IDNLearn.com provides a collaborative environment for finding and sharing answers. Find the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
The effect of the force of the fifth lumbar vertebra can be resolved into two forces which produce the same effect
- The force acting on the fifth vertebra is approximately 2.648·w
- The direction of the force is approximately 31.5°
Reason:
Given parameters in a similar question are;
Weight of head, w₁ = 0.07·w, location (distance from weight) = 0.72 m, angle formed with vertebra = 60°
Weight of arms, w₂ = 0.12·w, location = 0.48 m, angle = 60°
Weight of trunk, w₃ = 0.46·w, location, 0.36 m, angle = 60°
Force of muscle = [tex]F_M[/tex], location = 0.48 m, angle = 12°
At equilibrium, we have, ∑M = 0, therefore;
0.48×sin(30°)×cos(18°) ×[tex]F_M[/tex] - 0.48×cos(30°)×sin(18°) ×[tex]F_M[/tex] = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃
Where;
cos(18°) ×[tex]F_M[/tex] = [tex]F_{Mx}[/tex]
sin(18°) ×[tex]F_M[/tex] = [tex]F_{My}[/tex]
Which gives;
(0.48×sin(30°)×cos(18°) - 0.48×cos(30°)×sin(18))×[tex]F_M[/tex] = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃
[tex]F_M = \dfrac{0.72 \times sin(60^{\circ}) \times w_1 + 0.48 \times sin(60^{\circ}) \times w_2 + 0.36 \times sin(60^{\circ}) \times w_3}{(0.48 \times sin(30^{\circ})\times cos(18^{\circ}) - 0.48\times cos(30^{\circ})\times sin(18^{\circ})) }[/tex]
Therefore;
[tex]F_M = \dfrac{0.72 \times sin(60^{\circ}) \times0.07\cdot w + 0.48 \times sin(60^{\circ}) \times 0.12 \cdot w + 0.36 \times sin(60^{\circ}) \times 0.46\cdot w}{(0.48 \times sin(30^{\circ})\times cos(18^{\circ}) - 0.48\times cos(30^{\circ})\times sin(18^{\circ})) }[/tex]
[tex]F_M = \dfrac{0.236944550476}{0.099797611593} \approx 2.374[/tex]
At equilibrium sum of forces, ∑F = 0
∑Fₓ = [tex]F_{Mx}[/tex] = cos(18°) ×[tex]F_M[/tex]
∴ ∑Fₓ = 2.374 × cos(18°) ≈ 2.258·w
[tex]\sum F_y[/tex] = [tex]F_{My}[/tex] + w₁ + w₂ + w₃
∴ [tex]\sum F_y[/tex] = sin(18°) ×[tex]F_M[/tex] + w₁ + w₂ + w₃
[tex]\sum F_y[/tex] ≈ 0.734·w + 0.07·w + 0.12·w + 0.46·w ≈ 1.384·w
[tex]Force \ on \ vertebra, \ F_v = \sqrt{\left(\sum F_x \right)^2 + \left(\sum F_y \right)^2}[/tex]
Therefore;
[tex]Force \ on \ vertebra, \ F_v = \sqrt{\left(2.258\right)^2 + \left(1.384 \right)^2} \approx 2.648[/tex]
The force acting on the fifth vertebra, [tex]F_v[/tex] ≈ 2.648·w
[tex]The \ direction \ of \ the \ force,\, \theta = tan^{-1} \left(\dfrac{F_{My}}{F_{Mx}} \right)[/tex]
[tex]\theta = tan^{-1} \left(\dfrac{1.384}{2.258} \right) \approx 31.5^{\circ}[/tex]
The direction of the force, θ ≈ 31.5°
Learn more here:
https://brainly.com/question/1858958
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.
