PBSet#12_Solutions_2023
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University of Guelph *
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Course
2270
Subject
Mathematics
Date
Jan 9, 2024
Type
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21
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Principles of Biomechanics
1
HK*2270
Problem Set #12
P
ART
A.
D
YNAMIC
E
QUILIBRIUM
Question 1.
Given the FBD of a foot in swing phase and the following information, solve for the reaction
forces and net joint moment at the ankle for a 1.76 m tall female with a mass of 57.2 kg.
Her foot length is 25% of her overall height
.
ax = 9.07 m/s
2
ay = -6.62 m/s
2
α = 21.69 rad/s
2
α
ay
ax
9.85 cm
8.60 cm
1.95 cm
0.8 cm
+y
+x
M
+
+y
+x
M
+
Ankle
Toe
3
Question 3. You are given the following
DYNAMIC
data about a long jumper.
For the Lower Leg & Foot combined segments
You are given the following
DYNAMIC
data about a long jumper.
For the Lower Leg & Foot combined segments
Accelerations from COM:
a
x
= - 7.20 m/s
2
a
y
= - 2.43 m/s
2
α
= - 3.20 rad/s
2
Moment of inertia about combined Lower Leg & Foot
COM
= 0.876 kg*m
2
Vertical Ground reaction force under the ball of foot = 162.5 N, oriented
48° below the horizontal
Mass of person = 64.3 kg
In absolute space:
Knee location to Lower Leg & Foot
COM
= 14.5 cm posterior; 11.5 cm above combined
COM
Heel of foot location to Lower Limb & Foot
COM
= 16.70 cm anterior and 8.70 cm below combined
COM
A. Complete the free body diagram of the situation described above in the space provided.
B. Solve for the joint reaction forces at the knee.
Free Body Diagram
Person’s mass: 64.3kg.
Mass proportion for the foot from anthropometric tables: 0.061
Mass of lower leg & foot: (0.061) * 64.3 kg = 3.9223 kg
Weight of lower leg & foot: 3.9223 kg * 9.81 = 38.4778 N
Since F
Gnd
= 162.5 N,
F
gnd-x
= 162.5 cos 48 = 108.734 N
F
gnd-y
= 162.5 sin 48 = 120.761 N
Σ
F
x
= ma
x
:
-F
gnd-x
+ R
Kx
= m
LL&oot *
a
x
- 108.734 + R
Kx
= (3.9223) * (- 7.20)
-108.734 + R
Kx
= - 28.241
F
Kx
= + 80.493 N
Σ
F
y
= ma
y
:
F
gnd-y
+ R
Ky
– Fcom = m
LL&foot *
a
y
120.761 + R
Ky
– 38.4778 = (
3.9223
) * (- 2.43)
82.283 + R
Ky
= - 9.531
F
Ky
= - 91.814 N
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4
C. Solve for the net joint moment at the knee.
D. Based on the moment calculated for this dynamic situation, is the knee tending to
FLEX
or
EXTEND
(circle one)
E. Thus the _______________________________ muscles must be the major contributors to the production
of the moment in this dynamic situation
KNEE FLEXORS
or
KNEE EXTENSORS
(circle one)
Σ
Mabout COM = ICOM
α
M
K
- M
Gx
+ M
Gy
– MR
Kx
+ MR
Ky
= I
com
α
M
K
- (F
Gx
)(0.0870) + (F
Gy
)(0.1670) - (R
Kx
)(0.1150) + (R
Ky
)(0.1450) = I
com
α
M
K
- (
108.734
)(0.0870) + (
120.761
)(0.1670) - (
80.493
)(0.1150) + (
91.814
)(0.1450) = (
0.876
) (
-3.20
)
M
K
- (
9.4599
) + (
20.1671
) - (9.2567) + (13.3130) =
-
2.8032
M
K
+ 14.7635=
-
2.8032
M
K
=
-
17.5667 N.m
7
Question 4. You are given the following
DYNAMIC
data about a lifting task.
For the combined Forearm & Hand
Accelerations from the COM:
a
x
= 2.10 m/s
2
a
y
=
3.54 m/s
2
α
= 5.67 rad/s
2
Moment of inertia about the Forearm & Hand
COM
= 0.078 kg*m
2
Force of the basket on the hand is 147 N, oriented 35° below the horizontal
Mass of person = 59.2 kg
In absolute space:
Elbow location to Forearm & Hand
COM
= 10.30 cm posterior and 12.40 cm above Forearm & Hand
COM
Basket location to Forearm & Hand
COM
= 9.34 cm anterior and 7.4 cm below Forearm & Hand
COM
A. Complete the free body diagram of the situation described above in the space provided.
B. Solve for the joint reaction forces at the elbow.
Free Body Diagram
Person’s mass: 59.2 kg.
