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 N. What is the tension in the back muscle?
A. 1500 N
B. 1700 N
C. 2700 N
D. 3500 N

Answer 1

Answer:

C) 2700 N

Explanation:

As per FBD of the given situation we 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

The tension force is acting at an angle of 12 degree

while both weight are acting perpendicular to the length

so here we have

$\tau_{clockwise} = \tau_{counterclockwise}$

$m_1g(\frac{L}{2}) + m_2g(L) = Tsin12(\frac{2L}{3})$

now we have

$36(9.81)(\frac{L}{2}) + 20(9.81)(L) = T sin12(\frac{2L}{3})$

T = 2700 N