Chapter 11, Problem 13A

To determine

(a)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.010in.+0.004in.$

Explanation of Solution

Given:

Thickness $=0.014in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.014in.$ is

  $\begin{array}{l}\Rightarrow 0.010+0.004\\ \Rightarrow 0.014in.\end{array}$

Conclusion:

The combination of leaves is $T=0.010in.+0.004in.$

To determine

(b)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015in.+0.010in.+0.008in.$

Explanation of Solution

Given:

Thickness $=0.033in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.033in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.010^{″}+0.008^{″}\\ \Rightarrow 0.033in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015in.+0.010in.+0.008in.$

To determine

(c)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015in.+0.006in.$

Explanation of Solution

Given:

Thickness $=0.021in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.021in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.006\\ \Rightarrow 0.021in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015in.+0.006in.$

To determine

(d)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015in.+0.012in.+0.008in.+0.003in.$

Explanation of Solution

Given:

Thickness $=0.038in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.038in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.012^{″}+0.008^{″}+0.003^{″}\\ \Rightarrow 0.038in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015in.+0.012in.+0.008in.+0.003in.$

To determine

(e)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.008in.+0.003in.$

Explanation of Solution

Given:

Thickness $=0.011in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.011in.$ is

  $\begin{array}{l}\Rightarrow 0.008^{″}+0.003^{″}\\ \Rightarrow 0.011in.\end{array}$

Conclusion:

The combination of leaves is $T=0.008in.+0.003in.$

To determine

(f)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015^{″}+0.012^{″}+0.010^{″}+0.003^{″}+0.002^{″}.$

Explanation of Solution

Given:

Thickness $=0.042in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.042in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.012^{″}+0.010^{″}+0.003^{″}+0.002^{″}\\ \Rightarrow 0.042in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015^{″}+0.012^{″}+0.010^{″}+0.003^{″}+0.002^{″}.$

To determine

(g)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015^{″}+0.012^{″}+0.002^{″}.$

Explanation of Solution

Given:

Thickness $=0.029in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.029in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.012^{″}+0.002^{″}\\ \Rightarrow 0.029in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015^{″}+0.012^{″}+0.002^{″}.$

To determine

(h)

The smallest combination of gauge leaves thickness.

Answer to Problem 13A

  $T=0.015^{″}+0.012^{″}+0.010^{″}+0.008^{″}+0.004^{″}.$

Explanation of Solution

Given:

Thickness $=0.049in.$

Calculation:

In Figure a customary fleer gauge is given with different leaf thickness.

Thus, the smallest combination of leaves that will give a thickness of $0.049in.$ is

  $\begin{array}{l}\Rightarrow 0.015^{″}+0.012^{″}+0.010^{″}+0.008^{″}+0.004^{″}\\ \Rightarrow 0.049in.\end{array}$

Conclusion:

The combination of leaves is $T=0.015^{″}+0.012^{″}+0.010^{″}+0.008^{″}+0.004^{″}.$