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Course
MEM30012A
Subject
Mechanical_engineering
Date
May 17, 2024
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STUDENT
STUDENT – KNOWLEDGE ASSESSMENT TASK
Task Number
1B of 3
Task Name
Assignment PART B
National unit/s code
MEM30012A National unit/s title
Apply Mathematical Techniques in a Manufacturing Engineering or Related Environment
National qualification code
MEM60112/ 22479VIC
National qualification title
Advanced Diploma of Engineering (Mechanical)
Advanced Diploma of Engineering (Mechanical / Aeronautical) Advanced Diploma of Engineering Technology
(Civil Engineering Design)
MEM60112
RMIT Program code
C6130 / C6131/ C6162
RMIT Course code
MATH5268C
Section A – Assessment Information
Assessment duration and/or due date
Week 16-17, Refer Canvas for Exact Date.
Task instructions
Summary and Purpose of Assessment
This assignment in the unit of competency covers the application of the concepts of mathematics
to appropriate and simple engineering
situations within the individual's area of engineering expertise.
You must demonstrate an understanding of:
Use two-dimensional geometry to solve practical problems
Use trigonometry to solve practical problems
Perform basic statistical calculations
What and where: This is an individual knowledge-based (problem solving) assessment on Apply Mathematical Techniques in a Manufacturing Engineering or Related Environment. You need to complete this assessment task before its due date. This assessment will take place outside your class time.
How: You will be assessed according to the criteria outlined in the assessment below. Please see below “Additional Instructions to Students”.
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Conditions for assessment
1. Students must submit an electronic copy of this assignment via RMIT CANVAS before its due date.
2. Students found in breach of test conditions can be charged with academic misconduct, have their results cancelled, be excluded from the program and receive other penalties. Penalties can also apply if a student’s test material is copied by others.
3. Plagiarism is the presentation of the work, idea or creation of another person as though it is one’s own. It is a form of cheating and is a very serious academic offence that may lead to expulsion from the University. Plagiarised material can be drawn from, and presented in, written, graphic and visual form, including electronic data, and oral presentations. Plagiarism occurs when the origin of the material used is not appropriately cited. 4. RMIT special consideration is to enable you to maintain your academic progress despite adverse circumstances. The process for special consideration can be found at https://www.rmit.edu.au/students/student-essentials/assessment-and-exams/assessment/special-
consideration
5. Students with a disability or long-term medical or mental health condition can apply for adjustments to their study and assessment conditions (Reasonable Adjustments and Equitable Assessment Arrangements) by registering with Equitable Learning Services (ELS)
at https://www.rmit.edu.au/students/support-and-facilities/student-support/equitable-learning-services
Additional Instructions:
1.
Performance requirement: a.
Must attempt all the questions b.
Satisfactory (S) performance
– successfully complete
at least 1 question
correctly from each group. There are all together 9 Groups (A to I). c.
Not Yet Satisfactory (NYS) performance
– unable to successfully complete at least 1 question
correctly in each group. 2. You are eligible to resubmit the assessment one more time
within allowed time frame, if your first attempt result is NYS.
3. Result from this assessment will count towards the final result of competent only if the result is satisfactory (S)
4.
Late submission approval will be in line with RMIT policy Application for extension of submittable work (7 calendar days or less) form.
5.
Special consideration for assessment Application for special consideration form.
6.
If a student’s result is NYS: feedback and agreed deadline date for the student to demonstrate competency will be advised.
7.
The date(s) must be within enrolment dates for this competency otherwise the student will be given a final result of NYS and the student may need to re-enrol/ repeat this course.
Instructions on submitting Knowledge Assessment You are required to submit this assessment task in CANVAS assignment submission page before its due date.
Equipment/resources students must supply (if applicable):
Equipment/resources to be provided by RMIT or the workplace (if applicable):
Writing pens/pencil, ruler and the like
Scientific calculator or similar
Computer and RMIT Internet Access
RMIT classrooms
Microsoft Office Suite
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STUDENT
Section B – Marking Guide:
Two Dimensional Geometry
Group A
Q1:
Express the angle “
θ
” in radian between the horizontal position and the wings of the airplane?
(PC 3.1)
Satisfact
ory
response
Ye
s
☐
N
o
☐
A:
Angle between the two axes is 90°.
Given angle is 63°.
Find θ.
Subtract given angle from 90°.
