Chapter 46, Problem 53A

To determine

(a)

To perform the mathematical operation on formula $\left(1\right)$ for $D_o$ as indicated.

Answer to Problem 53A

The formula $\left(1\right)$ that is $FW=\sqrt{D_o^2-D^2}$ can be solved for $D_o$ as $D_o=\sqrt{{\left(FW\right)}^2+D^2}$.

Explanation of Solution

Given information:

Given that, the dimensions shown in the figure can be found using these two formulas.

  $\left(1.\right)FW=\sqrt{D_o^2-D^2}\left(2.\right)D_o=2C-d+2a$

  

Calculation:

We have been given the formula, $\left(1.\right)FW=\sqrt{D_o^2-D^2}and\left(2.\right)D_o=2C-d+2a$ to determine the dimensions shown in the figure

We need to solve the formula $\left(1\right)$ for $D_o$.

Formula $\left(1\right)$ states that,

  $\Rightarrow FW=\sqrt{D_o^2-D^2}$

Taking square of both sides, we get

  $\Rightarrow {\left(FW\right)}^2=D_o^2-D^2$

Adding $D^2$ on both sides, we get

  $\Rightarrow {\left(FW\right)}^2+D^2=D_o^2$

Taking square root of both sides, we get

  $\Rightarrow D_o=\sqrt{{\left(FW\right)}^2+D^2}$

Hence, the formula $\left(1\right)$ that is $FW=\sqrt{D_o^2-D^2}$ can be solved for $D_o$ as $D_o=\sqrt{{\left(FW\right)}^2+D^2}$.

To determine

(b)

To solve formula $\left(1\right)$ for $D$.

Answer to Problem 53A

The formula $\left(1\right)$ that is $FW=\sqrt{D_o^2-D^2}$ can be solved for $D$ as $D=\sqrt{{\left(FW\right)}^2+D_o{}^2}$.

Explanation of Solution

Given information:

Given that, the dimensions shown in the figure can be found using these two formulas.

  $\left(1.\right)FW=\sqrt{D_o^2-D^2}\left(2.\right)D_o=2C-d+2a$

  

Calculation:

We have been given the formula, $\left(1.\right)FW=\sqrt{D_o^2-D^2}and\left(2.\right)D_o=2C-d+2a$ to determine the dimensions shown in the figure

We need to solve the formula $\left(1\right)$ for $D$.

Formula $\left(1\right)$ states that,

  $\Rightarrow FW=\sqrt{D_o^2-D^2}$

Taking square of both sides, we get

  $\Rightarrow {\left(FW\right)}^2=D_o^2-D^2$

Adding $D_o^2$ on both sides, we get

  $\Rightarrow {\left(FW\right)}^2+D_o^2=D^2$

Taking square root of both sides, we get

  $\Rightarrow D=\sqrt{{\left(FW\right)}^2+D_o{}^2}$

Hence, the formula $\left(1\right)$ that is $FW=\sqrt{D_o^2-D^2}$ can be solved for $D$ as $D=\sqrt{{\left(FW\right)}^2+D_o{}^2}$.

To determine

(c)

To perform the mathematical operation on formula $\left(2\right)$ for $d$.

Answer to Problem 53A

The formula $\left(2\right)$ that is $D_o=2C-d+2a$ can be solved for $d$ as $d=2a+2C-D_o$.

Explanation of Solution

Given that, the dimensions shown in the figure can be found using these two formulas.

  $\left(1.\right)FW=\sqrt{D_o^2-D^2}\left(2.\right)D_o=2C-d+2a$

  

Calculation:

We have been given the formula, $\left(1.\right)FW=\sqrt{D_o^2-D^2}and\left(2.\right)D_o=2C-d+2a$ to determine the dimensions shown in the figure

We need to solve the formula $\left(2\right)$ for $d$.

Formula $\left(2\right)$ states that,

  $\Rightarrow D_o=2C-d+2a$

Adding $d$ on both sides, we get

  $\Rightarrow D_o+d=2C+2a$

Subtracting $D_o$ on both sides, we get

  $\Rightarrow d=2C+2a-D_o$

Hence, the formula $\left(2\right)$ that is $D_o=2C-d+2a$ can be solved for $d$ as $d=2a+2C-D_o$.

To determine

(d)

To solve formula $\left(2\right)$ for $a$.

Answer to Problem 53A

The formula $\left(2\right)$ that is $D_o=2C-d+2a$ can be solved for $a$ as $a=\frac{D_o+d-2C}{2}$.

Explanation of Solution

Given that, the dimensions shown in the figure can be found using these two formulas.

  $\left(1.\right)FW=\sqrt{D_o^2-D^2}\left(2.\right)D_o=2C-d+2a$

  

Calculation:

We have been given the formula, $\left(1.\right)FW=\sqrt{D_o^2-D^2}and\left(2.\right)D_o=2C-d+2a$ to determine the dimensions shown in the figure

We need to solve the formula $\left(2\right)$ for $a$.

Formula $\left(2\right)$ states that,

  $\Rightarrow D_o=2C-d+2a$

Adding $d$ on both sides, we get

  $\Rightarrow D_o+d=2C+2a$

Subtracting $2C$ on both sides, we get

  $\Rightarrow D_o+d-2C=2a$

Dividing by $2$ on both sides, we get

  $\Rightarrow a=\frac{D_o+d-2C}{2}$

Hence, the formula $\left(2\right)$ that is $D_o=2C-d+2a$ can be solved for $a$ as $a=\frac{D_o+d-2C}{2}$.