Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

6.5 Numerical versus Symbolic Solutions 143

6 6..5 5 NNuummeerriiccaall vveerrssuuss SSyymmbboolliicc SSoolluuttiioonnss

When you take your engineering classes, you need to be aware of two important things: (1) under-

standing the basic concepts and principles associated with that class, and (2) how to apply them to

solve real physical problems (situations). In order to gain an understanding of the basic concepts,

you need to study carefully the statement of governing laws and the derivations of engineering for-

mulas and their limitations. After you have studied the underlying concepts, you then need to apply

them to physical situations by solving problems. After studying a certain concept initially, you may

think that you completely understand the concept, but it is through the application of the concept

(by doing the homework problems) that you really can test your understanding.

Moreover, homework problems in engineering typically require either a numerical or a

symbolic solution. For problems that require numerical solution, data is given. In contrast, in

the symbolic solution, the steps and the final answer are presented with variables that could be

substituted with data, if necessary. The following example will demonstrate the difference

between numerical and symbolic solutions.

Example 6.4 Determine the load that can be lifted by the hydraulic system shown. All of the necessary infor-

mation is shown in Figure 6.3.

The general relationship among force, pressure, and area is explained in detail in Chapter 10.

At this time, don’t worry about understanding these relationships. The purpose of this example

is to demonstrate the difference between a numerical and a symbolic solution. The concepts that

are used to solve this problem are: F 1 m 1 g,F 2 m 2 g,and F 2 (A 2 /A 1 )F 1 , whereFdenotes

force,mis mass,gis acceleration due to gravity (g9.81 m/s

2

), andArepresents area.

Numerical Solution

We start by making use of the given data and substituting them into appropriate equations as

follows.

F 1 m 1 g 1 100 kg 2 1 9.81 m/s

2

2 981 N

Load

m 2 ?

m 1  100 kg

R 1  5 cm A 1 A 2 R 2  15 cm

■Figure 6.3

The hydraulic system of

Example 6.4.

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

圀圀圀⸀夀䄀娀䐀䄀一倀刀䔀匀匀⸀䌀伀䴀圀圀圀⸀夀䄀娀䐀䄀一倀刀䔀匀匀⸀䌀伀䴀