Chapter 1, Problem 1.53P

A sidewalk is to be constructed around a swimming pool that measures (10.0 ± 0.1) m by (17.0 ± 0.1) m. If the sidewalk is to measure (1.00 ± 0.01) m wide by (9.0 ± 0.1) cm thick, what volume of concrete is needed and what is the approximate uncertainty of this volume?

To determine

The required volume of concrete and approximate uncertainty of the volume.

Answer to Problem 1.53P

The required volume of concrete is $5.2{\text{m}}^3$ and approximate uncertainty of the volume is $3.0\%$.

Explanation of Solution

Given info: The length of swimming pool is $\left(10.0\pm 0.1\right)\text{m}$ and the width of swimming pool is $\left(17.0\pm 0.1\right)\text{m}$ and the width of the side walk is $\left(1.00\pm 0.01\right)\text{m}$ and the thickness of the sidewalk is $\left(9.0\pm 0.1\right)\text{cm}$.

The length of the side walk in long side is,

$\begin{array}{c}{l}_1=17.0\text{m}+1.00\text{m}+1.00\text{m}\\ \text{=19}\text{.0}\text{m}\end{array}$

The total length in small side is,

$l_2=10.0\text{m}$

Formula to calculate the volume of rectangle in long side is,

$V_1=l_1\times b\times h$ (1)

Here,

$b$ is the width of sidewalk.

$h$ is the thickness of sidewalk.

Substitute $\text{19}\text{.0}\text{m}$ for $l_1$, $1.00\text{m}$ for $b$ and $9.00\text{cm}$ for $h$ in the equation (1).

$\begin{array}{c}{V}_1=\text{19}\text{.0}\text{m}\times 1.00\text{m}\times \left(9.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =1.71{\text{m}}^3\end{array}$

Thus, the volume of rectangle in long side is $1.71{\text{m}}^3$.

Formula to calculate the volume of rectangle in small side is,

$V_2=l_2\times b\times h$ (2)

Substitute $10.0\text{m}$ for $l_2$, $1.00\text{m}$ for $b$ and $9.00\text{cm}$ for $h$ in the equation (1).

$\begin{array}{c}{V}_2=10.0\text{m}\times 1.00\text{m}\times \left(9.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.90{\text{m}}^3\end{array}$

Thus, the volume of rectangle in small side is $0.90{\text{m}}^3$.

Write the expression for total volume of the concrete needed,

$V=2\left(V_1+V_2\right)$

Substitute $1.71{\text{m}}^3$ for $V_1$ and $0.90{\text{m}}^3$ for $V_2$ in the above equation.

$\begin{array}{c}V=2\left(1.71{\text{m}}^3+0.90{\text{m}}^3\right)\\ =5.2{\text{m}}^3\end{array}$

Thus, the total volume of concrete required is $5.2{\text{m}}^3$.

The uncertainty of the volume is the sum of uncertainty in each dimension.

The total length with uncertainty in the long side is,

$\begin{array}{c}{l}_1{}^{\prime }=\left(17.0\pm 0.1\right)\text{m}+\left(1.00\pm 0.01\right)\text{m}+\left(1.00\pm 0.01\right)\text{m}\\ =19.12\text{m}\end{array}$

The total length with uncertainty in small side is,

$\begin{array}{c}{l}_2{}^{\prime }=\left(10.0\pm 0.1\right)\text{m}\\ \text{=10}\text{.1}\text{m}\end{array}$

Formula to calculate the volume with uncertainty in long side is,

$V_1{}^{\prime }=l_1{}^{\prime }\times b^{\prime }\times h^{\prime }$ (3)

Here,

$b^{\prime }$ is the width of sidewalk with uncertainty.

$h^{\prime }$ is the thickness of sidewalk with uncertainty.

Substitute $19.12\text{m}$ for $l_1{}^{\prime }$, $1.01\text{m}$ for $b^{\prime }$ and $9.1\text{cm}$ for $h^{\prime }$ in the equation (3).

$\begin{array}{c}{V^{\prime }}_1=19.12\text{m}\times 1.01\text{m}\times \left(9.1\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =1.75{\text{m}}^3\end{array}$

Thus, the volume with uncertainty in long side is $1.75{\text{m}}^3$.

Formula to calculate the volume with uncertainty in small side is,

$V_2{}^{\prime }=l_2{}^{\prime }\times b^{\prime }\times h^{\prime }$ (4)

Substitute $\text{10}\text{.1}\text{m}$ for $l_2{}^{\prime }$, $1.01\text{m}$ for $b^{\prime }$ and $9.1\text{cm}$ for $h^{\prime }$ in the equation (4).

$\begin{array}{c}{V}_2{}^{\prime }=\text{10}\text{.1}\text{m}\times 1.01\text{m}\times \left(9.1\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =0.92{\text{m}}^3\end{array}$

Thus, the volume with uncertainty in small side is $0.92{\text{m}}^3$.

Write the expression for total volume with uncertainty,

$V^{\prime }=2\left({V^{\prime }}_1+V_2{}^{\prime }\right)$

Substitute $1.75{\text{m}}^3$ for $V_1{}^{\prime }$ and $0.92{\text{m}}^3$ for $V_2{}^{\prime }$ in the above equation.

$\begin{array}{c}{V}^{\prime }=2\left(1.75{\text{m}}^3+0.92{\text{m}}^3\right)\\ =5.34{\text{m}}^3\end{array}$

The total volume with uncertainty is $5.34{\text{m}}^3$.

Write the expression for the uncertainty in the volume,

$\text{Uncertainty}=\left(\frac{V^{\prime }-V}{V}\right)\times 100\%$

Substitute $5.34{\text{m}}^3$ for $V^{\prime }$ and $5.2{\text{m}}^3$ for $V$ in the above equation.

$\begin{array}{c}\text{Uncertainty}=\left(\frac{5.34{\text{m}}^3-5.2{\text{m}}^3}{5.2{\text{m}}^3}\right)\times 100\%\\ \approx 3.0\%\end{array}$

Thus, the uncertainty in volume is $3.0\%$.

Conclusion:

Therefore, the required volume of concrete is $5.2{\text{m}}^3$ and approximate uncertainty of the volume is $3.0\%$.