Chapter 13, Problem 89P

(a)

To determine

Volume of the balloon

Answer to Problem 89P

  $69.67{\text{ m}}^3$

Explanation of Solution

Given:

Density of air $={\rho }_a=1.3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Density of helium $={\rho }_{He}=0.18\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Mass of skin of balloon $=m_b=1.5\text{ kg}$

Weight of skin of balloon $=w_b$

Weight of the load $=w_L=750\text{ N}$

Force of buoyancy $=F_B$

Volume of the balloon $=V_b$

Formula Used:

Weight is given as

  $w=mg$

Force of buoyancy is given as

  $F_B=\rho Vg$

Calculation:

Weight of the skin of the balloon is given as

  $\begin{array}{l}{w}_b=m_{b}g\\ w_b=(1.5)(9.8)\\ w_b=14.7\text{ N}\end{array}$

Weight of the helium in the balloon in downward direction is given as

  $w_{He}={\rho }_{He}{V}_{b}g$

Force of buoyancy on the balloon in upward direction is given as

  $F_B={\rho }_a{V}_{b}g$

Using equilibrium of force in vertical direction, we get the force equation as

  $\begin{array}{l}{F}_B=w_b+w_{He}+w_L\\ {\rho }_a{V}_{b}g=14.7+{\rho }_{He}{V}_{b}g+750\\ (1.3)V_b(9.8)=(0.18)V_b(9.8)+764.7\\ (12.74)V_b=(1.764)V_b+764.7\\ (12.74-1.764)V_b=764.7\\ V_b=\frac{764.7}{10.976}\\ V_b=69.67{\text{ m}}^3\end{array}$

Conclusion:

Hence,the volume of the balloon comes out to be $69.67{\text{ m}}^3$ .

(b)

To determine

The initial acceleration of the balloon.

Answer to Problem 89P

  $5.2{\text{ ms}}^{-2}$

Explanation of Solution

Given:

Density of air $={\rho }_a=1.3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Density of helium $={\rho }_{He}=0.18\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

Mass of skin of balloon $=m_b=1.5\text{ kg}$

Weight of skin of balloon $=w_b$

Weight of the load $=w_L=900\text{ N}$

Mass of the load $=m_L$

Force of buoyancy $=F_B$

Volume of the balloon $=V_b=2(69.67)=139.34{\text{ m}}^3$

Acceleration of the balloon $=a$

Formula Used:

Weight is given as

  $w=mg$

Force of buoyancy is given as

  $F_B=\rho Vg$

Mass is given as

  $\begin{array}{l}\text{mass =}\frac{\text{weight}}{g}\\ \text{m = }\frac{w}{g}\end{array}$

According to Newton’s second law, we have

  $F_{net}=ma$

Calculation:

Weight of the skin of the balloon is given as

  $\begin{array}{l}{w}_b=m_{b}g\\ w_b=(1.5)(9.8)\\ w_b=14.7\text{ N}\end{array}$

Weight of the helium in the balloon in downward direction is given as

  $w_{He}={\rho }_{He}{V}_{b}g$

Force of buoyancy on the balloon in upward direction is given as

  $F_B={\rho }_a{V}_{b}g$

Mass of the load is given as

  $\begin{array}{l}{m}_L=\frac{w_L}{g}\\ m_L=\frac{900}{9.8}\\ m_L=91.8\text{ kg}\end{array}$

Using the forces in vertical direction, the force equation as

  $\begin{array}{l}{F}_B-w_b-w_{He}-w_L=(m_b+m_{He}+m_L)a\\ F_B-w_b-w_{He}-w_L=(m_b+{\rho }_{He}{V}_b+m_L)a\\ {\rho }_a{V}_{b}g-w_b-{\rho }_{He}{V}_{b}g-w_L=(m_b+{\rho }_{He}{V}_b+m_L)a\\ (1.3)(139.34)(9.8)-14.7-(0.18)(139.34)(9.8)-900=(1.5+(0.18)(139.34)+91.8)a\\ 614.7=(118.4)a\\ a=5.2{\text{ ms}}^{-2}\end{array}$

Conclusion:

Hence, the initial acceleration of the balloon comes out to be $5.2{\text{ ms}}^{-2}$ .