Question #ff26c

Feb 5, 2015

The mass of the wall will be $\text{27300 lbs}$.

You're dealing with a pretty straightforward unit conversion problem. Like with most such problems, there are many ways in which to reach the desired result, which in this case is mass expressed in pounds.

I'll show you one of those ways. So, since density is given to you in ${\text{g/cm}}^{3}$, you can convert it to ${\text{kg/m}}^{3}$ by using conversion factors

$1.74 {\text{g"/"cm"^3 * (10^(6)"cm"^3)/("1 m"^3) * ("1 kg")/("1000 g") = "1740 kg/m}}^{3}$

Now you need to determine the volume of the wall. Since you're basically dealing with a rectangular prism, the volume will be equal to

$V = L \cdot l \cdot h = {\text{36.5 ft" * "7.5 ft" * "0.917 ft" = "251.0 ft}}^{3}$

You need however to express this volume in ${\text{m}}^{3}$, since that is the unit used in the expression of density. So,

${\text{251.0 ft"^3 * ("0.02832 m"^(3))/("1 ft"^3) = "7.11 m}}^{3}$

At this point, you have what you need to determine the mass of the wall in $\text{kg}$.

$\rho = \frac{m}{V} \implies m = \rho \cdot V = 1740 \text{kg"/"m"^3 * "7.11 m"^3 = "12371.4 kg}$

All you need to do from this point on is to convert the mass to pounds

$\text{12371.4 kg" * ("2.2046 lbs")/("1 kg") = "27274 lbs}$, or $\text{27300 lbs}$ rounded to two sig figs.