Muscle Moments

1
Muscle Moments

• Determining the force required from a muscle to
  maintain static equilibrium.

2
3 Equations for Solving Moment Problems
3
Muscle Moment Example
Fm
Fm ? Arm/hand mass 3 kg Shot mass 6 kg
r1 .02 m
A
mg
Fshot
r2 0.3 m
r3 0.6 m
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)
2205 N .02
How much less force would the biceps have to
generate if the shot was moved 10 cm closer?
5
Using the same 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 in the hand if
the bicep moment arm was increased by 1 cm?
6
Determine the joint reaction force from Slide 5
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
7
Given a maximum force of 6000 N in the bicep.
How much weight can a person with a lower arm
length of 55 cm lift? How much weight can a
person with a lower arm length of 65 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.
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
8
(No Transcript)
9
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
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)(position of
center of gravity) 0 CofG (381)(1.5)
0.58 from hands or 0.92 from feet
981 (1.5 0.58)
10
(No Transcript)