90° - 63° = 27°
Converting angle from degrees to radians using conversion factor 1° = π 180
27° = π x27
180
Simplify:
= 3π 20
= 3 x 3.14
20
= 0.417 radians.
Q2:
Convert 0.749 radians into equivalent nearest degree.
(PC 3.1)
Satisfact
ory
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response
Ye
s
☐
N
o
☐
A:
0.749 x 180
°
Π
= 0.749 x 180
°
3.14
= 134.82
°
3.14
= 42.9°
To the Nearest degree:
= 43°
Group B
Q3:
A piece of metal has been cut in the shape of a sector of a circle as shown below. What is the area of this piece of metal?
(PC 3.2)
Satisfactor
y response
Yes
☐
No
☐
A: Converting 120° into radians.
= 120 x π
180
= 2.0947
Area of the piece of metal = A = ½ r
2
θ
= ½ x 12
2
x 2.0947 = ½ x 144 x 2.0947 = 301.637
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2
= 150.8 cm
2
Q4:
A structural supporting member is made in the shape shown in below. What is the
outside
perimeter and shaded area
of the structural member?
(PC 3.2)
Satisfactor
y response
Yes
☐
No
☐
A:
P = 4 x a
P = 4 x 4”
P = 18”in
Area of shaded region = 5” x 4” = 20”
4.75” x 3.75” = 17.8125
= 20 – 17.8125
= 2.1875 in
2
Group C
Q5:
The liquid oxygen tank in the external tank of the space shuttle resembles a combination of a hemisphere, a cylinder, and a cone. Find the volume of the liquid oxygen tank.
(PC 3.3)
Satisfactor
y response
Yes
☐
No
☐
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A:
Volume of hemisphere
= 2/3 x π x r
3
= 2/3 x 3.14 x 4.2
3
= 2/3 x 3.14 x 74.088
= 2/3 x 232.64
= 465.27/3
= 155.17 Volume of Cylinder = π x r
2
x h
= 3.14 x 4.2
2
x 4
= 3.14 x 17.64 x 4
= 3.14 x 70.56
= 221.67
Volume of Cone = 1/3 x π x r
2
x h
= 1/3 x 3.14 x 17.64 x 8.1
= 1/3 x 3.14 x 142.884
= 1/3 x 448.656
= 149.63
Sum of all volumes = 155.17 + 221.67 + 149.63 = 526.47m
3
Q6:
Find the total surface area of (exterior and interior) of the PVC pipe shown.
(PC 3.3)
Satisfactor
y response
Yes
☐
No
☐
A:
Exterior surface area:
= 2
x π x r
2
+ 2 x π x r x h
=
2 x π x (5)
2
+ 2 x π x (5)(3) = 2 x π x (25) + 2 x π x (15)
= 50 x π + 30 x π = 80π in
2. 8.94 x 3.142 = 28.09. 251.36
Interior Surface area:
= 2
π x r
2
+ 2 x π x r x h
= 2 π x (1.6)
2
+ 2 x π x (1.6) (3)
= 2 π x (2.56) + 2 x π x (4.8)
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=5.12 π + 9.6π
= 14.72π. 46.25
Add the exterior and interior surface areas to find the total surface area:
80
π
in
2 + 14.72
π
in
2
= 94.72
π
in
2
Group D
Q7:
The figure below shows five equally spaced holes are drilled on an 80 mm pitch circle diameter
metal plate. Calculate their co-ordinates relative to axes 0
x
and 0
y
in (i) Polar form, (ii) Cartesian form.
(PC 3.4)
Satisfactor
y response
Yes
☐
No
☐
A:
Given: Diameter of the circle = 80mm
Radius = r = 40mm
The hole 2 lies on the y-axis
Its argument, θ = π/2
Its polar co-ordinates are (r,θ) = (40,π/2)
Angle between any two consecutive holes = 2
π/5
Argument of hole 1 = π/2 – 2π/2 = π/10
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Argument of hole 3 = π/2 + 2π/5 = 9π/10
Argument of hole 4 = 2π/10 + 2π/5 = 13π/10
Argument of hole 5 = 13π/10 + 2π/5 = 17π/10
The polar co-ordinates of hole 1= (40, π/10)
hole 2= (40, π/2)
hole 3= (40, 9π/10)
hole 4= (40, 13π/10)
hole 5= (40, 17π/10)
cartesian coordinates;
if polar coordinates are (r,
θ)
then the cartesian coordinates are (x, y) = (r cos
θ, r sin θ)
the cartesian coordinates of hole 1: = {40cos (
π/10), 40sin(π/10)}
= {40(0.9510565), 40 (0.30901070)}
= (38.042, 12.361).
