Chapter 1.10, Problem 74E

a .

To determine

how doubling the width affects the beam’s maximum load.

Answer to Problem 74E

New maximum load is double of the original load.

Explanation of Solution

Concept Used:

For the variables x , y and z , the statement z varies jointly as x and y can be translated as:

  $y=kxy$

Where k is a non-zero constant called the constant of proportionality.

For the variables x and y , the statement y varies inversely as x can be translated as:

  $y=\frac{k}{x}$

Where k is a non-zero constant called the constant of proportionality.

Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.

Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.

Using above definitions, the given statement can be translated as:

  $L=\frac{kw{d}^2}{l}$

Here, the expression $L=\frac{kw{d}^2}{l}$ gives the original maximum load of beam.

Now, find the new maximum load of beam $L_1$ , when width is doubled, that is replacing w by $2w$ .

  $\begin{array}{c}{L}_1=\frac{k\left(2w\right)d^2}{l}\\ =2\left(\frac{kw{d}^2}{l}\right)\\ =2L\end{array}$

Conclusion:

New maximum load is double of the original load.

b.

To determine

how doubling the depth affects the beam’s maximum load.

Answer to Problem 74E

New maximum load is four times the original load.

Explanation of Solution

Concept Used:

For the variables x , y and z , the statement z varies jointly as x and y can be translated as:

  $y=kxy$

Where k is a non-zero constant called the constant of proportionality.

For the variables x and y , the statement y varies inversely as x can be translated as:

  $y=\frac{k}{x}$

Where k is a non-zero constant called the constant of proportionality.

Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.

Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.

Using above definitions, the given statement can be translated as:

  $L=\frac{kw{d}^2}{l}$

Here, the expression $L=\frac{kw{d}^2}{l}$ gives the original maximum load of beam.

Now, find the new maximum load of beam $L_2$ , when depth is doubled, that is replacing d by $2d$ .

  $\begin{array}{c}{L}_2=\frac{kw{\left(2d\right)}^2}{l}\\ =4\left(\frac{kw{d}^2}{l}\right)\\ =4L\end{array}$

Conclusion:

New maximum load is four times the original load.

c.

To determine

how halving the length affects the beam’s maximum load.

Answer to Problem 74E

New maximum load is double of the original load.

Explanation of Solution

Concept Used:

For the variables x , y and z , the statement z varies jointly as x and y can be translated as:

  $y=kxy$

Where k is a non-zero constant called the constant of proportionality.

For the variables x and y , the statement y varies inversely as x can be translated as:

  $y=\frac{k}{x}$

Where k is a non-zero constant called the constant of proportionality.

Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.

Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.

Using above definitions, the given statement can be translated as:

  $L=\frac{kw{d}^2}{l}$

Here, the expression $L=\frac{kw{d}^2}{l}$ gives the original maximum load of beam.

Now, find the new maximum load of beam $L_3$ , when length is halved, that is replacing l by $\frac{l}{2}$ .

  $\begin{array}{c}{L}_3=\frac{kw{d}^2}{\frac{l}{2}}\\ =2\left(\frac{kw{d}^2}{l}\right)\\ =2L\end{array}$

Conclusion:

New maximum load is double of the original load.

d.

To determine

how halving the width and doubling the length affects the beam’s maximum load.

Answer to Problem 74E

New maximum load is one-fourth of the original load.

Explanation of Solution

Concept Used:

For the variables x , y and z , the statement z varies jointly as x and y can be translated as:

  $y=kxy$

Where k is a non-zero constant called the constant of proportionality.

For the variables x and y , the statement y varies inversely as x can be translated as:

  $y=\frac{k}{x}$

Where k is a non-zero constant called the constant of proportionality.

Now, it is given that, the maximum load that a horizontal beam can safely support varies jointly as the width of the beam and the square of its depth and inversely as the length of the beam.

Let w denotes the width of the beam, d denotes the depth of the beam, l denotes the length of the beam and L denotes the beam’s maximum load.

Using above definitions, the given statement can be translated as:

  $L=\frac{kw{d}^2}{l}$

Here, the expression $L=\frac{kw{d}^2}{l}$ gives the original maximum load of beam.

Now, find the new maximum load of beam $L_4$ , when width is halved and the length is doubled, that is replace w by $\frac{w}{2}$ and l by $2l$ .

  $\begin{array}{c}{L}_4=\frac{k\left(\frac{w}{2}\right)d^2}{2l}\\ =\frac{kw{d}^2}{4l}\\ =\frac{1}{4}\left(\frac{kw{d}^2}{l}\right)\\ =\frac{1}{4}L\end{array}$

Conclusion:

New maximum load is one-fourth of the original load.