Assignment 1
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University of Guelph *
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Course
2270
Subject
Mechanical Engineering
Date
Dec 6, 2023
Type
Pages
5
Uploaded by DeanHeat18314
Assignment #1
Goal:
The goal of Assignment #1 is to complete Biomechanical analyses of static
equilibrium
.
The Task:
Solve for
Net Joint Forces and Moments; Solve for Muscle Forces & Bone
on Bone Forces; and finally solve for Compression & Shear forces.
Given the questions provided:
a)
solve the questions asked in the space provided
b)
scan in your answers and submit these as a .pdf document.
NOTE:
Only a .pdf copy will be accepted. All other versions uploaded will NOT be graded and
will receive a ZERO.
There are several free software available that can compile multiple .jpg files into a single
.pdf; these are very easy to use (e.g.
https://jpg2pdf.com/
or Adobe Scanner)
Due on Wednesday October 11 at 9:00 AM by
CourseLink Drop box
This assignment is worth 8% of your overall grade in course
Penalty of -5% off of the total assignment for each day this assignment is late
.
For example:
You deposit your assignment at 9:30 am Oct 11, 2023
Even
if you get
35/35
(100%) on the assignment
è
you will be graded
33.25/35
and
receive
7.6 out of 8%
for this component of the course
Assignments handed in
³
7 days late will not be graded.
A grade of zero will be recorded.
Student Name: ___________________________
Student ID#: ______________________
1
HK*2270 Principles of Human Biomechanics
Assignment #1
PLEASE SOLVE THE QUESTIONS BELOW IN THE SPACE PROVIDED.
Question 1. (20 marks)
_____________
Question 2. (6 marks)
_____________
Question 3. (9 marks)
_____________
Total (35 marks): ________
1. (20 marks) You are a sport biomechanist who has been asked to perform a biomechanical
analysis of a paralympic skier as they turn during a race.
Note:
his right leg is pushing off the
snow to change direction.
Draw a free body diagram of the skier’s right knee (red dot) and calculate the
Net Joint Forces and Moment at his right knee. Assume Static Equilibrium.
His body mass is 80 kg
The lower leg is 6% of his body mass
The foot is 2% of his body mass
His lower leg/foot is oriented 55 degrees
(from right horizontal)
The ground reaction force acting under the right foot
acts in
XY space
and is located 20 cm to the left and 32
cm below the knee joint (from his perspective)
From the knee:
Lower leg COM is located 17 cm along the extended leg
Prosthetic foot COM is located 28 cm along the
extended leg
150⁰
F
ground
= 640N
Knee
55⁰
+y
+x
M
+
Space for calculations:
(a) Free Body Diagram
morgan
brewster
1229518
-
i
..
3
Y
X
Yi
,
x
SOH
CAH
TOA
A
I
1300
x
1880
⑱
>
380
3
⑧
-
158
:
-
e
stee
ing
H
x
7gx
weight
of
leg
:
weight
of
foot
We
=
0
02
x80
x
9
81
W
,
0
06
x
80
x
9
81
We=
15
.
696
N
W
,
47
088N
↓
firm-foot
from-leg
Student Name: ___________________________
Student ID#: ______________________
2
(e) The right knee is being held in a static position by a __________ (
Choose one:
flexion;
extension; hyperextension; abduction; adduction; circumduction
) moment about the knee.
(f) What is the plane of motion for the action identified in 1(e)?
_______________________
(g) What is the axis of motion identified in 1(e)?
___________________________________
(h) Provide a definition of the movement identified in 1(e) in the space below.
