Chapter 9, Problem 102P

(a)

To determine

The weight of beach ball including the air inside it.

Answer to Problem 102P

The weight of beach ball is $1.4\text{N}$.

Explanation of Solution

Radius of ball is $20.0\text{cm}$, mass of ball is $0.10\text{kg}$, and the air density is $1.3\text{kg}/{\text{m}}^{\text{3}}$.

Write the equation for the weight of ball with air inside it.

$W=m_{b}g+\frac{4}{3}\pi r^3{\rho }_{a}g$

Here, $W$ is the weight of ball with air, $m_b$  is the mass of ball, $g$ is the acceleration due to gravity, $r$ is the radius of ball, and ${\rho }_a$ is the density of air.

Conclusion:

Substitute $0.10\text{kg}$ for $m_b$, $9.8{\text{m}/\text{s}}^2$ for $g$, $20.0\text{cm}$ for $r$, $3.14$ for $\pi$, and $1.3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_a$ in the above equation to find $W$.

$\begin{array}{c}W=\left(\left(0.10\text{kg}\right)\left(9.8{\text{m}/\text{s}}^2\right)\right)+\left(\frac{4}{3}\left(3.14\right){\left(20.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}^3\left(1.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(\left(9.8{\text{m}/\text{s}}^2\right)\right)\right)\\ =0.98\text{N+0}\text{.12}\text{N}\\ \text{=1}\text{.4}\text{N}\end{array}$

Therefore, the weight is $1.4\text{N}$.

(b)

To determine

The buoyant force on beach ball.

Answer to Problem 102P

The buoyant force is $0.43\text{N}\text{in}\text{upward}\text{direction}$.

Explanation of Solution

Radius of ball is $20.0\text{cm}$, mass of ball is $0.10\text{kg}$, and the air density is $1.3\text{kg}/{\text{m}}^{\text{3}}$.

Write the equation for buoyant force on each ball.

$F_B={\rho }_{a}g\left(\frac{4}{3}\pi r^3\right)$

Here, $F_B$ is the buoyant force.

Conclusion:

Substitute $9.8{\text{m}/\text{s}}^2$ for $g$, $20.0\text{cm}$ for $r$, $3.14$ for $\pi$, and $1.3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_a$ in the above equation to find $F_B$.

$\begin{array}{c}{F}_B=\left(1.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.8{\text{m}/\text{s}}^2\right)\frac{4}{3}\left(3.14\right){\left(20.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}^3\\ =0.43\text{N}\end{array}$

Buoyant force will acts in opposite direction of weight. That is directed upwards.

Therefore, the buoyant force is $0.43\text{N}\text{in}\text{upward}\text{direction}$.

(c)

To determine

The acceleration of ball at the top of its path.

Answer to Problem 102P

The acceleration is $6.8\text{m}/{\text{s}}^{\text{2}}\text{in}\text{downward}\text{direction}$.

Explanation of Solution

Radius of ball is $20.0\text{cm}$, mass of ball is $0.10\text{kg}$, and the air density is $1.3\text{kg}/{\text{m}}^{\text{3}}$.

The free body diagram is shown below in figure 1.

Write the equation net force on the ball.

$ma=F_B-mg$

Here, $a$ is the acceleration of ball at the top of its path and $m$ is the mass of ball with air.

Rewrite the above equation in terms of $a$.

$\begin{array}{c}a=\frac{F_B-mg}{m}\\ =\frac{F_B}{m}-g\end{array}$

Write the equation to find $m$.

$m=m_b+{\rho }_a\left(\frac{4}{3}\pi r^3\right)$

Rewrite the equation for $a$ by substituting the above relation for $m$ and ${\rho }_{a}g\left(\frac{4}{3}\pi r^3\right)$ for $F_B$.

$\begin{array}{c}a=\frac{F_B-mg}{m}\\ =\frac{{\rho }_{a}g\left(\frac{4}{3}\pi r^3\right)}{m_b+{\rho }_a\left(\frac{4}{3}\pi r^3\right)}-g\\ =\frac{g}{\left(\frac{m_b}{{\rho }_a\left(\frac{4}{3}\pi r^3\right)}+1\right)}-g\\ =g\left(1+{\left(\frac{m_b}{{\rho }_a\left(\frac{4}{3}\pi r^3\right)}\right)}^{-1}-1\right)\end{array}$

Conclusion:

Substitute $9.8{\text{m}/\text{s}}^2$ for $g$, $0.10\text{kg}$ for $m_b$, $20.0\text{cm}$ for $r$, $3.14$ for $\pi$, and $1.3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_a$ in the above equation to find $F_B$.

$\begin{array}{c}a=\left(9.8{\text{m}/\text{s}}^2\right)\left(1+{\left(\frac{0.10\text{kg}}{\left(1.3\text{kg}/{\text{m}}^{\text{3}}\right)\frac{4}{3}\left(3.14\right){\left(20.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}^3}\right)}^{-1}-1\right)\\ =\left(9.8{\text{m}/\text{s}}^2\right)\left(-0.694\right)\\ =-6.8\text{m}/{\text{s}}^{\text{2}}\end{array}$

Negative value indicates that the acceleration is directed in downward direction.

Therefore, the acceleration is $6.8\text{m}/{\text{s}}^{\text{2}}\text{in}\text{downward}\text{direction}$.