A person bending forward to lift a load \"with his back\" (Figure a) rather than

Question

A person bending forward to lift a load "with his back" (Figure a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magritude of the forces involved, and to understand why back problems are common among humans consider the model shown in Figure b, of a person bending forward to lift a Wo-185-N object. The spine and upper body are represented as a uniform horizontal rod of weight W- 370 N plvoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0 ector spinelis ains the position of Back muscle Pivot R, (o) Find the tension in the back muscle. kN (b) Find the compressional force in the spine. (Enter the magnitude.) KN

Explanation / Answer

Given:
Wo=185N
Wb=370N
Angle=12
Length of the spine/rod = L
Axis of rotation = the left-most point on the rod / the pivot at the base of the spine
Wb distance from the axis of rotation = (1/2)L
T distance from the axis of rotation = (2/3)L

Formulas:
Fx=0
RxTcos(12)=0
After solving Part A, plug T into this equation and solve for Rx to get the compressional force in the spine.
Fy=0
RyWb+Tsin(12)=0

=0
For each individual torque:
=RF or =RFsin()

Wb(L/2)+(2/3L)sin(12)WoL=0
Factor out L, which is unknown, to get:
1/2Wb+(23)sin(12)Wo=0
Solve this equation for T for the answer to
Part A.

0 = -(1/2)*370+(2/3)T *sin12 - 185
0 = -185+(2/3)*(0.1386)T - 185
0 = -185 + 0.0924T -185
270 = 0.0924T
T = 2.922kN

Part B:
Rx2.922cos(12)=0
Rx2.858=0
Rx=2.858kN

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