Chapter 3, Problem 9AP

Summary Introduction

To determine: The density of different objects and rank them in the order of their densities.

Answer to Problem 9AP

The density of the object A is $12,244\text{lb}/{\text{ft}}^{\text{3}}$, object B is $212,376\text{lb}/{\text{ft}}^{\text{3}}$, object C is $64\text{lb}/{\text{ft}}^{\text{3}}$, object D is $20\text{lb}/{\text{ft}}^{\text{3}}$, and object E is $93,714\text{lb}/{\text{ft}}^{\text{3}}$. Hence, the objects are arranged in the order of higher densities to lower as $D_{\text{B}}>D_{\text{E}}>D_{\text{A}}>D_{\text{C}}>D_{\text{D}}$.

Explanation of Solution

Calculation:

Write an expression to find the density of object A.

$D_{\text{A}}=\frac{\text{Mass}}{\text{Volume}}$

Here, $D_{\text{A}}$ is the volume of the object A.

Substitute 50 kg for mass and $15.00{\text{in}}^{\text{3}}$ for volume to find the density of the object A.

$\begin{array}{c}{D}_{\text{A}}=\frac{\text{50}\text{kg}}{\text{15}\text{.00}{\text{in}}^3}\\ =\frac{\text{50}\text{kg}\left(\frac{2.20462\text{lb}}{1\text{kg}}\right)}{\text{15}\text{.00}{\text{in}}^3\left(\frac{{\left(0.0833\text{ft}\right)}^3}{{\left(1\text{in}\right)}^3}\right)}\\ =\frac{110.2\text{lb}}{0.009{\text{ft}}^{\text{3}}}\\ \approx 12,244\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Therefore, the density of the object A is $12,244\text{lb}/{\text{ft}}^{\text{3}}$.

Write an expression to find the density of object B.

$D_B=\frac{\text{Mass}}{\text{Volume}}$

Here, $D_{\text{B}}$ is the volume of the object B.

Substitute 90 lb for mass and $12.00{\text{cm}}^{\text{3}}$ for volume to find the density of the object B.

$\begin{array}{c}{D}_{\text{B}}=\frac{\text{90}\text{lb}}{\text{12}\text{.00}{\text{cm}}^3}\\ =\frac{\text{90}\text{lb}}{\text{12}\text{.00}{\text{cm}}^3\left(\frac{{\left(\text{0}\text{.0328}\text{ft}\right)}^3}{{\left(1\text{cm}\right)}^3}\right)}\\ =\frac{90\text{lb}}{0.004{\text{ft}}^{\text{3}}}\\ \approx 212,376\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Therefore, the density of the object B is $212,376\text{lb}/{\text{ft}}^{\text{3}}$.

Write an expression to find the density of object C.

$D_C=\frac{\text{Mass}}{\text{Volume}}$

Here, $D_{\text{C}}$ is the volume of the object C.

Substitute 3 slugs for mass and $1.50{\text{ft}}^{\text{3}}$ for volume to find the density of the object C.

$\begin{array}{c}{D}_{\text{C}}=\frac{\text{3}\text{slugs}}{\text{1}\text{.50}{\text{ft}}^3}\\ =\frac{\text{3}\text{slugs}\left(\frac{32.174\text{lb}}{1\text{slugs}}\right)}{\text{1}\text{.50}{\text{ft}}^3}\\ =\frac{96.522\text{lb}}{\text{1}\text{.50}{\text{ft}}^3}\\ \approx 64\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Therefore, the density of the object C is $64\text{lb}/{\text{ft}}^{\text{3}}$.

Write an expression to find the density of object D.

$D_{\text{D}}=\frac{\text{Mass}}{\text{Volume}}$

Here, $D_{\text{D}}$ is the volume of the object D.

Substitute 450 N for mass and $0.14{\text{m}}^{\text{3}}$ for volume to find the density of the object D.$\begin{array}{c}{D}_{\text{D}}=\frac{\text{450}\text{N}}{0.14{\text{m}}^3}\\ =\frac{\text{450}\text{N}\left(\frac{0.2248\text{lb}}{1\text{N}}\right)}{0.14{\text{m}}^3\left(\frac{{\left(\text{3}\text{.28084}\text{ft}\right)}^3}{1{\text{m}}^{\text{3}}}\right)}\\ =\frac{101.16\text{lb}}{4.944{\text{ft}}^3}\\ \approx 20\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Therefore, the density of the object D is $20\text{lb}/{\text{ft}}^{\text{3}}$.

Write an expression to find the density of object E.

$D_{\text{E}}=\frac{\text{Mass}}{\text{Volume}}$

Here, $D_{\text{E}}$ is the volume of the object E.

Substitute 45 kg for mass and $30.00{\text{cm}}^{\text{3}}$ for volume to find the density of the object E.

$\begin{array}{c}{D}_{\text{E}}=\frac{\text{45}\text{kg}}{\text{30}\text{.00}{\text{cm}}^3}\\ =\frac{\text{45}\text{kg}\left(\frac{2.20462\text{lb}}{1\text{kg}}\right)}{\text{30}\text{.00}{\text{in}}^3\left(\frac{{\left(\text{0}\text{.0328}\text{ft}\right)}^3}{{\left(1\text{cm}\right)}^3}\right)}\\ =\frac{99.2\text{lb}}{0.001{\text{ft}}^{\text{3}}}\\ \approx 93,714\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Therefore, the density of the object E is $93,714\text{lb}/{\text{ft}}^{\text{3}}$.

Conclusion

Therefore, the density of the object A is $12,244\text{lb}/{\text{ft}}^{\text{3}}$, object B is $212,376\text{lb}/{\text{ft}}^{\text{3}}$, object C is $64\text{lb}/{\text{ft}}^{\text{3}}$, object D is $20\text{lb}/{\text{ft}}^{\text{3}}$, and object E is $93,714\text{lb}/{\text{ft}}^{\text{3}}$. Hence, the objects are arranged in the order of higher densities to lower as $D_{\text{B}}>D_{\text{E}}>D_{\text{A}}>D_{\text{C}}>D_{\text{D}}$.