Chapter 20, Problem 44A

To determine

(a)

Length of pin extension in (a).

Answer to Problem 44A

Length of pin extension is $9.925\text{mm}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.25$ and diameter of the pin is $7.94\text{mm}$.

Concept used:

Relation between number of times pin extension is given as follows:

  $N=\frac{L}{D}$ ........ (1)

Here, number of times pin extension is $N$, length of the pin extension is $L$ and diameter of the pin is $D$.

Calculation:

Substitute $1.25$ for $N$ and $7.94\text{mm}$ for $D$ in equation (1).

  $\begin{array}{c}1.25=\frac{L}{7.94}\\ L=9.925\text{mm}\end{array}$

Thus, Length of pin extension is $9.925\text{mm}$.

Conclusion:

Length of pin extension is $9.925\text{mm}$.

To determine

(b)

Length of pin extension in (b).

Answer to Problem 44A

Length of pin extension is ${\frac{5}{8}}^{\prime \text{}\prime }$.

Explanation of Solution

Given:

The number of times the pin extension is $1\frac{1}{4}$ and diameter of the pin is ${\frac{1}{2}}^{\prime \text{}\prime }$.

Calculation:

Substitute $1\frac{1}{4}$ for $N$ and ${\frac{1}{2}}^{\prime \text{}\prime }$ for $D$ in equation (1).

  $\begin{array}{c}1\frac{1}{4}=\frac{L}{{\frac{1}{2}}^{\prime \text{}\prime }}\\ L=\frac{5}{4}\times \frac{1}{2}\\ L={\frac{5}{8}}^{\prime \text{}\prime }\end{array}$

Thus, Length of pin extension is ${\frac{5}{8}}^{\prime \text{}\prime }$.

Conclusion:

Length of pin extension is ${\frac{5}{8}}^{\prime \text{}\prime }$.

To determine

(c)

Diameter of the pin in (c).

Answer to Problem 44A

Diameter of the pin is ${\frac{3}{5}}^{\prime \text{}\prime }$.

Explanation of Solution

Given:

The number of times the pin extension is $1\frac{1}{4}$ and length of the pin is ${\frac{3}{4}}^{\prime \text{}\prime }$.

Calculation:

Substitute $1\frac{1}{4}$ for $N$ and ${\frac{3}{4}}^{\prime \text{}\prime }$ for $L$ in equation (1).

  $\begin{array}{c}1\frac{1}{4}=\frac{{\frac{3}{4}}^{\prime \text{}\prime }}{D}\\ D=\frac{{\frac{3}{4}}^{\prime \text{}\prime }}{\frac{5}{4}}\\ D={\frac{3}{5}}^{\prime \text{}\prime }\end{array}$

Thus, Diameter of the pin is ${\frac{3}{5}}^{\prime \text{}\prime }$.

Conclusion:

Diameter of the pin is ${\frac{3}{5}}^{\prime \text{}\prime }$.

To determine

(d)

Diameter of the pin in (d).

Answer to Problem 44A

Diameter of the pin is $6.349\text{mm}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.375$ and length of the pin is $8.73\text{mm}$.

Calculation:

Substitute $1.375$ for $N$ and $8.73\text{mm}$ for $L$ in equation (1).

  $\begin{array}{c}1.375=\frac{8.73}{D}\\ D=\frac{8.73}{1.375}\\ D=6.349\text{mm}\end{array}$

Thus, Diameter of the pin is $6.349\text{mm}$.

Conclusion:

Diameter of the pin is $6.349\text{mm}$.

To determine

(e)

Length of pin extension in (e).

Answer to Problem 44A

Length of pin extension is $20.15\text{mm}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.25$ and diameter of the pin is $16.12\text{mm}$.

Concept used:

Relation between number of times pin extension is given as follows:

  $N=\frac{L}{D}$ ........ (1)

Here, number of times pin extension is $N$, length of the pin extension is $L$ and diameter of the pin is $D$.

