Basic Engineering Mathematics, Fifth Edition

Solving simple equations 77

8.

2

a

=

3

8

9.

1

3 n

+

1

4 n

=

7

24

10.

x+ 3

4

=

x− 3

5

+ 2

11.

3 t

20

=

6 −t

12

+

2 t

15

−

3

2

12.

y

5

+

7

20

=

5 −y

4

13.

v− 2

2 v− 3

=

1

3

14.

2

a− 3

=

3

2 a+ 1

15.

x

4

−

x+ 6

5

=

x+ 3

2

16. 3

√

t= 9

17. 2
    √
    y= 5


18. 4=

√(

3

a

)

+ 3

19.

3

√

x

1 −

√

x

=− 6

20. 10= 5

√(

x

2

− 1

)

21. 16=

t^2

9

22.

√(

y+ 2

y− 2

)

=

1

2

23.

6

a

=

2 a

3

24.

11

2

= 5 +

8

x^2

11.3 Practical problems involving

simple equations

There are many practical situations in engineering in
which solving equations is needed. Here are some
worked examples to demonstrate typical practical
situations

Problem 17. Applying the principle of moments

to a beam results in the equation

F× 3 =( 7. 5 −F)× 2

whereFis the force in newtons. Determine the

value ofF

Removing brackets gives 3 F= 15 − 2 F

Rearranging gives 3 F+ 2 F= 15

i.e. 5 F= 15

Dividing both sides by 5 gives

5 F

5

=

15

5

from which, force,F= 3 N.

Problem 18. A copper wire has a lengthLof

1.5km, a resistanceRof 5and a resistivity of

17. 2 × 10 −^6 mm. Find the cross-sectional area,a,

of the wire, given thatR=

ρL

a

SinceR=

ρL

a

then

5 =

( 17. 2 × 10 −^6 mm)( 1500 × 103 mm)

a

.

From the units given,ais measured in mm^2.

Thus, 5a= 17. 2 × 10 −^6 × 1500 × 103

and a=

17. 2 × 10 −^6 × 1500 × 103

5

=

17. 2 × 1500 × 103

106 × 5

=

17. 2 × 15

10 × 5

= 5. 16

Hence, the cross-sectional area of the wire is5.16mm^2.

Problem 19. PV=mRTis the characteristic gas

equation. Find the value of gas constantRwhen

pressureP= 3 × 106 Pa, volumeV= 0 .90m^3 ,

massm= 2 .81kg and temperatureT=231K

Dividing both sides ofPV=mRTbymTgives

PV

mT

=

mRT

mT

Cancelling gives

PV

mT

=R

Substituting values gives R=

(

3 × 106

)

( 0. 90 )

( 2. 81 )( 231 )