Given When a person lifts an object the spine is flexed appr

Given: When a person lifts an object the spine is flexed approximately 35^o. The force produced by the weight of the upper body (W) is 510 N. The force produced by the weight of the object (P) is 150 N.

L_P (distance from C to P) = 0.4 m

L_W (distance from C to W) = 0.25 m

L_E (distance from C to E) = 0.05 m

Find: The magnitude of force E on the erector spinae muscles, the comnpressive force C, and the shear force on the disc S.

Solution

balancing torque around c

E*0.05 = w*0.25*sin(55) + p*0.4*sin(55)

=>E*0.05 = 153.59

=>E = 3071.82 N

also

Balancing forces parallel to backbone,

S = wcos(55) + p*cos(55)

=> S = 378.56 N

and

Balancing forces perpendicular to backbone

C = E+wcos(35) + p*cos(35)

=> C = 3612.46