Mass proportion for the forearm & hand from anthropometric tables: 0.022
Mass of forearm & hand: (0.022) * 59.2 kg = 1.3024 kg
Weight of forearm & hand: 1.3024 kg * 9.81 = 12.777 N
Since F
basket
= 147 N,
F
basket-x
= 147 cos 35 =
120.415 N
F
basket-y
= 147 sin 35 = 84.316 N
Σ
F
x
= ma
x
:
- F
basket-x
+ R
Ex
= m
f+h *
a
x
- 120.415 + R
Ex
= (1.3024)
*
(2.10)
- 120.415 + R
Ex
= (2.735)
R
Ex
= 123.150 N
Σ
F
y
= ma
y
:
F
basket-y
+ R
Ey
- W
f+h
= m
f+h *
a
y
84.316 + R
Ey
– 12.777 = (1.3024)
*
(3.54)
71.539 + R
Ey
= (4.6105)
R
Ey
= - 66.929 N
8
C. Solve for the net joint moment at the elbow.
D. Based on the moment calculated for this dynamic situation, is the elbow tending to
FLEX
or
EXTEND
(circle one)
E. Thus the _______________________________ muscles must be the major contributors to the production
of the moment in this dynamic situation
BICEPS
or
TRICEPS
(circle one)
Σ
Mabout COM = ICOM
α
M
E
+ M
Basket-x
+ M
Basket-y
+ MR
Ex
+ MR
Ey
= I
com
α
M
E
- (
M
Basket-x
)(0.074) + (
M
Basket-
y
)(0.0934) - (R
Ex
)(0.1240) + (R
Ey
)(0.1030) = I
com
α
M
E
- (
120.415
)(0.074) + (
84.316
)(0.0934) - (
123.150
)(0.1240) + (
66.929
)(0.1030) = (
0.078
) (5.67)
M
E
- (
8.9107
) + (
7.875
) - (
15.2706
) + (
6.8937
) =
0.44226
M
E
- 9.4126 =
0.44226
M
E
=
+9.855 N.m
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9
Question 5. You are given the following
DYNAMIC
data about a figure skater
For the combined Forearm & Hand
You are given the following
DYNAMIC
data about a figure skater.
For the Foot segment
Accelerations from COM:
a
x
= 4.20 m/s
2
a
y
= - 4.37 m/s
2
α
= 7.89 rad/s
2
Ground reaction force under the ball of foot = 519 N, oriented 127° below the horizontal
Mass of person = 51.0 kg; height of person = 156 cm
moment of inertia about Foot
COM
= 0.0006176 kg*m
2
In absolute space:
Ankle joint location to Foot
COM
= 8.10 cm posterior; 4.9 cm above combined
COM
Distance from ground reaction force to Foot
COM
= 5.7 cm anterior and 4.3 cm below Foot
COM
A.
Complete the free body diagram of the situation described above in the space provided.
B. Solve for the joint reaction forces at the ankle.
Free Body Diagram
Σ
F
x
= ma
x
:
F
gnd-x
+ R
Ax
= m
foot *
a
x
(519 cos 53) + R
Ax
= (0.7395) * (4.20)
(312.34) + R
Ax
= 3.1059
R
Ax
= - 309.23 N
Σ
F
y
= ma
y
:
F
gnd-y
+ R
Ay
– F
com
= m
foot *
a
y
(519 sin 53) + R
Ay
– (0.7395 * 9.81) = (0.7395) * (- 4.37)
414.492 + R
Ay
- 7.254 = - 3.232
407.238 + R
Ay
=
- 3.232
R
Ay
= - 410.47 N
10
C. Solve for the net joint moment at the ankle.
D. Based on the moment calculated for this dynamic situation, is the ankle tending to
PLANTARFLEX
or
DORSIFLEX
(circle one)
ΣM
@COM
= I
COM
α
M
A
+ M
Gx
+ M
Gy
+ MR
Ax
+ MR
Ay
= I
com
α
M
A
+ (F
Gx
)(0.043) + (F
Gy
)(0.057) + (R
Ax
)(0.049) + (R
Ay
)(0.057) = I
com
α
M
A
+ (
519 cos 53
)(0.043) + (
519 sin 53
)(0.057) + (
309.23
)(0.049) + (
410.47
)(0.0810) = (
0.0006176
)
(
7.89
)
M
A
+ (312.34)(0.043) + (414.492)(0.057) + (
309.23
)(0.049) + (
410.47
)(0.0081) = (
0.0006176
)
(
7.89
)
M
A
+ (
13.4306
) + (
23.626
) + (15.152) + (33.248) =
0.004873
M
A
+ 85.46=
0.004873
M
A
=
-
85.46 N.m
11
Question 6. You are given the following
DYNAMIC
data about a long distance runner.
