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 magnitude 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 aWo = 210–N object. The spine and upper body are represented as a uniform horizontal rod of weight Wb = 415 N pivoted 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°.

Find the tension in the back. kN

Find the compression force on the spine. kN

Explanation / Answer

Given:

Wo=210N
Wb=415N
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(L2)+(23L)sin(12)WoL=0
Factor out L, which is unknown, to get:
12Wb+(23)sin(12)Wo=0
Solve this equation for T for the answer to Part A.

Part A:

0=(1/2)415+(2/3)Tsin12210
0=180+(2/3)(.20791)T210
0=207.5+.13861T210
417.5=.13861T
T=3.01kN

Part B:

Rx3.01cos(12)=0
Rx2.54=0
Rx=2.54kN

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.