Chapter 5.1, Problem 42PS

Solution

(a)

To determine: The change in $P$ when the width and length of the beam are doubled.

There will be no change in $P$.

Given information:

The load $P$ (in pounds) that can be safely supported by a horizontal beam varies jointly with the beam's width $W$ and the square of its depth $D$ and inversely with its unsupported length $L$.

Formula used:

Two variables $x\text{ and }y$ show inverse variation if they are related as below:

  $y=\frac{a}{x}$.

Two variables $x\text{ and }y$ show direct variation if they are related as below:

  $y=ax$.

  $z$ varies jointly with $x\text{ and }y$ is shown as:

  $z=axy$.

Where, $a$ is a constant.

Explanation:

The load $P$ varies jointly with the beam's width $W$ and the square of its depth $D$ , so

  $P=aW{D}^2$ …… (1)

The load $P$ varies inversely with its unsupported length $L$ , so

  $P=\frac{a}{L}$ …… (2)

From (1) and (2),

  $P=\frac{aW{D}^2}{L}$

Now, if the width and length of the beam are doubled, then

  $\begin{array}{c}P=\frac{a2W{D}^2}{2L}\\ =\frac{aW{D}^2}{L}\end{array}$

Therefore, there will be no change in $P$.

(b)

To determine: The change in $P$ when the width and depth of the beam are doubled.

The value of $P$ will increase by eight times.

Given information:

The load $P$ (in pounds) that can be safely supported by a horizontal beam varies jointly with the beam's width $W$ and the square of its depth $D$ and inversely with its unsupported length $L$.

Formula used:

Two variables $x\text{ and }y$ show inverse variation if they are related as below:

  $y=\frac{a}{x}$.

Two variables $x\text{ and }y$ show direct variation if they are related as below:

  $y=ax$.

  $z$ varies jointly with $x\text{ and }y$ is shown as:

  $z=axy$.

Where, $a$ is a constant.

Explanation:

The load $P$ varies jointly with the beam's width $W$ and the square of its depth $D$ , so

  $P=aW{D}^2$ …… (1)

The load $P$ varies inversely with its unsupported length $L$ , so

  $P=\frac{a}{L}$ …… (2)

From (1) and (2),

  $P=\frac{aW{D}^2}{L}$

Now, if the width and depth of the beam are doubled, then

  $\begin{array}{c}P=\frac{a\left(2W\right){\left(2D\right)}^2}{L}\\ =\frac{8aW{D}^2}{L}\end{array}$

Therefore, the value of $P$ will increase by eight times.

(c)

To determine: The change in $P$ when all the three dimensions are doubled.

The value of $P$ will increase by four times.

Given information:

The load $P$ (in pounds) that can be safely supported by a horizontal beam varies jointly with the beam's width $W$ and the square of its depth $D$ and inversely with its unsupported length $L$.

Formula used:

Two variables $x\text{ and }y$ show inverse variation if they are related as below:

  $y=\frac{a}{x}$.

Two variables $x\text{ and }y$ show direct variation if they are related as below:

  $y=ax$.

  $z$ varies jointly with $x\text{ and }y$ is shown as:

  $z=axy$.

Where, $a$ is a constant.

Explanation:

The load $P$ varies jointly with the beam's width $W$ and the square of its depth $D$ , so

  $P=aW{D}^2$ …… (1)

The load $P$ varies inversely with its unsupported length $L$ , so

  $P=\frac{a}{L}$ …… (2)

From (1) and (2),

  $P=\frac{aW{D}^2}{L}$

Now, if all the three dimensions are doubled, then

  $\begin{array}{c}P=\frac{a\left(2W\right){\left(2D\right)}^2}{2L}\\ =\frac{4aW{D}^2}{L}\end{array}$

Therefore, the value of $P$ will increase by four times.

(d)

To determine: The ways a beam can be modified if the safe load it is required to support is increased by a factor of $4$.

The unsupported length will be multiplied by the same factor of $4$.

Given information:

The load $P$ (in pounds) that can be safely supported by a horizontal beam varies jointly with the beam's width $W$ and the square of its depth $D$ and inversely with its unsupported length $L$.

Formula used:

Two variables $x\text{ and }y$ show inverse variation if they are related as below:

  $y=\frac{a}{x}$.

Two variables $x\text{ and }y$ show direct variation if they are related as below:

  $y=ax$.

  $z$ varies jointly with $x\text{ and }y$ is shown as:

  $z=axy$.

Where, $a$ is a constant.

Explanation:

Consider the equation $P=\frac{aW{D}^2}{L}$.

The factors of $4=1,2,\text{ and }4$.

If the safe load it is required to support is increased by $2$ , then

  $\begin{array}{l}2P=\frac{aW{D}^2}{L}\\ P=\frac{aW{D}^2}{2L}\end{array}$

If the safe load it is required to support is increased by $4$ , then

  $\begin{array}{l}4P=\frac{aW{D}^2}{L}\\ P=\frac{aW{D}^2}{4L}\end{array}$

Therefore, if the safe load it is required to support is increased by a factor of $4$ then the unsupported length will be multiplied by the same factor of $4$.