Question

A cord is a volume of cut wood equal to a stack 8 ftlong, 4 ft wide, and 4 ft high. How many cords of wood are in1.6 m³?

Answer 1

Concepts and reason

The concepts used to solve the problem are the unitary method and proper conversions of units to standard system.

Initially, use the concept of unit and dimensions to convert the given parameters into the standard notations.

Finally, use the unitary methods to find the total number of cords and check every stage by using the dimensional analysis to avoid any mistakes.

Fundamentals

The expression for the volume of a cord is given below:

$V = lwh$

Here, the volume is $V$, the length is $l$, the width is $w$, and the height is $h$.

The expression for the number of cords is given below:

$N = \frac{{{V_{{\rm{total}}}}}}{V}$

Here, the number of cords is $N$ and the total volume of wood is ${V_{{\rm{total}}}}$.

The expression for the volume of a cord is given below:

$V = lwh$

Substitute $8\,{\rm{ft}}$ for $l$, $4\,{\rm{ft}}$ for $w$ and $4\,{\rm{ft}}$ for $h$.

$\begin{array}{c}\\V = \left( {8\,{\rm{ft}}} \right)\left( {4\,{\rm{ft}}} \right)\left( {4\,{\rm{ft}}} \right)\\\\ = 128\,{\rm{f}}{{\rm{t}}^3}\\\end{array}$

Convert the foot into SI units as follows:

$\begin{array}{c}\\1\,{\rm{ft}} = {\rm{30}}{\rm{.48}}\,{\rm{cm}}\left( {\frac{{0.01\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)\\\\ = 0.3048\,{\rm{m}}\\\\1\,{\rm{f}}{{\rm{t}}^3} = {\left( {0.3048} \right)^3}\,{{\rm{m}}^3}\\\\ = 0.0283\,{{\rm{m}}^{\rm{3}}}\\\end{array}$

Now, the volume of cord can be expressed in SI unit as follows:

$\begin{array}{c}\\128{\rm{f}}{{\rm{t}}^3} = 128{\rm{f}}{{\rm{t}}^3}\left( {\frac{{0.0283\,{{\rm{m}}^{\rm{3}}}}}{{1\,{\rm{f}}{{\rm{t}}^3}}}} \right)\\\\ = 3.6224\,{{\rm{m}}^{\rm{3}}}\\\end{array}$

The volume of the cord is found out to be $3.6224\,{{\rm{m}}^{\rm{3}}}$ in SI units.

The expression for the number of cords is given below:

$N = \frac{{{V_{{\rm{total}}}}}}{V}$

Substitute $3.6224\,{{\rm{m}}^{\rm{3}}}$ for $V$ and $1.6\,{{\rm{m}}^{\rm{3}}}$ ${V_{{\rm{total}}}}$.

$\begin{array}{c}\\N = \frac{{\left( {1.6\,{{\rm{m}}^{\rm{3}}}} \right)}}{{\left( {3.6224\,{{\rm{m}}^{\rm{3}}}} \right)}}\\\\ = 0.442\\\end{array}$

Ans:

The number of cords from the given volume is ${\bf{0}}{\bf{.442}}$.