The cartesian coordinates of hole 2:
= {40cos (π/2), 40sin (π/2)}
= {40(0), 40(1)}
= (0, 40).
The cartesian coordinates of hole 3:
= {40cos (9 π/10), 40sin (9 π/10)}
= {40(-0.9510565), 40(0.3090170)}
= ( -38.042, 12.361).
The cartesian coordinates of hole 4:
= (40cos (13 π/10), 40sin 913 π)}
= {40 (0.587785), 40(-0.89017)}
= (23.511, -32.381)
Q8:
(i)
Convert (25, 36) Cartesian coordinate into equivalent polar coordinate.
Satisfactor
y response
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(ii)
Express (4.5, 5.16 rad) polar coordinate in Cartesian co-
ordinates.
(PC 3.4)
Yes
☐
No
☐
A:
-
To convert Cartesian coordinates (25,36) into polar coordinates, we can use the following formulas:
r = √x
² + y
² θ = tan
-1
(y/x)
Given the Cartesian coordinates (25, 36), we can calculate the polar coordinates as follows:
r = √
25
2
+ 36
2
r = √
625+1,296
r = 1,921
r ≈ 43.83 θ=tan
-1
(3625)
≈55.22°
≈0.964
Therefore, the equivalent polar coordinates are approximately (43.83,0.964).
-
To express the polar coordinates (4.5,5.16rad) in Cartesian coordinates.
Polar coordinates (r,
𝜃
) = (4.5,295.65
°
) Since 5.16 radians = 295.65 degrees Use the following formulas:
x=rcos
𝜃
y=rsin
𝜃
The polar coordinates (4.5,295.650), we can calculate the Cartesian coordinates as follows:
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STUDENT
X = 4.5×cos (295.65
°
) ≈ 4.5×0.433 ≈ 1.9485 y = 4.5×sin (295.650) ≈ 4.5×−0.9014 ≈ −4.0563
Therefore, the equivalent Cartesian coordinates are approximately (1.97, −4.04).
Trigonometry
Group E
Q9:
A car weighing 1500N is on a 20
°
hill. Find the components of the car’s weight parallel “
W
x
” and perpendicular “
W
y
” to the road.
(PC 4.1)
Satisfactor
y response
Yes
☐
No
☐
A:
Given that:
𝜃 = 20°
Explanation:
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Parallel component of weight to the road force Wx = Wsin
𝜃
Perpendicular component of weight to the force Wy = Wcos
𝜃
Wx=1,500×sin20°
=0.342×1,500
=513N
Wy=1,500×cos20
°
=1,500×0.9397
=
1410N
Therefore Wy = 1409.55N and Wx = 513N.
Q10:
A surveyor is to determine the height of a tower. The transit is
positioned at a horizontal distance of 35 meters from the foot of the
tower. An angle of elevation of 58
°
is read in sighting the top of the
tower. The height from the ground to the transit telescope is 1.70
meters. This is shown in the diagram below. Determine the height “
h
” of
the tower. Satisfactor
y response
Yes
☐
No
☐
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STUDENT
(PC 4.1)
A:
Distance from tower to man = 35m
Angle of elevation = 58°
Tan 58° = opposite side/adjacent side
Tan 58 = x/35
1.60033 = x/35
x = 1.60033(35)
x = 56.0117
total height = x +1.7
h = 57.71
total height of tower = h = 57.71m
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STUDENT
Group F
Q11:
A highway entrance ramp rises 11.4 m in a horizontal distance of 55 m. What is the measure of the angle in degree that the ramp makes with the
horizontal? Give the answer in the nearest whole number.
(PC 4.2)
Satisfactor
y response
Yes
☐
No
☐
A:
We know that, tan
𝜃 = Ramp rise/Horizontal Distance
Where, Ramp rise = 11.4 m Horizontal Distance = 55 m tan
𝜃
=11.4/55
𝜃
=11.71°
𝜃
= 12 °
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STUDENT
Q12:
shape of a right triangle measure 25.50 cm and 28.18 cm as shown in the diagram below. What are the measures of the two acute angles “
θ
” and “
α
” of the piece in degrees?
