Solutions to Analytical Engineering Mathematics 1 Homework

This assignment provides solutions to several analytical engineering mathematics problems. Task 1 involves calculations related to a crank mechanism, including determining the distance traveled by a point on the mechanism, calculating angles, and finding distances moved. Task 2 covers angular velocity, linear velocity, and sinusoidal current calculations, including amplitude, periodic time, frequency, phase angle, and current values at specific times. Task 3 focuses on trigonometric identities and solving equations involving sine, cosine, sinh, and cosh functions. The solution uses trigonometric identities and algebraic manipulation to solve for unknown variables. Several references are cited to support the methods and concepts used in the solutions. Desklib offers a wide array of solved assignments and study resources for students.

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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