Kinesiology 201 Solutions
Simple Biomechanical Models
Tony Leyland
School of Kinesiology
Simon Fraser University
1. a) It is not easy to draw the joint
reaction force as you can only guess
at its direction. However, it is usually
large so as to counteract the muscle
force. And as the muscle force is
usually a much lager forece than the
external laods and segment weights,
the joint reaction force is generally
pretty much close to being in the
opposite direction of the muscle
force (slightly modified by the
gravitational and other external
forces of course).
Fquads
Fjoint
FKin-Com
Mgshank
Fdistal (due to mg foot)
b. Σ Torques = 0 Torquemachine + Torquegravity+ Torquemuscle = 0
Mass of shank and foot = 4.27 kg
Weight = mg = -41.8887 N
Therefore torque due to body segment weight at angle θ = cosθ x 41.8887
Note: If you assume the shank is aligned similar to that shown in the diagram above
(and the photograph), then the torques due to the machine and gravity are positive
(anticlockwise) and hence the torque due to quadriceps action is negative. However, the
quantity of the torque is the crucial factor and the directional component unimportant as
it would change if the subject faced the other way.
60o
c.
230 + (41.887 x 0.3cos60 o)+ TQ60 = 0
TQ60 = -236.28 Nm
Torque = force x perpendicular distance (moment arm).
Muscle torque = 230 + 6.28 = 236.28 Nm (torque due to gravity at 60o = 6.28 Nm)
FQ
MQ
30o
0.09
Therefore: FQsin30(0.09) = MQ
FQ (0.045) = 236 Nm
FQ = 5244 N
60o
FQ 30o
FQ is in fact vertical as the limb is at 60o flexion
1
We should now determine the value of the force exerted on this limb by the Kin-Com
machine (FKC) and redraw the free-body diagram.
5244
FKC = FKC/0.25 = 230/0.25 = 920 N
920
ΣFy = 0 (solve for y-component of joint force FJY)
mg + FKCsin30 + FQ + FJY = 0
-41.89 – 920(0.5) + 5244 + FJY = 0
∴
FJY = -4742 N
This makes sense as the quads are pulling up.
You should be
able to see why
the vertical
component of the
Kin-Com force is
FKCsin30o
41.89
ΣFx = 0 (solve for y-component of joint force FJX)
FKCcos30 + FJX = 0
920(0.866) + FJX = 0
∴
FJX = -796.7 N
This makes sense as the action would be from right-to-left.
∴
Fjoint
FJ = √{(-796.7)2 + (-4742)2} = 4808 N = at θ = tan-1(-4742/-796.7) = tan-1(5.95) = 80.5o
This vector lies in the third quadrant so add 180o
∴ θ = 180 + 80.5 = 260.5o
∴ FJ = 4808 N at an angle θ = 261o
d. As we are assuming the moment arm is constant, the ratio of the torques will be the
same as the ratio of the forces.
10o
72.37 / 112.37 = 0.64
60o
236.28 / 256.28 = 0.92
40o
159.63 / 189.63 = 0.84
90o
100 / 120.00 = 0.83
e. The greatest reduction in strength is clearly towards full extension. “The vastus
medialis extends most forcibly in the last 10o to 20o of extension.” Thompson, C.W.
Manual of Structural Kinesiology (11th Ed.) Times Mirror/Mosby, 1989.
2. a) Maximal Voluntary Contraction (MVC) = 3600N
Should not exceed 5% of MVC
∴ Maximum tension (force) = 0.05 x 3600 = 180 N
b) The free-body diagram is opposite (R is the unknown
joint reaction force). I chose the system to be the
head and helmet. If the individual were on a bike
travelling at speed air resistance would be a
substantial force and should be included.
c) This is a static situation so ΣΜ = 0.
Take moments about C7 so one of your unknown values
(R and Fm) does not produce a moment about C7 (R) and
can be eliminated from this equation.
mghelmet
mghead
Fm
R
Mass of head/neck = 8% of total body mass = 0.08 x 70 = 5.6 kg (mg =54.94 N)
2
Centre of gravity of head and neck = 60% of segment length = 0.6 x 0.28 = 0.168 m
Moment arms for the three forces:
Fm ⇒ 0.04 m (given in question)
mghelmet ⇒ 0.24cos55o = 0.138 m
mghead ⇒ 0.0.168cos55o = 0.0964 m
35
o
55
o
In summing the moments we set Fm to be 180 N (the maximum allowable muscle force)
and solve for what we want to find out; namely the maximum allowable helmet mass.
ΣΜ = 0
Mhead + Mhelemt + Mmuslce = 0
{5.6 x (-9.81) x 0.0964} + {Masshelmet x (-9.81) x 0.138} + (180 x 0.04) = 0
-5.294 – 1.354Masshelmet + 7.2 = 0
Masshelmet = 1.906/1.354 = 1.408 kg
As most helmets on the market these days are 400 grams or so you can see there is
unlikely to be a problem with fatiguing the neck extensor muscles.
d) Compression along segment longitudinal axis (FC).
We must first calculate Fm as it is no longer 5% of MVC. It is now the force needed to
keep the system in equilibrium when the helmet is 0.4 kg.
ΣΜ = 0
Mhead + Mhelmet + Mmuscle = 0
{5.6 x (-9.81) x 0.0964} + {0.4 x (-9.81) x 0.138} + (Fm x 0.04) = 0
-5.294 – 0.5416 + 0.04 Fm = 0
o
Fm = 5.84/0.04 = 145.89 N
35
The line of action of the muscle was given as 35o
from the horizontal and the segment lies at 55o from
the horizontal. So the angle between the muscle’s
line of action and the segment is 20o.
