Chapter 8.2, Problem 11P

To determine

To calculate: The load the beam can support when turned on its side.

Answer to Problem 11P

The beam can support $378\text{ kg}$ of load when turned on its side.

Explanation of Solution

Given information:

The following variation:

The load a beam with a rectangular cross section can support is jointly proportional to the beam’s width and square of its depth and inversely proportional to is length.

When width is $3\text{ cm}$ and depth is $5\text{ cm}$ , the beam can support a load of $\text{630 kg}$ .

Calculation:

Let L be load, w be width and d be depth. Then, L varies jointly as w and $d^2$ ,

  $L=kw{d}^2$

where k represents proportionality constant and length.

Substituting $L=630\text{ kg}$ , $w=3\text{ cm}$ and $d=5\text{ cm}$ ,

  $\begin{array}{l}L=kw{d}^2[\text{Equation]}\\ 630=k\left(3\right){\left(5\right)}^2[\text{Substitute]}\\ 630=k\left(3\right)\left(25\right)[\text{Find square]}\\ 630=75k[\text{Multiply]}\\ k=8.4[\text{Divide both sides by 75]}\end{array}$

Therefore, an equation of the given variation is $L=8.4w{d}^2$

When beam is turned on its side, $w=5\text{ cm}$ and $d=3\text{ cm}$ . Substituting this value in $L=8.4w{d}^2$ ,

  $\begin{array}{l}L=8.4w{d}^2[\text{Equation]}\\ L=8.4\left(5\right){\left(3\right)}^2[\text{Substitute]}\\ L=8.4\left(5\right)\left(9\right)[\text{Find square]}\\ L=378[\text{Simplify]}\end{array}$