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Transcript
A two dimensional back model
Adopted from Chaffin et al. (1977).
Assumptions:
• Lifting is occurring a the mid-sagittal plane
• The erector spinae muscle is the sole provider of the counterbalancing moment of lift
• No other back muscles are significantly active
We want to estimate:
• FMUSc - tension in the back muscle (erector spinae)
• FCOMP –compressive force acting on spine
• FSHEAR - shear force acting on spine
Given:
Load in hand = 225 N = (225*2.2/9.81 = ~50 lb)
Upper body weight above L5/S1 joint = 350 N
E = Distance of erector spinae muscle from the center of spine = 6.5 cm
h = distance of the load from the center of spine at L5/S1= 30 cm
b = distance of the upper body center of gravity from center of spine at L5/S1 joint =20 cm
α = Upper body angle with horizontal = 55ο
Solution: To find out the internal forces, we cut the upper body by a section perpendicular to
spinal axis at L5/S1 joint and show all internal and external forces acting on the upper body.
For rotational equilibrium at L5/S1 joint:
FMUSC*6.5 = 350*20+225*30
FMUSC = (350*20+225*30)/6.5
= 2,115 N
Resolving the 350 N vertical force along the spinal axis and normal to the spinal axis, the two
components are 287 N and 201 N, respectively.
Similarly, resolving the 225 N vertical force along the spinal axis and normal to the spinal axis,
the two components are 184 N and 129 N, respectively.
Then, the reaction forces
FCOMP = 2,115+287+184 = 2,586 N
and,
FSHEAR = 201+129 = 330 N
Note that:
(1) The spine compressive force FCOMP is about 2,586/225=11 times the load in
hand.
(2) If the person has to lean forward, distance b increases, FCOMP increases.
(3) If the load distance h increases, FCOMP increases.
(4) For occupational tasks, FCOMP is positively correlated to risk of back pain.