20804139 hnc mechanical engineering analytical methods assignment 2 trigonometric methods copy
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Steve Goddard
Contents
Topic PageTrigonometry 2
Cartesian and Polar Co-ordinates and Radian measure
9
Sinusoidal Functions 12Trigonometric Identities 16
Bibliography 19
Page 1 of 19
Steve Goddard
Analytical Methods – Assignment 2
Trigonometric Methods
Trigonometry
1. A 4.2m long ladder is placed against a perpendicular pylon with its foot 60cm from the pylon.
1.1 Determine how far up the pylon the ladder reaches
1.2 Calculate how far the top of the ladder rises when the foot of the ladder is moved 20cm towards the pylon
So the top of the ladder rises:
4.18 – 4.156 = 0.024m
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Steve Goddard
2. From a point on horizontal ground a surveyor measures the angle of elevation of a church spire as 20°. He moves 50m closer to the church and measures the angle of elevation as 25°. Calculate the height of the spire.
Equation 1
In triangle PQS
Hence
i.e.
Equation 2
In triangle PQR,
Hence , i.e.
Equating equation 2 and 2 gives:
From equation 2,
Height of the building h =
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20°RS
PP
Q
25°
50x
h
Steve Goddard
3. Solve the triangle ABC given that: and
So to find BC:
Tan 35 x 5mm = 3.501
And to find the hypotenuse (BA):
Page 4 of 19
35°
θ
5mmA
C
B
Steve Goddard
4. A ship, X, sails at a steady speed of 50km/hr in direction W 30° N (i.e. A bearing of 300° from port). At the same time another ship, Y, leaves port at a steady speed of 40km/hr in a direction N 20° E (i.e. a bearing of 20°). Determine their distance apart after 15hrs.
First of all I worked out how far each ship had traveled by multiplying their speed by time.
Ship 1 travels = 50Km/hr x 15 Hrs = 750Km
Ship 2 travels = 40Km/hr x 15Hrs = 600Km
Using the cosine rule: Cos C
So:
I then square-rooted this to find c
c = 875.362 Km
Page 5 of 19
C
B
600
750
b
a
c
A
Steve Goddard
5. An aero plane is sighted due east from a radar station at an elevation of 50° and a height of 5,000m Later it is sighted at an elevation of 45° and a height of 2500m in a direction of E 70° S.
5.1 If it is descending uniformly, find the angle of decent
Calculations
Firstly I worked out length AO and OB.
Next from these Lengths I worked out the hypotenuse of each triangle H1 and H2.
Page 6 of 19
A
O
B
H1 H2
H3
50
45
5000m
2500m
H1
H2
H3
701
H1
H2
H3
2500m
5000m
O
O
O
B
B
A
A
θ
a
b
c
Steve Goddard
Using the Cosine rule I calculated H3
=39313498.56
Square Root the answer to give H3
I need to work out θ so in order to simplify things these are my essential calculations.
To work out θ I used the SOH-CAH-TOA method and used SOH
23.498 is the angle of decent.
This problem can be solved a number of ways such as firstly solving the length AB using Pythagoras’s theorem and then using TOA to work out the angle.
Eg.
5.2 Find the speed of the aero plane if the time between observations is 40s
Speed = Distance over time but first of all I converted the units:
Page 7 of 19
6270.047
2500m
2500m
BA
θ
b
Steve Goddard
I calculated 40 seconds in hours
Hrs
From this I used the normal speed equation
Or keeping unconverted in seconds
6. Find the magnitude and direction of the resultant of two concurrent and coplanar forces of 30N and 50N when the angle between them is 55°. State the direction of the resultant relative to the 50N force.
.
Therefore
Resultant of 40.978 N at 36.848 degrees from the 50 N force.
Cartesian and Polar Co-ordinates and Radian Measure
7. Change the following to polar co-ordinates:
Illustrate your answer with a diagram (-2.2, 5.5)
Page 8 of 19
55°
30N 50N
x
Steve Goddard
From Pythagoras theorem
By trigonometric ratios or 1.19028 rad
Hence
Or rad
Hence the position of point P in polar co-ordinates form is (5.923, 111.801) or (5.923, 1.951)
8. Change the following to Cartesian co-ordinates:
Illustrate your answer with a diagram (6.4, 2.27 rad)
This corresponds to the length OA in the diagram.
