Share
4–107. A biomechanical model of the lumbar region of the human trunk is shown
ReportQuestion
Please briefly explain why you feel this question should be reported.
4–107. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form
Answer ( 1 )
Please briefly explain why you feel this answer should be reported.
Problem Statement:
We are tasked with determining the equivalent force and couple moment at a point \( O \) given multiple forces acting on a system.
Given Data:
– Force on Rectus: \( F_R = 35 \, \mathrm{N} \)
– Force on Oblique: \( F_O = 45 \, \mathrm{N} \)
– Force on Lumbar Latissimus Dorsi: \( F_L = 23 \, \mathrm{N} \)
– Force on Erector Spinae: \( F_E = 32 \, \mathrm{N} \)
– Distance for \( F_R \): \( d_R = 75 \, \mathrm{mm} = 0.075 \, \mathrm{m} \)
– Distance for \( F_E \): \( d_E = 15 \, \mathrm{mm} = 0.015 \, \mathrm{m} \)
– Distance for \( F_L \): \( d_L = 45 \, \mathrm{mm} = 0.045 \, \mathrm{m} \)
Part i: Determine the Resultant Force at Point \( O \)
The formula for the resultant force is:
\( F_R = F_R + F_O + F_L + F_E \)
Summing the forces:
\( F_R = 35 + 45 + 23 + 32 \)
Performing the calculation:
\( F_R = 135 \, \mathrm{N} \)
If the forces are duplicated symmetrically:
\( F_R = 2 \cdot 135 = 270 \, \mathrm{N} \)
Part ii: Determine the Couple Moment at Point \( O \)
The formula for the couple moment is:
\( M_O = \sum \mathbf{r} \times \mathbf{F} \)
Compute moments for individual forces:
For \( F_R \):
\( M_R = F_R \cdot d_R = 35 \cdot 0.075 \)
\( M_R = 2.625 \, \mathrm{N \cdot m} \)
For \( F_E \):
\( M_E = F_E \cdot d_E = 32 \cdot 0.015 \)
\( M_E = 0.48 \, \mathrm{N \cdot m} \)
For \( F_L \):
\( M_L = F_L \cdot d_L = 23 \cdot 0.045 \)
\( M_L = 1.035 \, \mathrm{N \cdot m} \)
Summing up the moments:
\( M_O = 2 \cdot (M_R + M_E + M_L) \)
Performing the calculation:
\( M_O = 2 \cdot (2.625 + 0.48 + 1.035) = 2 \cdot 4.14 = 8.28 \, \mathrm{N \cdot m} \)
Conclusion:
– Resultant Force at Point \( O \): \( F_R = 270 \, \mathrm{N} \)
– Couple Moment at Point \( O \): \( M_O = 8.28 \, \mathrm{N \cdot m} \)