Chapter 13, Problem 54SP

Determine the unbalanced force acting on an iron ball $\left(r=1.5\text{ cm},\rho =7.8{\text{ g/cm}}^{\text{3}}\right)$ when just released while totally immersed in (a) water and (b) mercury $\left(\rho =13.6{\text{ g/cm}}^{\text{3}}\right)$. What will be the initial acceleration of the ball in each case?

To determine

The unbalanced force and the initial acceleration of the iron ball when it is just released while totally immersed in water if $r=1.5\text{ cm}$ and ${\rho }_{ball}=7.8{\text{ g/cm}}^3$.

Answer to Problem 54SP

Solution:

$0.94\text{ N}$ down, $8.6{\text{ m/s}}^2$ down

Explanation of Solution

Given data:

The density of the iron ball is $7.8{\text{ g/cm}}^3$

The radius of the iron ball is $1.5\text{ cm}$.

Formula used:

The volume of a sphere is expressed as,

$V=\frac{4}{3}\pi r^3$

Here, $V$ is the volume of the sphere and $r$ is the radius of the sphere.

The density of a body is calculated by the formula,

$\rho =\frac{m}{V}$

Here, $\rho$ is the density of the body and $m$ is the mass of the body.

Archimedes’ principle states that the apparent upward force experienced by the solid immersed in a fluid is equal to the weight of the fluid displaced by it. This upward buoyant force on the solid is expressed as,

$F_B={\rho }_l{V}_{S}g$

Here, $F_B$ is the buoyant force, ${\rho }_l$ is the density of the liquid in which the solid is submerged, $V_S$ is the volume of the solid submerged in the liquid, and $g$ is the acceleration due to gravity.

The weight of a body is expressed as,

$W=mg$

Here, $W$ is the weight of the body.

Newton’s second law for a body is expressed as,

${\displaystyle \sum F}=ma$

Here, $a$ is acceleration of the body in the direction of net force and ${\displaystyle \sum F}$ is the net force on the body.

Explanation:

Draw a free body diagram for the iron ball when the ball is submerged in water.

Here, $W$ is the force exerted due to the weight of the iron ball and $F_{B1}$ is the buoyant force on the iron ball exerted by water.

Understand that the ball is in the shape of a sphere. Recall the formula for the volume of a sphere to calculate the volume of the iron ball.

$V=\frac{4}{3}\pi r^3$

Substitute $1.5\text{ cm}$ for $r$

$\begin{array}{c}V=\frac{4}{3}\pi {\left(1.5\text{ cm}\right)}^3\\ =\frac{4}{3}\pi \left(3.375{\text{ cm}}^3\right)\\ =14.14{\text{ cm}}^3\end{array}$

Recall the formula for the density of the iron ball.

$\begin{array}{c}\rho =\frac{m}{V}\\ m=\rho V\end{array}$

Substitute $14.14{\text{ cm}}^3$ for $V$ and $7.8{\text{ g/cm}}^3$ for $\rho$

$\begin{array}{c}m=\left(7.8{\text{ g/cm}}^3\right)\left(14.14{\text{ cm}}^3\right)\\ =110.29\text{ g}\end{array}$

Recall the formula for the weight of the iron ball.

$W=mg$

The standard value of acceleration due to gravity is $9.81{\text{ m/s}}^2$. Substitute $110.29\text{ g}$ for $m$ and $9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}W=\left(110.29\text{ g}\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(110.29\text{ g}\right)\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\left(9.81{\text{ m/s}}^2\right)\\ =1.082\text{ N}\end{array}$

Understand that the ball is completely submerged in water. Therefore, write the expression for the volume of a ball submerged in water.

$V_S=V$

Substitute $14.14{\text{ cm}}^3$ for $V$

$V_S=14.14{\text{ cm}}^3$

Recall the expression for the buoyant force exerted by water on the iron ball.

$F_{B1}={\rho }_w{V}_{S}g$

Here, ${\rho }_w$ is the density of water.

The standard value of $g$ is $9.81{\text{ m/s}}^2$. The standard value of the density of water is $1000{\text{ kg/m}}^3$. Therefore, substitute $1000{\text{ kg/m}}^3$ for ${\rho }_w$, $14.14{\text{ cm}}^3$ for $V_S$, and $9.81{\text{ m/s}}^2$ for $g$.

$\begin{array}{c}{F}_{B1}=\left(1000{\text{ kg/m}}^3\right)\left(14.14{\text{ cm}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(1000{\text{ kg/m}}^3\right)\left(14.14{\text{ cm}}^3\right)\left(\frac{{10}^{-6}{\text{ m}}^3}{1{\text{ cm}}^3}\right)\left(9.81{\text{ m/s}}^2\right)\\ =0.139\text{N}\end{array}$

Write the expression for net force on the ball along the vertical direction by taking the upward force as positive and the downward force as negative.

${\displaystyle \sum F}=F_{B1}-W$

Substitute $0.139\text{N}$ for $F_{B1}$ and $1.082\text{ N}$ for $W$

$\begin{array}{c}{\displaystyle \sum F}=0.139\text{N}-1.082\text{ N}\\ \simeq -\text{0}\text{.94 N}\end{array}$

Here, the negative sign indicates that the force acts in the downward direction.