This corresponds to the length AB in the diagram.
Thus (-4.119, 4.898) in Cartesian coordinates corresponds to the polar coordinates (6.4, 2.27 rad)
9. Convert the following between degrees and radians as appropriate:
9.1 30°
1° = 0.0174
So:
Page 9 of 19
P
5.5
r
2.2
Y
X0
a
θ
0
B
A
R= 6.4
θ=2.27 rad
Steve Goddard
30 x 0.0174 = 0.523
9.2 90°
90 x 0.0174 = 1.566
9.3
Rad
1.047 x 57.295 = 59.987°
9.4 0.838 rads
0.838 x 57.295 = 48.0132°
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Steve Goddard
10. A train is traveling at 108km/h and has wheels of diameter 80cm. Determine:
10.1 The angular velocity of the wheels in both rad/s and rpm
Linear velocity km/h
Radius of the wheel mm
m
From the equation , from which,
Angular velocity
Rad/sAnd
Where is in rev/s
Meaning angular speed rev/s
Rev/min
Rev/min
10.2 The number of revolutions made by one of the wheels if the speed remains constant for 2.7km and there is no slipping
Since then the time taken to travel 2.7km i.e. 2700m at a constant speed of
30m/s is given by:
Time t =
Since the wheel is rotating at 716.19 rev/min, then in 90/60 minutes it makes:
Revolutions
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Steve Goddard
Sinusoidal Functions
11. Solve the following in the range 0° to 360°:
312.47)678.0(1 Cos
The solutions of Cos-1 (-0.678), between 0 & 360 degrees are: 132.69 and 227.31
180 – 47.312 = 132.69 and 180 + 47.312 = 227.31
12. The voltage, v, in an alternating current circuit at any time, t, (seconds) is given by:
Determine:
12.1 The amplitude, periodic time, frequency and phase angle (in degrees)
Amplitude = 100 V
Angular Velocity,
Hence periodic time, T =
Or
Frequency, f =
Phase Angle = 0.785rad =
= Lagging v = 100 Sin (200πt)
12.2 The voltage when t = 0
12.3 The voltage when t = 5ms
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Steve Goddard
12.4 The time when the voltage first reaches 50v
When v = 50 then 50 =
Hence
Hence when v = 50v
Time, t = seconds
13. A complex voltage waveform, v, is comprised of a 141.4v rms fundamental voltage at a frequency of 200 Hz, a 40% third harmonic
component leading the fundamental voltage at zero time by and a
20% fifth harmonic component lagging the fundamental by .
13.1 Write down an expression for the voltage, v
Voltage = 141.4 V (rms). So the maximum value or amplitude is: = 200
If the fundamental frequency is 200Hz then angular velocity
Rads/s
Hence fundamental voltage of
The third harmonic component has an amplitude equal to 40% of 200 V i.e. 80
The frequency of the third harmonic component is Hz so:
Hence third harmonic voltage is represented by
The fifth harmonic has an amplitude equal to 20% of 200 V i.e. 40
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Resulting complex waveform for V over one cycle of the fundamental waveform
-300
-200
-100
0
100
200
300
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (ms)
Vo
ltag
e
Fundamental Voltage
Third Harmonic
Fifth Harmonic
Total Voltage
Steve Goddard
The frequency of the fifth harmonic component is:
So:
Hence the voltage of the fifth harmonic is shown as
Overall
13.2 Plot the resulting complex waveform for v over one cycle of the fundamental waveform.
The data used for this graph can be found on the next page.
