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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
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