EBK MATHEMATICS FOR MACHINE TECHNOLOGY
EBK MATHEMATICS FOR MACHINE TECHNOLOGY
8th Edition
ISBN: 9781337798396
Author: SMITH
Publisher: CENGAGE LEARNING - CONSIGNMENT
Question
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Chapter 2, Problem 15A
To determine

Evaluate the length of A, B, C, D, E and F of the part.

Expert Solution & Answer
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Answer to Problem 15A

The length of A, B, C, D, E and F of the part are 1116in, 5964in, 1916in, 1116in, 31164in and 1316in respectively.

Explanation of Solution

Given:

All the dimensions are shown in below Fig:

EBK MATHEMATICS FOR MACHINE TECHNOLOGY, Chapter 2, Problem 15A , additional homework tip  1

Concept used:

Mark all the corner points as k, l, m, n, o, p, q, r, s, t, u, v, w, x, y and z.

EBK MATHEMATICS FOR MACHINE TECHNOLOGY, Chapter 2, Problem 15A , additional homework tip  2

  A=pq+rs   .... (1)

Here, distance A is the sum of distance pq and rs as marked in above Fig.

  B=qr+st   .... (2)

Here, distance B is the sum of distance qr and st as marked in above Fig.

  C=uv+wx+yz    .... (3)

Here, distance C is the sum of distance uv, wx and yz as marked in above Fig.

  D=kl+mn   .... (4)

Here, distance D is the sum of distance kl and mn as marked in above Fig.

  E=B+C+D...... (5)

Here, distance E is the sum of distance B, C and D as marked in above Fig.

  F=zk+ml+no    .... (6)

Here, distance F is the sum of distance zk, ml and no as marked in above Fig.

Calculation:

EBK MATHEMATICS FOR MACHINE TECHNOLOGY, Chapter 2, Problem 15A , additional homework tip  3

Substitute 12in for pq and 916in for rs in equation (1).

  A=12+916A=1716A=1116in

Substitute 38in for qr and 3564in for st in equation (2).

  B=38+3564B=24+3564B=5964in

Substitute 3132in for uv, 18in for wx and 1532in for yz in equation (3).

  C=3132+18+1532C=31+4+1532C=2516inC=1916in

Substitute 14in for kl and 716in for mn in equation (4).

  D=14+716D=4+716D=1116in

Substitute 5964in for B, 2516in for C and 1116in for D in equation (5).

  E=5964+2516+1116E=59+100+4464E=20364inE=31164in

Substitute 2164in for zk, 516in for ml and 1164in for no in equation (6).

  F=2164+516+1164F=21+20+1164F=5264inF=1316in

Thus, the length of A, B, C, D, E and F of the part are 1116in, 5964in, 1916in, 1116in, 31164in and 1316in respectively.

Conclusion:

The length of A, B, C, D, E and F of the part are 1116in, 5964in, 1916in, 1116in, 31164in and 1316in respectively.

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