Spreadsheet Formula
A1 = TimeB1 = Fundamental Voltage = 200*SIN(400*PI()*A3*0.001)C1 = Third Harmonic = 80*SIN(1200*PI()*A3*0.001+PI()/4) D1 = Fifth Harmonic = 40*SIN(2000*PI()*A3*0.001-PI()/3) E1 = Total Voltage = B3+C3+D3
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Steve Goddard
Time (ms)
Fundamental voltage
Third Harmonic
Fifth Harmonic
Total Voltage
0 0 56.56854249 -34.64101615 21.927526340.1 25.06664671 73.42037005 -16.26946572 82.217551040.2 49.73797743 79.96052483 8.316467633 138.01496990.3 73.62491054 75.27046152 29.72579302 178.62116510.4 96.35073482 60.00888557 39.78087581 196.14049620.5 117.5570505 36.31923998 34.64101615 188.51730660.6 136.9094212 7.528665065 16.26946572 160.7075520.7 154.1026486 -22.31928848 -8.316467633 123.46689240.8 168.8655851 -49.03256429 -29.72579302 90.107227790.9 180.9654105 -68.85936216 -39.78087581 72.32517252
1 190.2113033 -79.01506725 -34.64101615 76.555219861.1 196.4574501 -78.07334096 -16.26946572 102.11464351.2 199.6053457 -66.16644594 8.316467633 141.75536741.3 199.6053457 -44.96667023 29.72579302 184.36446851.4 196.4574501 -17.45145931 39.78087581 218.78686661.5 190.2113033 12.5147572 34.64101615 237.36707661.6 180.9654105 40.72331326 16.26946572 237.95818951.7 168.8655851 63.21240099 -8.316467633 223.76151851.8 154.1026486 76.82349485 -29.72579302 201.20035041.9 136.9094212 79.64495717 -39.78087581 176.7735025
2 117.5570505 71.28052194 -34.64101615 154.19655622.1 96.35073482 52.90494923 -16.26946572 132.98621832.2 73.62491054 27.09903362 8.316467633 109.04041182.3 49.73797743 -2.512860726 29.72579302 76.950909732.4 25.06664671 -31.77183125 39.78087581 33.075691282.5 -6.43149E-14 -56.56854249 34.64101615 -21.92752632.6 -25.06664671 -73.42037005 16.26946572 -82.2175512.7 -49.73797743 -79.96052483 -8.316467633 -138.014972.8 -73.62491054 -75.27046152 -29.72579302 -178.6211652.9 -96.35073482 -60.00888557 -39.78087581 -196.140496
3 -117.5570505 -36.31923998 -34.64101615 -188.5173073.1 -136.9094212 -7.528665065 -16.26946572 -160.7075523.2 -154.1026486 22.31928848 8.316467633 -123.4668923.3 -168.8655851 49.03256429 29.72579302 -90.10722783.4 -180.9654105 68.85936216 39.78087581 -72.32517253.5 -190.2113033 79.01506725 34.64101615 -76.55521993.6 -196.4574501 78.07334096 16.26946572 -102.1146433.7 -199.6053457 66.16644594 -8.316467633 -141.7553673.8 -199.6053457 44.96667023 -29.72579302 -184.3644683.9 -196.4574501 17.45145931 -39.78087581 -218.786867
4 -190.2113033 -12.5147572 -34.64101615 -237.3670774.1 -180.9654105 -40.72331326 -16.26946572 -237.9581894.2 -168.8655851 -63.21240099 8.316467633 -223.7615184.3 -154.1026486 -76.82349485 29.72579302 -201.200354.4 -136.9094212 -79.64495717 39.78087581 -176.7735034.5 -117.5570505 -71.28052194 34.64101615 -154.1965564.6 -96.35073482 -52.90494923 16.26946572 -132.9862184.7 -73.62491054 -27.09903362 -8.316467633 -109.0404124.8 -49.73797743 2.512860726 -29.72579302 -76.95090974.9 -25.06664671 31.77183125 -39.78087581 -33.0756913
5 1.2863E-13 56.56854249 -34.64101615 21.92752634
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Steve Goddard
Trigonometric Identities
14. Show that: and explain where this would be
useful
If
And
Then:
This equation is useful with integrating when calculating RMS values.
15. Solve the following in the range:
Rearranging the equation to make the answer equal to zero would give me:
Now that the equation is in this format I can easily apply the quadratic formula:
So: a = 5, b = 3 and c = -4
+ Sinx = 0.6433- Sinx = -1.2433
Therefore the values in the range are:
X = 40°038’ and 139°962
16. Solve the following for values:
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Steve Goddard
Or 1.043 radians
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Steve Goddard
17. Express the equation:
17.1 In the form R Sin (t+α) and hence
Equating co-efficients:
17.2 Solve, for values
Page 18 of 19
Steve Goddard
Bibliography
www.google.com
John Bird – Higher Engineering Mathematics
Page 19 of 19
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