For the entire leg
Accelerations from COM:
a
x
= - 2.78 m/s
2
a
y
= 6.89 m/s
2
α
= 9.76 rad/s
2
Ground reaction force under the foot = 1275 N, oriented 56° to the horizontal
Mass of person = 65.2 kg; height of person = 179 cm
Moment of inertia about LowerExtremity
COM
= 0.0951 kg*m
2
In absolute space:
Hip joint location to LowerExtremity
COM
= 4.8 cm to the runner’s right and 18.6 cm below hip
Ground reaction force to LowerExtremity
COM
= 3.2 cm to the runner’s left and 35.6 cm below COM
A.
Complete the free body diagram of the situation described above in the space provided.
B. Solve for the joint reaction forces at the ankle.
Free Body Diagram
m
foot =
Body mass *LowerExtremity
MassProportion
= 65.2*0.161 = 10.4972 kg
Σ
F
x
= ma
x
:
F
gnd-x
+ R
Hx
= m
foot *
a
x
(1275 sin 34) + R
Hx
= (10.4972) * (-2.78)
(712.971) + R
Hx
= -29.1822
R
Hx
=
- 742.153
N
Σ
F
y
= ma
y
:
F
gnd-y
+ R
Ay
– F
com
= m
foot *
a
y
(1275 cos 34) + R
Ay
– (10.4972 * 9.81) = (10.4972) * (6.89)
1057.023 + R
Ay
- 102.978 = 72.326
954.045 + R
Ay
=
72.326
R
Hy
=
- 881.719
N
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12
C. Solve for the net joint moment at the hip.
D. Based on the moment calculated for this dynamic situation, is the hip tending to
ABDUCT
or
ADDUCT
(circle one)
ΣM
@COM
= I
COM
α
M
H
+ M
Gndx
+ M
Gndy
+ MR
Hx
- MR
Hy
= I
com
α
M
H
+ (F
Gx
)(
0.356
) + (F
Gy
)(
0.032)
+ (R
Hx
)(
0.186
) - (R
Hy
)(
0.048
) = I
com
α
M
H
+ (
1275 sin 34
)(
0.356
) + (
1275 cos 34
)(
0.032
) + (
742.153
)(
0.186
) - (
881.719
)(
0.048
) = (
0.0951
)
(
9.76
)
M
H
+ (
712.97
)(
0.356
) + (
1057.023
)(
0.032
) + (
742.153
)(
0.186
) - (
881.719
)(
0.048
) = (
0.0951
)
(
9.76
)
M
H
+ (
253.82
) + (
33.82
) + (138.04) - (42.32) =
0.9282
M
H
+ 383.4 =
0.9282
M
H
=
-382.4
N.m
Principles of Biomechanics
11
HK*2270 Problem Set #12
P
ART
B.
E
NERGY
,
W
ORK AND
P
OWER
1.
How much mechanical energy does a sprinter, with a mass of 70 kg, posses whio is running with a
linear velocity of 12.00 m/s and has their center of mass located 115 cm above the ground? (ignore
rotational energy)
2.
A 60 kg peson in a wheelchair has a velocity of 2.0 m/s. Ignoring potential and rotatational energy,
how much energy does the individual posses?
3.
During a complex springboard dive, 4 distinct positions were observed and the resulting total body
moment of inertia was calculated.
At take off I
o
= 19.8 kg*m
2
In a pike position I = 5.9 kg*m
2
In a tuck position I = 3.8 kg*m
2
At entry I = 19.8 kg*m
2
a.
If the initial take off angular velocity was 5.0 rad/s, how much angular momentum does the diver
have at take off?
b.
Determine the angular velocity for the pike, tuck and entry positions.
4.
A 71.356 kg man climbs 20 stairs of the CN Tower in Toronto (each 0.2 m high) in a period of 8 sec.
a. Calculate the work done
b. Calculate the average power
c. Calculate the change in potential energy
5.
A 50 kg woman starts from rest and performs a vertical jump. She leaves the ground with a vertical
velocity of 2.0 m/s.
a. What is the magnitude of the impulse?
b. What is her maximum kinetic energy?
c. What is the maximum potential energy?
d. When do the maximum kinetic and potential energies occur?
6.
A ball carrier (80 kg) hits a defensive player (135 kg) who is standing motionless in a vertical
position. The ball carrier is leaning forward with an angle of 80 degrees above the horizontal and
exerts a force of 1780 N in that direction. The ball carrier is running with a velocity of 8.0 m/s. What
will be the result of the impact?
7.
