Analytical Engineering Mathematics 1 - Engineering Mathematics

This document contains solutions for tasks related to Crame mechanism, angle of lap, and trigonometric identities. It also includes references for further reading.

ANALYTICAL ENGINEERING MATHEMATICS 1
ENGINEERING MATHEMATICS
By (name)
Name of the class
Professor
Name of the school
City and state
Date

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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 Q travel∈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 sin140
4.2 = 0.137740
Q1 = arcsine 0.1377 = 7.9170 , angle O P1 Q1 = 180 - (140 + 7,917) = 32.080

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Distance O Q1 , OQ 1
sin P 1 = P 1 Q1
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 (100π t + 0.274) A
Amplitude = 120 A
Angular velocity ω = 100 π
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.

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(d) The value of current when t = 0
I = 120 Sin (100 πt + 0.274)
=120sin (0 + 0.274) = 120 sin 0.274 = 0.573 A ans
(e) Value of current when t =60 m/s
I = 120 sin (100πt 0.274)
= 120 sin (100 π 60
603 + 0,274) =120 sin (0.361266) rad convert to degree 0.361266 x 180
π
= 20.70 there for I = 120 sin 20.7 = 42.42 A ans
(f) The time when current first reaches 80 A , When I = 80 A then
80 = 120 sin (100πt + 0.274) 80
120 = sin (100π t + 0.274)
Hence (100 πt + 0.274) = arcsine 80
120 = 41.810 or 0.7297 rad
100 πt + 0.274 = 0.7297
100 πt = 0.729722- 0.274 = 0.455722
100 π t = 0.4557222
T = 0.455722
100 π = 1.45 x 10−3sec
When current is maximum, I = 120 A I = 120 Sin (100 π t + 0.274) there for
120 = sin (100 π t + 0.274) divide both sides with 120
1 = sin (100 πt + 0.274) = 100 πt + 0.274 = arcsine 1 = 90 or 1,571 rad
Therefor 100 πt = 1.571 – 0.274 = 1.927
T = 4.128 x 10−3x 1000
= 13.44 m/s

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Task 3
Draw a right angle triangle and apply SOH CAH TOA to solve the problem. Where r =
hypotenuse, y = opposite and x = adjacent.
Sin∅ = y
r make y the subject of the formula.
y =r sin∅
Cos∅ = x
r make x the subject of the formula.
X= r cos ∅
And finally tan ∅ = rsin∅
rcos ∅ = sin ∅
cos ∅
When you use Pythagoras theorem to the right angle above.
r2 = y2 + x2 substitute the value of y and x
r2 =r2 sin2 + r2 cos2 divide both sides with r2
The answer will be.
1 =sin2 ∅ +cos2 ∅
Then solve 10 cos2 ∅ +3sin∅ - A =0 from the trigonometric identity make sin2 ∅ the subject of
the formula for the trigonometric identity above, which is equal to
sin2 ∅ =1 -cos2 ∅ then substitute it into the equation
10cos2 + 3(1 - cos2 ) =A
10cos2 ∅ + 3 - 3cos2 ∅ = A
10cos2 ∅ - 3cos2 ∅ + 3 = A
Therefor A= 7cos2 ∅ + 3
(b) 5 sinh2 x - 3cosh x - A = 0
Change x to be ∅ therefor 5 sinh2 ∅ - 3 cosh ∅ - A = 0
Use trigonometric identity sinh2 ∅ - cosh2 ∅ = 1

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Make sinh2 ∅ the subject of the formula of the trig identity = 1 + cosh2 ∅ , then
substitute to the equation above.
5(1+ 5 cosh2 ) - 3 cosh ∅ - A = 0 , 5 + 5cosh2 ∅ - 3 cosh ∅ - A = 0
Let cosh ∅ be = y therefor 5 + 5 y2 3y - A = 0
5Y 2 - 3Y + 5 = A

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Reference
1 (2009). Maple and Mathematica. [Place of publication not identified],
Springer Vienna. http://dx.doi.org/10.1007/978-3-211-99432-0.
2BARNES-SVARNEY, P. L., & SVARNEY, T. E. (2012). The handy math answer
book. Canton, MI, Visible Ink Press.
http://literati.credoreference.com/content/title/viphamath.
3 MCLESTER, J., & ST. PIERRE, P. (2008). Applied biomechanics: concepts and
connections. Belmont, CA, Thompson Wadsworth.
4BAKER, D. (2002). Key maths GCSE. Cheltenham, Nelson Thornes.
5 Bird, 2006, 5th ed , Higher Engineering Mathematics, Burlington : Elsevier
Ltd.
6 Bird, J, 2014, Higher Engineering Mathematics, New York: Routledge.

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