Chapter 36, Problem 16AR

To determine

(a)

The combination of gage blocks

Answer to Problem 16AR

The combination of gage blocks are first $0.1004^{″}$, second $0.128^{″}$ and third $0.150^{″}$

Explanation of Solution

Write the expression for determine the combination of gage blocks.

  $S=S_1-S_2$  ......(I)

Here, Size of first block is $S_1$ and Size of second block is $S_2$

Calculation:

First block required one $0.1004^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $0.3784^{″}$ for $S_1$ and $0.1004^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=0.3784^{″}-0.1004^{″}\\ =0.2780^{″}\end{array}$

Second block required is one $0.1280^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $0.2780^{″}$ for $S_1$ and $0.1280^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=0.2780^{″}-0.1280^{″}\\ =0.1500^{″}\end{array}$

Third block required is one $0.1500^{″}$ from " $19$ of blocks $0.050^{″}$ series".

Substitute $0.1500^{″}$ for $S_1$ and $0.1500^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=0.1500^{″}-0.1500^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1004^{″}$, Second $0.1280^{″}$ and third $0.1500^{″}$.

To determine

(b)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1006^{″}$, second $0.148^{″}$ and third $0.30^{″}$ and fourth $2.000^{″}$.

Explanation of Solution

Calculation:

First block required one $0.1006^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $2.5486^{″}$ for $S_1$ and $0.1006^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.5486^{″}-0.1006^{″}\\ =2.300^{″}\end{array}$

Second block required is one $0.148^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $2.4480^{″}$ for $S_1$ and $0.148^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.4480^{″}-0.148^{″}\\ =2.300^{″}\end{array}$

Third block required is one $0.30^{″}$ from " $19$ of blocks $0.050^{″}$ series".

Substitute $2.300^{″}$ for $S_1$ and $0.30^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.300^{″}-0.30^{″}\\ =2.000^{″}\end{array}$

Fourth block required is one $2.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $2.000^{″}$ for $S_1$ and $2.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.000^{″}-2.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1006^{″}$, Second $0.148^{″}$, third $0.30^{″}$ .and fourth $2.000^{″}$

To determine

(c)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1002^{″}$, second $0.106^{″}$ and third $0.50^{″}$ and fourth $1.000^{″}$

Explanation of Solution

Calculation:

First block required one $0.1002^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $1.7062^{″}$ for $S_1$ and $0.1002^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=1.7062^{″}-0.1002^{″}\\ =1.606^{″}\end{array}$

Second block required is one $0.106^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $1.606^{″}$ for $S_1$ and $0.106^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=1.606^{″}-0.106^{″}\\ =1.50^{″}\end{array}$

Third block required is one $0.50^{″}$ from " $19$ of blocks $0.050^{″}$ series".

Substitute $1.50^{″}$ for $S_1$ and $0.50^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=1.50^{″}-0.50^{″}\\ =1.000^{″}\end{array}$

Fourth block required is one $1.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $1.000^{″}$ for $S_1$ and $1.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=1.000^{″}-1.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1002^{″}$, Second $0.106^{″}$, third $0.50^{″}$ .and fourth $1.000^{″}$

To determine

(d)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1007^{″}$, second $0.146^{″}$ third $0.40^{″}$ ,fourth $4.000^{″}$ and fifth $1.000^{″}$

Explanation of Solution

Calculation:

First block required one $0.1007^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $5.6467^{″}$ for $S_1$ and $0.1007^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=5.6467^{″}-0.1007^{″}\\ =5.5460^{″}\end{array}$

Second block required is one $0.146^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $5.5460^{″}$ for $S_1$ and $0.1460^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=5.5460^{″}-0.1460^{″}\\ =5.400^{″}\end{array}$

Third block required is one $0.40^{″}$ from " $19$ of blocks $0.050^{″}$ series".

Substitute $5.400^{″}$ for $S_1$ and $0.400^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=5.400^{″}-0.400^{″}\\ =5.000^{″}\end{array}$

Fourth block required is one $4.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $5.000^{″}$ for $S_1$ and $4.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=5.000^{″}-4.000^{″}\\ =1.000^{″}\end{array}$

Fifth block required is one $1.000^{″}$ from "4 blocks of $1.000^{″}$ series

Substitute $.000^{″}$ for $S_1$ and $1.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=1.000^{″}-1.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1007^{″}$, Second $0.146^{″}$, third $0.40^{″}$, fourth $4.000^{″}$ and fifth $1.000^{″}$.

To determine

(e)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1001^{″}$, second $0.140^{″}$ third $0.110^{″}$ ,fourth $0.60^{″}$ and fifth $2.000^{″}$

Explanation of Solution

Calculation:

First block required one $0.1001^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $3.0901^{″}$ for $S_1$ and $0.1001^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=3.0901^{″}-0.1001^{″}\\ =2.9900^{″}\end{array}$

Second block required is two $0.140^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $2.9900^{″}$ for $S_1$ and $0.140^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.9900^{″}-2\times 0.140^{″}\\ =2.7100^{″}\end{array}$

Third block required is one $0.110^{″}$ from " $49$ of blocks $0.001^{″}$ series".