(PC 4.2)
Satisfactor
y response
Yes
☐
No
☐
A:
-
Cos
𝜃 = ½ (25.5/28.18)
-
𝜃
= ½ cos
-1 (25.5/28.18)
-
𝜃
= 12°
-
sin
α = 25.5/28.18
-
α = sin
-1 (25.5/28.18)
-
α = 64.81°
Group G
Q13:
A crank mechanism of a petrol engine is shown below. Arm OA
is 10.0 cm long and rotates clockwise about O
. The connecting rod AB
is 30.0 cm long and end B
is constrained to move horizontally. For the position
shown below determine: (i)
The angle ABO
in degree
(ii)
The length OB
Hint: Use Sine Law.
(PC 4.3)
Satisfactor
y response
Yes
☐
No
☐
A:
(i)
Given that, OA = 10cm and AB = 30cm
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STUDENT
Using sine law
= a/sin A = b/sinB = c/sinC
In ABO:
Let angleB = 𝜃
10/sin𝜃 = 30/sin50
sin𝜃 = 10/30 x sin50
°
sin𝜃 = 0.255
On taking inverse of sin to find the angle.
𝜃 = sin
-1 (0.255)
𝜃 = 14.79
° = 15°
(ii)
The sum of all angles in a triangle = 180°
AngleA = 180 – (15 + 50)
AngleA = 115.21°
Using sin law
a/sin115.21° = 30/sin50°
a = (30/sin50°) x sin 115.21°
a = 35.43cm
OB = 35.43cm
The angle ABO
in degree = 15
°
The length OB = 35.4cm
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STUDENT
Q14:
A room 8.0m wide has a span roof which slopes at 33
°
on one side and 40
°
on the other as shown below. Find the length of the roof slope AB
. Hint: Use Sine Law.
(PC 4.3)
Satisfactor
y response
Yes
☐
No
☐
A:
-
Recognize that the sum of the angles in a triangle is 180
°
. Therefore, for the triangle formed by the roof and the ground, we have
𝑚 ∠𝐴
+
𝑚 ∠𝐵
+
𝑚 ∠𝐶
=180
°
Given that
𝑚 ∠𝐴
=33
°
and
𝑚 ∠𝐶
=40
°
calculate
𝑚 ∠𝐵
using the sum of angles in a triangle:
𝑚 ∠𝐵
=180
°
−33
°
−40
°
𝑚 ∠𝐵
=107
°
Apply the Law of Sines to find the length of the roof slope AB:
𝑏 /
sin
∠𝐵 = 𝑐 /
sin
∠𝐶
𝐴 𝐶 /
sin
∠𝐵 = 𝐴 𝐵 /
sin
∠𝐶
8.0/sin107
°
=
𝐴 𝐵 /
sin40
°
Solve for AB:
𝐴 𝐵 ⋅
sin107
° = 8.0
⋅
sin40
°
𝐴 𝐵
=8.0
⋅
sin40
°/
sin107
°
Calculate AB:
𝐴 𝐵
≈5.38
m
So, the length of the roof slope AB is approximately 5.4m.
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STUDENT
Group H
Q15:
A metal frame in the shape of an oblique triangle is to be fabricated. One side of the frame is 2.4 meters long. One end of the second side, which is 1.80 meters long, is to be fastened to an end of the 2.4 meters side at an angle of 58.008 as shown in the diagram below. Compute the required length of the third side of the frame. Hint: use Cosine law
.
(PC 4.4)
Satisfactor
y response
Yes ☐
No ☐
A:
Identify the given sides of the triangle and the angle between them. The sides are
𝑎
=1.8 m
and
𝑏
=2.4m
, and the angle between them is
𝐶
=58.00
Use the Cosine Law to find the length of the third side c, which is not given directly. The
Cosine Law states that
𝑐
2
=
𝑎
2
+
𝑏
2
−2
𝑎 𝑏
cos(
𝐶
)
Substitute the given values into the Cosine Law formula:
𝑐
2
= 1.8
2 + 2.4
2 – 2 x 1.8 x 2.4 x cos(58.00
∘
)
Calculate the value of
𝑐
2
:
𝑐
2
=3.24 + 5.76 – 2 x 1.8 x 2.4 x 0.5299
𝑐
2
=9.00 − 4.679088
𝑐
2
= 4.320912
Take the square root of
𝑐
2
to find the length of the third side c:
𝑐 = √
4.320912
𝑐 ≈ 2.10m
Therefore, the required length of the third side of the frame is approximately 2.10 meters.