Compression =
145.89 N
o
35
o
3.94 N
54.94 N
FC = (3.924 + 54.94)cos35o + 145.89cos20o
FC = (58.864 x 0.819) + (145.89 x 0.94)
FC = 48.22 + 137.092 = 185.3 N
3. If we draw a free-body diagram of the head we
see that the resultant contact force (FC) must
counteract the weight of the head (mg) and the
muscle force (FM).
FM
FMY + mg + FCY = 0
-34.415 – 49.05 + FCY = 0
∴ FCY = 83.465 N
35o
mg
Weight of head = 5 x g = 5 x (-9.81) = -49.05 N
Vertical = FMY = 60 x sin35o = -34.415 N
Components of FM
Horizontal = FMX = 60 x cos35o = 49.149 N
ΣFy = 0
20
FC
Where: FCY = vertical component of FC
3
ΣFx = 0
∴
FMX + FCX = 0
49.149 + FCX = 0
∴ FCY = -49.149 N
Where: FCX = horizontal component of FC
FC = √{(83.465)2 + (-49.149)2} = 96.81 N = 96.9 N
-59.5o
θ = tan-1(83.465/-49.149) = tan-1(-1.698) = -59.5o
120.5o
∴ θ = 180 - 59.5 = 120.5o
This vector lies in the second quadrant so add 180o
∴ FC = 96.9 N at an angle θ = 120o
Redraw spine and forces to help visualise
the angles between vectors and the
compression and shear components.
145o
25o
96.9 x cos25o = 87.8 N
96.9 x sin 25o = 40.95 N
Compression
Shear
∴ Compression = 87.8 N
Compression
120o
Shear
Shear = 41 N
4. The first step is to redraw the vectors, as they must be put tip-to-tale. The law of
cosines will allow us to calculate the magnitude of the resultant vector.
FQ
The angle θ is 180o – α = 180o – 117o = 63o
α
R
θ
R2 = FQ2 + FT2 – 2(FQ)(FT)cosθ
R2 = 15002 + 15002 – 2(15002)cos63o
R2 = 2250000 + 2250000 – (4500000)(0.454)
R2 = 4500000 – 2042957
R = √2457043 = 1567.5 N
FT
F/Q
To get the angle for this vector we can use the fact that the lower triangle above is an
isosceles triangle (two angles the same). This means that the vector FC splits the angle
between FQ and FT in two (58.5o each). As FT lies at 37o to the horizontal we can
calculate with some basic geometry that the vector FC is at 21.5o to the right horizontal.
FQ
37
o
37
o
o
58.5
FC
58.5
21.5o
o
58.5
o
10
o
63
o
o
21.5
FT
Therefore: FC = 1568 N at 21.5o
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b) Compression at the patellofemoral joint is the vector sum of tension in the
quadriceps and the patella tendon. As the two diagrams below show, in extension
this vector sum is low. As flexion increases compression increase for two reasons:
1) the change in orientation of the force vectors; and 2) increased tension in the
tendons due to the increased quadriceps activity required to maintain body position.
FQ
FQ
Fc
Fc
FT
FT
Note: In our example in part a) 1568 N would represent 2.3 times the body weight of a
70 kg individual. Reilly & Martens (1972) estimated compressive forces at the
patellofemoral joint of up to 7.6 times body weight during the squat exercise.
5. a) ΣFy = 0
Re - 25 - 300 = 0 Re = 325N
Therefore NET force at elbow would be: Re = 325N
b)
We need to break muscle force of 2000 N into vertical and horizontal components.
ΣFy = 0
Joint reaction force(y) + Fflexors(sin120o) - 25 - 300 = 0
Fyj + 2000(0.866) - 25 - 300 = 0
Fyj + 1732 - 25 - 300 = 0
Fyj + 1407 = 0
Fyj = - 1407 N
ΣFx = 0
Joint reaction force(x) + Fflexors(cos120o) = 0
Fxj + 2000(-0.5) = 0
Fxj - 1000 = 0
Fxj = 1000 N
θ 1000 N
Vector reconstruction:
-1407 N
R
10002 + (-1407) 2 = (Joint reaction force) 2 [Pythagorean theorem]
Joint reaction force (R) = 1726 N
Angle of R = θ = arctan (-1407/1000) = -54.6o
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6. a) 3.70 kg [1]
b) To calculate the answer in c) this first step
(calculate the moment arms) is needed anyway.
o
30
Upper arm 14cos 30 = 12.1 cm [1]
Forearm (32.2cos 30) + 10.9 = 38.8 cm [1]
Hand 27.9 + 25.3 + 9.2 = 62.4 cm [1]
c) The masses of the segments must now be multiplied by g to get the force. Note that
the three forces will result in negative moments about the shoulder axis (clockwise)
Moments:
0.121•2.1•-9.81 = -2.49 Nm
0.388•1.2•-9.81 = -4.57 Nm
0.624•0.4•-9.81 = -2.45 Nm
Total = -9.51 Nm
[4]
Therefore NET moment at the shoulder due to arm weight = -9.51 Nm
Note that if I had asked what the net muscle moment must be to maintain the arm in this
position the answer would have been +9.51 Nm (anti-clockwise muscle moment).
d) To resist this clockwise moment and keep the arm in this static position an anticlockwise moment would be generated by the muscles (as we look at the diagram).
Active muscles would therefore be the shoulder flexors; pectoralis major (upper
fibres - clavicular head), anterior deltoid, biceps brachii, coracobrachialis. The
later two muscles being assistors rather than prime movers. [3]
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