Analytical Engineering Mathematics 1 - Engineering Mathematics

ANALYTICAL ENGINEERING MATHEMATICS 1
ENGINEERING MATHEMATICS
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2
Task1 A=¿ cramechanism shown in Fig. A4.1 comprises arm OP, of length 0.90m, which rotates
anti
(a) If ∠ POR isinitially zero ,how far does end Qtravel ∈1 4 revolution (b) If ∠POR is
initially 40◦ find the angle between the connecting rod and the horizontal and the length
OQ
(c) Find the distance Q moves (correct to the nearest cm) when ∠POR
changes from 40◦ to 140
(a)Circumference ¿ 2 π r = 2π 0.9 = 5.655m.
1 revolution = 5.655m
Therefor 1
4 rev = ? 1
4 x5.655 = 1.41 m ans
(b) Apply sine rule
PQ
sinO = PO
sinQ , 4.2
sin 40 = 0.9
sin Q therefor sin Q = sin 40 X 0.9
4.20 = 0.137740 ,
Q = arcsine 0.13774 = 7.9170
(c) Angle OPQ = 180 - (40 + 7.917) = 132.080
OQ
sin P = PQ
sin 0 , OQ
sin 132.08 = 4.2
sin 40 , OQ = 4.2 sin 132.08
sin 40 = 4.850m
Apply sine rule
P 1 QI
sinO = P1 O
sin Q1 , 4.2
sin 40 = 0.9
sin Q1 , sin Q1 = 0.90 sin 140
4.2 = 0.137740
Q1 = arcsine 0.1377 = 7.9170 , angle O P1 Q1 = 180 - (140 + 7,917) = 32.080

3
Distance O Q1 , OQ 1
sin P 1 = P 1 Q 1
SINO , OQ 1
sin 32.08 = 4.2
sin 140 , O Q1 =
4.2 sin 32.08
sin 140 = 3.470 m
Distanced moved = 4.850 - 3.470 = 1.379 m , = 138 cm
Task 2
Q1 (a) angle of lap S = 140mm = 0.14m
But S = r ∅ r = 180
2 = 90 mm = 0.09 m
Therefor ∅ = s
r = 0.14
0.09 = 1.56 rad
(a) Angular velocity = ω
2 π but ω = 300
There for
(c) Linear velocity = ω x r = 300 x 0.09 = 27 m/s
Q 2 I = 120 sin (100 π t + 0.274) A
Amplitude = 120 A
Angular velocity ω = 100π
Periodic time, T = 2 π
ω = 2 π
100 π = 1
50
= 0.02 s
Frequency, f = 1
T = 1
0.02 = 50 HZ
Phase angle θ 0.274 rad = (0.274 x 180
π ) degree = 15.70 degrees.