Chapter 18, Problem 44A

To determine

(a)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=9.925\text{mm}$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.250,D=7.940mm$

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.250}{1}=\frac{L}{7.940}\\ \Rightarrow L=9.925mm\end{array}$

To determine

(b)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=0.625"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1\frac{1}{4},D=\frac{1\text{'}\text{'}}{2}$

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1\frac{1}{4}}{1}=\frac{L}{\frac{1"}{2}}\\ \Rightarrow L=0.625"\end{array}$

To determine

(c)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $D=0.6"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1\frac{1}{4},L=\frac{3\text{'}\text{'}}{4}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1\frac{1}{4}}{1}=\frac{L}{\frac{1"}{2}}=\frac{1\frac{1}{4}}{1}=\frac{\frac{3"}{4}}{D}\\ \Rightarrow D=0.6"\end{array}$

To determine

(d)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $$

  $D=6.349mm$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.250,L=8.730\text{mm}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.375}{1}=\frac{8.730}{D}\\ \Rightarrow D=6.349mm\end{array}$

To determine

(e)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=20.15mm$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.250,D=16.120\text{mm}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.250}{1}=\frac{L}{16.120}\\ \Rightarrow L=20.15mm\end{array}$

To determine

(f)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $D=0.75"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.375,L=1.032\text{'}\text{'}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.375}{1}=\frac{1.032"}{D}\\ \Rightarrow D=0.75"\end{array}$

To determine

(g)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=1.094"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.250,D=0.875\text{'}\text{'}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.250}{1}=\frac{L}{0.875"}\\ \Rightarrow L=1.094"\end{array}$

To determine

(h)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=5.52"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $h=1.500,D=3.680\text{mm}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.500}{1}=\frac{L}{3.680}\\ \Rightarrow L=5.52"\end{array}$

To determine

(i)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $D=0.25"$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.125,L=0.281\text{'}\text{'}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.125}{1}=\frac{0.281"}{D}\\ \Rightarrow D=0.25"\end{array}$

To determine

(j)

The missing value for each dimension in the problem.

Answer to Problem 44A

  $L=7.5mm$

Explanation of Solution

Given information:

The given relationship of the gear is $\frac{N}{1}=\frac{L}{D}$, $N=1.000,D=7.500\text{mm}$.

Calculation:

As it's known that from the given relation and dimensions

  $\begin{array}{l}\frac{N}{1}=\frac{L}{D}\\ \Rightarrow \frac{N}{1}=\frac{L}{D}=\frac{1.0}{1}=\frac{L}{7.50}\\ \Rightarrow L=7.5mm\end{array}$