basic_engineering_mathematics0

52 Basic Engineering Mathematics
(b) Find the value of R 3 given that R T = 3 , R 1 = 5
and R 2 = 10 .
5. Five pens and two rulers cost 94p. If a ruler costs 5p
more than a pen, find the cost of each.
6. Ohm’s law may be represented by I = V /R, where I is
the current in amperes, V is the voltage in volts and R is
the resistance in ohms. A soldering iron takes a current
of 0.30A from a 240V supply. Find the resistance of the
element.
7.5 Further practical problems involving
simple equations
Problem 22. The extension x m of an aluminium tie bar
of length l m and cross-sectional area A m 2 when carrying
a load of F newtons is given by the modulus of
elasticity E = Fl/Ax. Find the extension of the tie bar (in
mm) if E = 70 × 10 9 N/m 2 , F = 20 × 10 6 N, A = 0.1m 2
and l = 1.4m.
Problem 24. A painter is paid £4.20 per hour for a basic
36 hour week, and overtime is paid at one and a third times
this rate. Determine how many hours the painter has to work
in a week to earn £212.80
Basic rate per hour = £4.20;
overtime rate per hour = 1 1 × £4.20 = £5.60
3
Let the number of overtime hours worked = x
Then (36)(4.20) + (x)(5.60) = 212.80
151.20 + 5.60x = 212.80
5.60x = 212.80 − 151.20 = 61.60
x = 61.60
5.60 = 11
Thus 11 hours overtime would have to be worked to earn
£212.80 per week.
Hence the total number of hours worked is 36 + 11, i.e.
47 hours.
E = Fl/Ax, hence 70 × 10 9 N m = (20 ×106 N)(1.4m)
2 (0.1m 2 )(x)
(the unit of x is thus metres)
70 × 10 9 × 0.1 × x = 20 × 10 6 × 1.4
x = 20 × 106 × 1.4
70 × 10 9 × 0.1
Cancelling gives:
x = 2 × 1.4
7 × 100 m = 2 × 1.4 × 1000 mm
7 × 100
Hence the extension of the tie bar, x = 4mm.
Problem 23. Power in a d.c. circuit is given by
P = V 2
where V is the supply voltage and R is the
R
circuit resistance. Find the supply voltage if the circuit
resistance is 1.25 and the power measured is 320 W.
Since P = V 2
R
then 320 = V 2
1.25
(320)(1.25) = V 2
i.e. V 2 = 400
Supply voltage,
V = √ 400 = ±20 V
Problem 25. A formula relating initial and final states of
pressures, P 1 and P 2 , volumes V 1 and V 2 , and absolute temperatures,
T 1 and T 2 , of an ideal gas is P 1V 1
= P 2V 2
. Find
T 1 T 2
the value of P 2 given P 1 = 100 × 10 3 , V 1 = 1.0, V 2 = 0.266,
T 1 = 423 and T 2 = 293
Since P 1V 1
= P 2V 2
then (100 × 103 )(1.0)
= P 2(0.266)
T 1 T 2 423
293
‘Cross-multiplying’ gives:
(100 × 10 3 )(1.0)(293) = P 2 (0.266)(423)
P 2 = (100 × 103 )(1.0)(293)
(0.266)(423)
Hence P 2 = 260 × 10 3 or 2.6 × 10 5
Problem 26. The stress f
√
in a material of a thick cylinder
(
can be obtained from D ) f + p
d = . Calculate the stress,
f − p
given that D = 21.5, d = 10.75 and p = 1800
Since
√ ( )
√ (
D f + p
d = then 21.5
) f + 1800
f − p 10.75 = f − 1800
i.e.
√ ( ) f + 1800
2 =
f − 1800