Download Muscle Moments Lecture

Muscle Moments
Determining the force required
from a muscle to maintain static
equilibrium.
5/24/2017
Dr. Sasho MacKenzie - HK 376
1
3 Equations for Solving
Moment Problems
1
2
3
 M A  FX 1 r1  FY2 r2    0
 FX  FX 1  FX 2    0
 FY  FY1  FY2    0
5/24/2017
Dr. Sasho MacKenzie - HK 376
2
Muscle Moment Example
Fm
Fm = ?
Arm/hand mass = 3 kg
Shot mass = 6 kg
r1 = .02 m
A
r2 = 0.3 m
mg
Fshot
r3 = 0.6 m
5/24/2017
Dr. Sasho MacKenzie - HK 376
3
Determining the force in the biceps required to
hold the lower arm in static equilibrium.
MA = [(Fm)(r1)] – [mg(r2)] – [(Fshot)(r3)] = 0
MA = [(Fm)(.02)] – [(3)(9.81)(0.3)] – [(6)(9.81)(0.6)] = 0
Fm = [(3)(9.81)(0.3)] + [(6)(9.81)(0.6)] = 2205 N
.02
What would be the force in the biceps if the
shot was moved 10 cm closer to the elbow?
Fm = [(3)(9.81)(0.3)] + [(6)(9.81)(0.5)] = 1913 N (13%)
.02
5/24/2017
Dr. Sasho MacKenzie - HK 376
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Using the original diagram (Slide 3), how much
weight could be held in the hand if the maximum
isometric force of the bicep was 6000 N?
MA = [(Fm)(r1)] – [mg(r2)] – [(Fshot) (r3)] = 0
MA = [(6000)(.02)] – [(3)(9.81)(0.3)] – [(Fshot)(0.6)] = 0
Fshot = [(6000)(.02)] – [(3)(9.81)(0.3)] = 185 N = 41 lbs
0.6
How much more weight could be held if the
bicep moment arm was increased by 1 cm?
Fshot = [(6000)(.03)] – [(3)(9.81)(0.3)] = 285 N = 64 lbs
0.6
55% 
5/24/2017
Dr. Sasho MacKenzie - HK 376
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Determine the joint reaction force
Fc
Y
6000 N
A
X
29.4 N
285 N
Calculating the “x” component of the joint reaction force.
Fx = 0
There are no “x” components of force, therefore
there is no shear force at the elbow.
Calculating the “y” component of the joint reaction force.
Fy = 6000 – 29.4 – 285 + Fc = 0
Fc = – 5685.6 N
5/24/2017
Dr. Sasho MacKenzie - HK 376
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Given a maximum force of 6000 N in the bicep. How much
weight can a person with a lower arm length of 25 cm lift?
How much weight can a person with a lower arm length of
35 cm lift? Assume the weight of the arm is negligible and
the bicep force acts at an angle of 80 to the lower arm and
inserts .03 m from the elbow joint.
A
Y
6000 N
.03 m
X
?N
Lower arm length
MA = [(Fm)(sin )(r)] – [(Hand weight) (lower arm length)] = 0
Hand weight = [(Fm)(sin )(r)] / lower arm length =
5/24/2017
Dr. Sasho MacKenzie - HK 376
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mg
?m
Ffeet
Fhand = 600 N
1.5 m
The diagram above shows a 100 kg man in static
equilibrium at the top of a push-up. What is the normal
force acting at his feet (Ffeet)? And how far from his feet
is his center of gravity located? Confirm your answers
by using both the hands and feet as points to sum the
moments about.
5/24/2017
Dr. Sasho MacKenzie - HK 376
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Solution
• For no good reason, lets choose the hands as the
point of rotation.
• This means that we must find all the forces acting
at a distance from the hands and their respective
moment arms.
• Initially we only know the force due to gravity and
not the force at the feet.
• However, we can use the sum of the forces in the
Y direction to determine the forces at the feet.
Fy = Fhand + Ffeet - mg = 0
Ffeet = mg – Fhand = 981 - 600 = 381 N
5/24/2017
Dr. Sasho MacKenzie - HK 376
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Solution
Now we can sum the moments about the hands to
determine the position of the center of gravity from
the hands.
Mh = -[(381)(1.5)] + [(981)(CofG position)] = 0
CofG position = [(381)(1.5)]
981
5/24/2017
=
0.58 from hands
(1.5 – 0.58) = 0.92
Dr. Sasho MacKenzie - HK 376
10
biceps
10kg
CM
5cm
13cm
30cm
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Dr. Sasho MacKenzie - HK 376
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Dr. Sasho MacKenzie - HK 376
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