 kg/m3rhoair=1.16 kg/m3 . Neglect the mass of the balloon material and the volume of the package.

Answer 1

Answer:

d = 0.44 kg/m³

Explanation:

To determine the density of the gas into the balloon we need to consider the forces that are acting on the balloon in the x (Fx) and y (Fy) components:      

x: $\Sigma F_{x} = m \cdot a_{x} = 0$

The sum of the forces is zero because there is no force acting on the balloon in this component.  

y: $\Sigma F_{y} = - W_{b} - W_{p} + B = (m_{b} + m_{p}) \cdot a_{y}$ (1)

where $W_{b}$ and  $W_{p}$: are the weights of the balloon and the package, respectively, $m_{b}$  and $m_{p}$: are the masses of the balloon and the package, respectively, and B: buoyancy force.  

Knowing that:  

$W = m\cdot g$ (2)

$B = \rho \cdot g \cdot V$ (3)

where ρ: is the density of the air, g: gravitational acceleration and V: volume of the gas displaced.  

And introducing the weights of the balloon and the package from (2), and B from (3) on the equation (1), we can find the mass of the balloon:          

$- g (m_{b} + m_{p}) + \rho \cdot g \cdot V = (m_{b} + m_{p}) \cdot a_{y}$  

$m_{b} =\frac {\rho \cdot g \cdot V}{(a_{y} + g)} - m_{p}$        

$m_{b} = \frac {(1.16 \frac{kg}{m^{3}})(9.81 \frac {m}{s^{2}})(0.0664 m^{3})}{(9.81 \frac{m}{s^{2}} + 3.10 \frac {m}{s^{2}})} - 43.4 \cdot 10^{-3}Kg$    

$m_{b} = 0.015 kg$

Finally, with the mass of the balloon calculated we can determine the density (d) of the gas inside the balloon:

$d = \frac {m_{b}}{V} = \frac {0.015 kg}{0.0664 m^{3}} = 0.23 \frac{kg}{m^{3}}$

 

Have a nice day!