Question

A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle

Answer 1

T = 2689.6N

Step-by-step explanation:

Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree

while both weight are acting perpendicular to the length. Hence we have :

Torque ( clockwise) = Torque ( anticlockwise)

m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1

Where m1 = 36kg

m2 = 20kg

g = 9.81m/s^2

Theta = 12

Substituting into equation 1

36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)

353.16L/2+196.2L = T ×0.2079(2L/3)

176.58L+196.2L = T × 0.1386L

372.78L = 0.1386LT

T = 372.78L/0.1386L

T = 2689.6N