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STUDENT
Q16:
Find angle B
for the roof truss shown in the figure below. Hint: Use Cosine law
. (PC 4.4)
A:
Law of cosine:
For any triangle ABC
Consider the sides AB = C, BC = A, AC = B
The law of cosines defined as,
CosA = b
2
+ c
2
– a
2
/2bc, CosB = a
2
+ c
2 – b
2
/ 2ac, CosC = a
2
+ b
2
- c
2
/2ab
ABC sides – a = 11, b = 6, c = 9.
Finding angleB using law of cosine
CosB = 11
2
+ 9
2 – 6
2
/ 2 x 11 x 9
CosB = 121 + 81 + - 36/198
CosB = 0.8384
B = Cos
-1 (0.8384)
B = 33.03
B = 33
°
Basic statistics
Group I
Q17:
The lengths of forty nails measured to the nearest centimetre are:
4,5,3,7,3,3,5,4,6,8,8,8,6,1,7,3,6,4,6,3
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RMIT Classification: Trusted
STUDENT
4,6,8,9,5,6,5,2,3,2,6,4,8,1,5,7,6,3,5,5
Calculate
i.
Mean
ii.
Median
iii.
Mode
iv.
Standard deviation and v.
Interpret the data on the graphical representation with an aid of any computer software (computational analysis)
(PC 7.1, 7.2)
A:
Arranging data in ascending order:
1,1,2,2,3,3,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,7,7,7,8,8,8,8,8,9
Mean = sum of all values divided by number of values:
=1+1+2+2+3+3+3+3+3+3+3+4+4+4+4+4+5+5+5+5+5+5+5+6+6+6+6+6+6+6+6+7+7+7+8+8+
8+8+8+9/40
= 200/40
Mean = 5
Median = middle value of the total given values.
Median =5 + 5/2 = 5
Median = 5
Mode = Value that appears the most often
Mode = 6
Standard deviation = s = √ sum of individual value – Mean/ total number of values -1
s =√ 164/40-1
s = 2.02
Q18:
A manufacturer of printed wring boards used in the assembly of personal computers is testing the quality of its soldering process. To test the strength of the solder joints, some of the soldered leads are pulled until they either break or come off the board. The pull strength measured in pounds for 30 tests are shown below. 59.0
58.0
55.0
63.0
62.0
64.0
62.0
60.0
64.0
60.0
55.0
68.0
66.0
65.0
65.0
64.0
53.0
68.0
66.0
66.0
69.0
59.0
66.0
66.0
68.0
61.0
68.0
62.0
64.0
65.0
By using the given data calculate:
[C6130 MEM30012A (MATH5268C) Student Assessment 2 ]
[ 1B of 3] [ 01/07/2022
]
Student knowledge assessment task
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FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0
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RMIT Classification: Trusted
STUDENT
Calculate
vi.
Mean
vii.
Median
viii.
Mode
ix.
Standard deviation and x.
Interpret the data on the graphical representation with an aid of any computer software (computational analysis)
(PC 7.1, 7.2)
A:
Arranging data in ascending order:
53.0, 55.0, 55.0, 58.0, 59.0,
59.0, 60.0, 60.0, 61.0, 62.0, 62.0, 62.0, 63.0, 64.0, 64.0, 64.0, 64.0, 65.0, 65.0, 65.0, 66.0, 66.0, 66.0, 66.0, 66.0, 68.0, 68.0, 68.0, 68.0, 68.0.
Mean = sum of all values divided by number of values:
= 53.0 + 55.0 + 55.0 +
58.0 +
59.0 + 59.0 + 60.0 + 60.0 + 61.0 + 62.0 + 62.0 + 62.0 + 63.0 + 64.0 + 64.0 +
64.0 + 64.0 + 65.0 + 65.0 + 65.0 + 66.0 + 66.0 + 66.0 + 66.0 + 66.0 + 68.0 + 68.0 + 68.0 + 68.0 + 68.0/30
Mean = 63.03
Median = middle value of the total given values.
Median = 64 + 64/2
= 128/2
Median = 64
Mode = Value that appears the most often
Mode = 66
Standard deviation = s = √ sum of individual value – Mean/ total number of values -1
S = 1831 – 63.03/30 – 1 S = 4.11
[C6130 MEM30012A (MATH5268C) Student Assessment 2 ]
[ 1B of 3] [ 01/07/2022
]
Student knowledge assessment task
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RMIT Classification: Trusted
STUDENT
Section C – Feedback to Student Has the student successfully completed the task?
Yes No
Feedback to student:
Assessor Name
Date
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