Substitute $2.7100^{″}$ for $S_1$ and $0.110^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.7100^{″}-0.110^{″}\\ =2.600^{″}\end{array}$

Fourth block required is one $0.60^{″}$ from " $19$ blocks of $.050^{″}$ series.

Substitute $2.600^{″}$ for $S_1$ and $0.600^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.600^{″}-0.600^{″}\\ =2.000^{″}\end{array}$

Fifth block required is one $2.000^{″}$ from "4 blocks of $1.000^{″}$ series.

Substitute $2.000^{″}$ for $S_1$ and $2.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=2.000^{″}-2.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1001^{″}$, second $0.140^{″}$ third $0.110^{″}$ ,fourth $0.60^{″}$ and fifth $2.000^{″}$

To determine

(f)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1009^{″}$ and second $0.1000^{″}$

Explanation of Solution

Calculation:

First block required one $0.1009^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $0.2009^{″}$ for $S_1$ and $0.1009^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=0.2009^{″}-0.1009^{″}\\ =0.1000^{″}\end{array}$

Second block required is two $0.1000^{″}$ from " $19$ blocks of $0.050^{″}$ Series".

Substitute $0.1000^{″}$ for $S_1$ and $0.1000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=0.1000^{″}-0.1000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1009^{″}$ and second $0.1000^{″}$

To determine

(g)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1005^{″}$, second $0.149^{″}$ third $0.140^{″}$ ,fourth $0.50^{″}$, fifth $4.000^{″}$ and sixth $3.000^{″}$

Explanation of Solution

Calculation:

First block required one $0.1001^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $7.8895^{″}$ for $S_1$ and $0.1005^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.8895^{″}-0.1005^{″}\\ =7.7890^{″}\end{array}$

Second block required is two $0.149^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $7.7890^{″}$ for $S_1$ and $0.149^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.7890^{″}-0.149^{″}\\ =7.6400^{″}\end{array}$

Third block required is one $0.140^{″}$ from " $49$ of blocks $0.001^{″}$ series".

Substitute $7.6400^{″}$ for $S_1$ and $0.140^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.6400^{″}-0.140^{″}\\ =7.500^{″}\end{array}$

Fourth block required is one $0.50^{″}$ from " $19$ blocks of $.050^{″}$ series.

Substitute $7.500^{″}$ for $S_1$ and $0.500^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.500^{″}-0.500^{″}\\ =7.000^{″}\end{array}$.

Fifth block required is one $4.000^{″}$ from "4 blocks of $1.000^{″}$ series.

Substitute $7.000^{″}$ for $S_1$ and $4.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.000^{″}-4.000^{″}\\ =3.000^{″}\end{array}$

Sixth block require is one $3.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $3.000^{″}$ for $S_1$ and $3.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=3.000^{″}-3.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1005^{″}$, second $0.149^{″}$ third $0.140^{″}$ ,fourth $0.50^{″}$, fifth $4.000^{″}$ and sixth $3.000^{″}$

To determine

(h)

The combination of gage blocks.

Answer to Problem 16AR

The combination of gage blocks are first $0.1004^{″}$, second $0.141^{″}$ third $0.140^{″}$ ,fourth $0.110$, fifth $0.600^{″}$ sixth $4.000^{″}$ and seventh $3.000^{″}$.

Explanation of Solution

Calculation:

First block required one $0.1004^{″}$ from " $9$ blocks of $0.0001^{″}$ series".

Substitute $8.0014^{″}$ for $S_1$ and $0.1004^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=8.0014^{″}-0.1004^{″}\\ =7.9910^{″}\end{array}$

Second block required is two $0.141^{″}$ from " $49$ blocks of $0.001^{″}$ Series".

Substitute $7.9910^{″}$ for $S_1$ and $0.141^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.9910^{″}-0.141^{″}\\ =7.8500^{″}\end{array}$

Third block required is one $0.140^{″}$ from " $49$ of blocks $0.001^{″}$ series".

Substitute $7.8500^{″}$ for $S_1$ and $0.140^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.8500^{″}-0.140^{″}\\ =7.7100^{″}\end{array}$

Fourth block required is one $0.110^{″}$ from " $49$ blocks of $0.001^{″}$ series.

Substitute $7.7100^{″}$ for $S_1$ and $0.110^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.7100^{″}-0.110^{″}\\ =7.600^{″}\end{array}$

Fifth block required is one $0.600^{″}$ from " $19$ blocks of $0.050^{″}$ series.

Substitute $7.600^{″}$ for $S_1$ and $0.600^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.6000^{″}-0.600^{″}\\ =7.000^{″}\end{array}$

Sixth block require is one $4.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $7.000^{″}$ for $S_1$ and $4.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=7.000^{″}-4.000^{″}\\ =3.000^{″}\end{array}$

Seventh block require is one $4.000^{″}$ from " $4$ blocks of $1.000^{″}$ series.

Substitute $3.000^{″}$ for $S_1$ and $3.000^{″}$ for $S_2$ in Equation (I).

  $\begin{array}{c}S=3.000^{″}-3.000^{″}\\ =0\end{array}$

Conclusion:

The combination of gage blocks are first $0.1004^{″}$, second $0.141^{″}$ third $0.140^{″}$ ,fourth $0.110$, fifth $0.600^{″}$ sixth $4.000^{″}$ and seventh $3.000^{″}$.