1. Muscle Moments
Determining the force required from a muscle to maintain static equilibrium.
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2. 3 Equations for Solving Moment Problems1 2 3
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Dr. Sasho MacKenzie - HK 376

3. Muscle Moment Example Fm Fm = ? Arm/hand mass = 3 kg Shot mass = 6 kgmg
Fshot
r2 = 0.3 m
r3 = 0.6 m
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Dr. Sasho MacKenzie - HK 376

4. 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)] = 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)] = N (13%)
.02
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Dr. Sasho MacKenzie - HK 376

5. 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% 
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Dr. Sasho MacKenzie - HK 376

6. Determine the joint reaction forceY X Fc
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 = – – Fc = 0
Fc = – N
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Dr. Sasho MacKenzie - HK 376

7. MA = [(Fm)(sin )(r)] – [(Hand weight) (lower arm length)] = 0
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 =
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Dr. Sasho MacKenzie - HK 376

8. 1.5 m mg
? m
Ffeet
Fhand = 600 N
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.
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Dr. Sasho MacKenzie - HK 376

9. Solution Fy = Fhand + Ffeet - mg = 0
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 = = 381 N
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Dr. Sasho MacKenzie - HK 376

10. Solution Mh = -[(381)(1.5)] + [(981)(CofG position)] = 0
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)] = from hands
(1.5 – 0.58) = 0.92
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Dr. Sasho MacKenzie - HK 376

11. biceps
10kg CM
5cm
13cm
30cm
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