Therefore, the force on the iron ball when it is just released while immersed in water is $\text{0}\text{.94 N}$ in the downward direction.

Recall the expression for Newton’s second law in terms of net force.

${\displaystyle \sum F}=ma$

Substitute $-\text{0}\text{.94 N}$ for ${\displaystyle \sum F}$ and $110.29\text{ g}$ for $m$

$\begin{array}{c}-\text{0}\text{.94 N}=\left(110.29\text{ g}\right)a\\ a=\frac{-\text{0}\text{.94 N}}{110.29\text{ g}}\\ =\frac{-\text{0}\text{.94 N}}{110.29\text{ g}\left(\frac{1\text{ g}}{0.001\text{ kg}}\right)}\\ =\frac{-\text{0}\text{.94 N}}{0.11\text{ kg}}\end{array}$

Further solve as,

$a\simeq -8.6{\text{ m/s}}^2$

Here, the negative sign indicates that the acceleration acts in the downward direction.

Conclusion:

The net force on the iron ball when it is just released while immersed in water is $0.94\text{ N}$ down and the initial acceleration is $8.6{\text{ m/s}}^2$ down.

To determine

The unbalanced force and the initial acceleration of the iron ball with $r=1.5\text{ cm}$ and $\rho =7.8{\text{ g/cm}}^3$ when it is just released while totally immersed in mercury.

Answer to Problem 54SP

Solution:

$0.80\text{ N}$ up, $7.3{\text{ m/s}}^2$ up

Explanation of Solution

Given data:

The density of the iron ball is $7.8{\text{ g/cm}}^3$

The radius of the iron ball is $1.5\text{ cm}$.

The density of mercury is $13.6{\text{ g/cm}}^3$.

Formula used:

Archimedes’ principle states that the apparent upward force experienced by the solid immersed in a fluid is equal to the weight of the fluid displaced by it. This upward buoyant force on the solid is expressed as,

$F_B={\rho }_l{V}_{S}g$

Here, $F_B$ is the buoyant force, ${\rho }_l$ is the density of the liquid in which the solid is submerged, $V_S$ is the volume of the solid submerged in the liquid, and $g$ is the acceleration due to gravity.

The expression of Newton’s second law for a body is expressed as,

${\displaystyle \sum F}=ma$

Here, $a$ is acceleration of the body in the direction of the net force and ${\displaystyle \sum F}$ is the net force on the body.

Explanation:

Draw a free body diagram for the iron ball when the ball is submerged in mercury.

Here, $W$ is the force exerted due to the weight of the iron ball and $F_{B2}$ is the buoyant force on the iron ball exerted by mercury.

Understand that the ball is completely submerged in mercury. Therefore, write the expression for the volume of a ball submerged in mercury. Hence,

$V_S=V$

Substitute $14.14{\text{ cm}}^3$ for $V$

$V_S=14.14{\text{ cm}}^3$

Recall the expression for buoyant force exerted by mercury on the iron ball.

$F_{B2}={\rho }_{l1}{V}_{S}g$

Here, ${\rho }_{l1}$ is the density of mercury.

Consider, the standard value of $g$ is $9.81{\text{ m/s}}^2$.

Substitute $13.6{\text{ g/cm}}^3$ for ${\rho }_{l1}$, $14.14{\text{ cm}}^3$ for $V_S$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{B2}=\left(13.6{\text{ g/cm}}^3\right)\left(14.14{\text{ cm}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(13.6{\text{ g/cm}}^3\right)\left(14.14{\text{ cm}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(192.304\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)\\ =1.88\text{ N}\end{array}$

Write the expression for the net force on the ball in the vertical direction by taking the upward force as positive and the downward force as negative.

${\displaystyle \sum F}=F_{B2}-W$

Substitute $1.88\text{ N}$ for $F_{B2}$ and $1.082\text{ N}$ for $W$

$\begin{array}{c}{\displaystyle \sum F}=1.88\text{ N}-1.082\text{ N}\\ \simeq \text{0}\text{.80 N}\end{array}$

Here, the positive sign indicates that the force acts in the upward direction.

Therefore, the force on the iron ball when it is just released while immersed in mercury is $\text{0}\text{.80 N}$ in the upward direction.

Recall the expression for Newton’s second law for acceleration in terms of net force.

${\displaystyle \sum F}=ma$

Substitute $0.80\text{ N}$ for ${\displaystyle \sum F}$ and $110.29\text{ g}$ for $m$

$\begin{array}{c}0.80\text{ N}=\left(110.29\text{ g}\right)a\\ a=\frac{0.80\text{ N}}{110.29\text{ g}}\\ =\frac{0.80\text{ N}}{110.29\text{ g}}\left(\frac{1\text{ g}}{0.001\text{ kg}}\right)\\ =\frac{0.80\text{ N}}{0.11\text{ kg}}\end{array}$

Further solve as,

$a\simeq 7.3{\text{ m/s}}^2$

Here, the positive sign indicates that the acceleration acts in the upward direction.

Conclusion:

The net force on the iron ball when it is just released while immersed in mercury is $0.80\text{ N}$ up and the initial acceleration is $7.3{\text{ m/s}}^2$ up.