(b)
∑
F
x
= 0:
(c)
∑
F
y
= 0:
(d)
∑
M = 0:
morgan
brewster
1229518
Fknee-x-Fgx
0
Finee-y
+
fgy-from-leg
-
fcom-foot
=
0
Finee
-x
=
fgx
Ficnee-y
=- fgy+fcom-leg
+
fom-foot
finee
x
6
40N
x
C0S38
·
Ficnee
-
y
=
-640NxS1n300
+
47
088N+
15
696N
Finee
-
x=
554
N
f
knee
-y
=
-320N
+
47
088N
+
15
696N
the
force
on
the
knee
In
the
x
direction
Is
SS4N
ht
finee-y
=-
23TN
rig
:
the
force
on
the
knee
In
the y
direction
Is
ISTN
down
Mic
-
Mcom-leg
-
Mcom-foot
+
Megy
-
Megx
Mic
=
+
Mcom-leg
+
Mcom-foot
-
Megy
+
Megx
Mis
=
Farm-leg
x d1
+
from-foot
xd1-Fgy
>
d1
"fgx
&1
M
,
x
=
47
88840
17m
+
15
696
+
0
28m
-- 257
x0
.
32m) - (554
=
0
28m)
Mx
=
8
00496N
-
m
+
4
39488N-m-(-82
.
24N
m)
+
/10
.
8N
M
Mic
=
20SNM
the
momentum
in
the
knee
is
205
N
m
counter
clockwise
adduction
frontal
plane
anterior
-
posterior
axis
adduction
is
the
movement
of
a
body
part
towards
the
bodies
midline
Your preview ends here
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Student Name: ___________________________
Student ID#: ______________________
3
2. (6 marks) One of your classmates decided to analyze the role of the steering arm during a
turning motion within a ski race. Given the information below, calculate the (a) muscle force
(Fsme) and (b) bone-on-bone forces acting at the elbow.
Your classmate calculated the F
elbow-x
= 178 N and F
elbow-y
= 273 and the M
elbow
= 84.1 Nm.
Note
that the single muscle equivalent acts at a distance (along dashed line) of 1.8 cm from the
elbow joint at 141 degrees from right horizontal; Arrow heads indicate direction of forces
applied.
(a) Muscle Force (Fsme)
(b) Bone-on-Bone Forces at the Elbow:
morgan
brewster
1229518
I
a
M
S
t
1
-
Jux
BNBx-fsmex
Jry
BNBy
Sme-y
IT8N
BNB
x
-5400C0s39
·
IT3N
=
BNBx+
S4WNSI
39:
178N
=
BNBx
(4194
593
IT3N
=
BNBy+
3398
33N
BNBx=
43
TON
BNBY=-3130N
Me
Fsme
-y
x
d1
H
-84
/Nm
Fsme-y
x
0
018m
0
390
1211
Esme-y
=
4672
IN
:
the
bone
on
bone
forces
in
the
X
direction
is
4370N
right
A
Isme=
Esme
-y
and
in
the
y
direction
3130N
down
Sin
39:
-sme
=
4672
2
N
Sin39
Fsme
=
5400N
:
Esme=
S400N
at
39
·
Clockwise
Student Name: ___________________________
Student ID#: ______________________
4
3. (9 marks) Your classmate has been given a task from the skier’s coach and has asked you
for help. They have asked you to calculate the total compression and shear forces at the skier’s
elbow. Be sure to use an appropriate,
anatomically relevant
axis system for these calculations.
Please use the angle and Bone-on-Bone forces provided in the Diagram below; please
note the direction of arrows for your calculations.
Diagram:
Compression Forces:
Shear Forces:
morgan
brewster
1229518
compression
-
-
Ve
shear
O
A
BrBy
Shear
4000N
42
ampression
⑦
B
Y
BrBx
55UON
a
18420
⊥
espear
Bett
Compx=
Buxx
SinG
COMPy
-
BuByxCOSt
Shearx=BNBx
x
COSO
-
Sheary=
BuByx
SinG
-
⑧
-COMPX=
5500NxSin42
-COmpy=
4000NxCOS42
-
Shearx=SS00NxCOS42
-
Sheary
4000NxS1n42
·
-COMPX=-504/N
Compx=
5041N
COMpy=-2973N
-
Shearx=-1200N
-Sheary=
2677N
Shear
=
2200N
Sheary=-2677N
Comptotal
=
COMDx
+
COMPy
Shear
total
=
Shearx+
Sheary
COMPtotal
=5041N
+12973N)
Sheartal
=
2200N
+1 -
2677N)
COMProtal
=
20G8N
Shearttal
=
-
47TN
:
total
compression
Is
2068N
:
total
Shear
1S-477N
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