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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.