A 800 N individual drops off a desk and hit the ground with a velocity of 5 m/s. Assuming no force
absorbtion (elastic collision), calculate the force of landing. Give your answers as a multiple of body
weight.
a. if he stops in 1/100 second
b. if he stops in ¼ second
c. why are these two answers different? i.e. What is happening?
8.
A 190 kg running back has a speed of 8.0 m/s. What is his, A. kinetic energy? B. potential energy?
9.
A 700 N diver at the peak of her dive achieves a peak height of 3.6 m above the water. What is her:
a. Kinetic Energy?
b. Potential Energy?
c. Kinetic energy at 1 m above the water?
d. Velocity when she hits the water?
Principles of Biomechanics
12
P
ART
B.
A
NSWERS
1. Mechanical energy? 5.83 kJ
2. How much energy? 120 J
3. a. 99.0 kg*m
2
b. Pike = 16.78 rad/s; Tuck = 26.05 rad/s; Entry = 5.0 rad/s
4. a. work done = 2800 J
b. average power = 350 W
c. change in potential energy = 2800 J
5. a. magnitude of the impulse?
100 N*s
b. maximum kinetic energy? 100 J
c. maximum potential energy? 100 J
d. maximum kinetic energy? Take off. Maximum potential energy? Peak height of jump
6. The defensive lineman will be pushed backwards with a velocity of 2.98 m/s
7.
a. Impact (landing) force = 51 times body weight,
b. Impact (landing) force = 2 times body weight.
c. Although the impulse is the same, duration is different, therefore the average force will be
reduced if the time duration is LONGER. The peak force will be attenuated.
8. KE: 6080 Joules; PE = 0 Joules
9.
a. Kinetic Energy? 0 Joules. Why? At peak height, all energy is PE and velocity = 0 m/s;
b. Potential Energy? 2520 Joules at peak height
c. Kinetic energy at 1 m above the water? 1820 Joules
d. Velocity when she hits the water? 8.4 m/s
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Principles of Biomechanics
1
EQUATION SHEET- HK*2270 PRINCIPLES OF BIOMECHANICS
v = Δd
a = Δv
Δt
Δt
_______________________________________________
v
f
= v
o
+ at
d
f
= d
o
+ v
o
t
d
f
= d
o
+ v
o
t + ½at
2
v
f
2
= v
o
2
+ 2a (d
f
- d
o
)
d
f
= d
o
+ ½ (v
o
+ v
f
)t
_______________________________________________
v
fx
= v
ox
d
x
= d
ox
+ v
ox
t
v
fy
= v
oy
+ gt
d
fy
= d
oy
+ v
oy
t + ½gt
2
v
fy
2
= v
oy
2
+ 2g(d
fy
- d
oy
)
d
fy
= d
oy
+ ½ (v
oy
+ v
fy
)t
_______________________________________________
θ =
s
1 radian = 180 degrees
≈ 57.3 degrees
r
π
ω
= dθ
α = dω
d
t
d
t
ω
f
= ω
o
+ αt
θ
f
= θ
o
+ ω
o
t + ½αt
2
ω
f
2
= ω
o
2
+ 2α (θ
f
- θ
o
)
θ
f
= θ
o
+ ½ (ω
o
+ ω
f
)t
_________________________________________________________________
v
t
= rω
a
t
= rα
a
r
= rω
2
and
a
r
= v
t
2
r
a = √a
t
2
+ a
r
2
_________________________________________________________________
Principles of Biomechanics
2
ANTHROPOMETRIC FIGURE AND TABLE
Proportions
used to calculate various body segment parameters
(adapted from Robertson, 2004).
Segment
p
Kcom
Rprox
Rdist
Hand
0.006
0.297
0.506
0.494
Forearm
0.016
0.303
0.430
0.570
Forearm and hand
0.022
0.468
0.682
0.318
Upper Arm
0.028
0.322
0.436
0.564
Whole Upper extremity
0.050
0.368
0.530
0.470
Foot
0.0145
0.475
0.500
0.500
Lower Leg
0.0465
0.302
0.433
0.567
Lower Leg and foot
0.061
0.416
0.606
0.394
Thigh
0.1
0.323
0.433
0.567
Whole Lower extremity
0.161
0.326
0.447
0.553
Head and neck
0.081
0.495
1.00
0.000
Trunk
0.497
0.500
0.500
0.500
Trunk, head and neck
0.578
0.503
0.660
0.370
p:
segment mass as proportion of total body mass
Kcom:
segment radius of gyration about segment center of mass as proportion of segment’s length
Rprox:
location of center of mass from proximal ends of segment as proportion of segment’s length
Rdist:
location of center of mass from distal ends of segment as proportion of segment’s length
0.039H
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