Calculation:

Substitute $1.25$ for $N$ and $16.12\text{mm}$ for $D$ in equation (1).

  $\begin{array}{c}1.25=\frac{L}{16.12}\\ L=20.15\text{mm}\end{array}$

Thus, Length of pin extension is $20.15\text{mm}$.

Conclusion:

Length of pin extension is $20.15\text{mm}$.

To determine

(f)

Diameter of the pin in (f).

Answer to Problem 44A

Diameter of the pin is $0.75054^{″}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.375$ and length of the pin is $1.032^{″}$.

Calculation:

Substitute $1.375$ for $N$ and $1.032^{″}$ for $L$ in equation (1).

  $\begin{array}{c}1.375=\frac{1.032}{D}\\ D=\frac{1.032^{″}}{1.375}\\ D=0.75054^{″}\end{array}$

Thus, Diameter of the pin is $0.75054^{″}$.

Conclusion:

Diameter of the pin is $0.75054^{″}$.

To determine

(g)

Length of pin extension in (g).

Answer to Problem 44A

Length of pin extension is $1.09375^{″}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.25$ and diameter of the pin is $0.875^{″}$.

Concept used:

Relation between number of times pin extension is given as follows:

  $N=\frac{L}{D}$ ........ (1)

Here, number of times pin extension is $N$, length of the pin extension is $L$ and diameter of the pin is $D$.

Calculation:

Substitute $1.25$ for $N$ and $0.875^{″}$ for $D$ in equation (1).

  $\begin{array}{c}1.25=\frac{L}{0.875}\\ L=1.09375^{″}\end{array}$

Thus, Length of pin extension is $1.09375^{″}$.

Conclusion:

Length of pin extension is $1.09375^{″}$.

To determine

(h)

Length of pin extension in (h).

Answer to Problem 44A

Length of pin extension is $5.52\text{mm}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.5$ and diameter of the pin is $3.68\text{mm}$.

Concept used:

Relation between number of times pin extension is given as follows:

  $N=\frac{L}{D}$ ........ (1)

Here, number of times pin extension is $N$, length of the pin extension is $L$ and diameter of the pin is $D$.

Calculation:

Substitute $1.5$ for $N$ and $3.68\text{mm}$ for $D$ in equation (1).

  $\begin{array}{c}1.5=\frac{L}{3.68}\\ L=5.52\text{mm}\end{array}$

Thus, Length of pin extension is $5.52\text{mm}$.

Conclusion:

Length of pin extension is $5.52\text{mm}$.

To determine

(i)

Diameter of the pin in (i).

Answer to Problem 44A

Diameter of the pin is $0.2497^{″}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.125$ and length of the pin is $0.281^{″}$.

Calculation:

Substitute $1.125$ for $N$ and $0.281^{″}$ for $L$ in equation (1).

  $\begin{array}{c}1.125=\frac{0.281}{D}\\ D=\frac{0.281}{1.125}\\ D=0.2497^{″}\end{array}$

Thus, Diameter of the pin is $0.2497^{″}$.

Conclusion:

Diameter of the pin is $0.2497^{″}$.

To determine

(j)

Length of pin extension in (j).

Answer to Problem 44A

Length of pin extension is $7.5\text{mm}$.

Explanation of Solution

Given:

The number of times the pin extension is $1.0$ and diameter of the pin is $7.5\text{mm}$.

Concept used:

Relation between number of times pin extension is given as follows:

  $N=\frac{L}{D}$ ........ (1)

Here, number of times pin extension is $N$, length of the pin extension is $L$ and diameter of the pin is $D$.

Calculation:

Substitute $1.0$ for $N$ and $7.5\text{mm}$ for $D$ in equation (1).

  $\begin{array}{c}1.0=\frac{L}{7.5}\\ L=7.5\text{mm}\end{array}$

Thus, Length of pin extension is $7.5\text{mm}$.

Conclusion:

Length of pin extension is $7.5\text{mm}$.