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 <u>2.648·w</u>
- The direction of the force is approximately <u>31.5°</u>
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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 = , location = 0.48 m, angle = 12°
At equilibrium, we have, ∑M = 0, therefore;
0.48×sin(30°)×cos(18°) × - 0.48×cos(30°)×sin(18°) × = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃
Where;
cos(18°) × =
sin(18°) × =
Which gives;
(0.48×sin(30°)×cos(18°) - 0.48×cos(30°)×sin(18))× = 0.72×sin(60°)×w₁ + 0.48×sin(60°)×w₂ + 0.36×sin(60°)×w₃
Therefore;
At equilibrium sum of forces, ∑F = 0
∑Fₓ = = cos(18°) ×
∴ ∑Fₓ = 2.374 × cos(18°) ≈ 2.258·w
= + w₁ + w₂ + w₃
∴ = sin(18°) × + w₁ + w₂ + w₃
≈ 0.734·w + 0.07·w + 0.12·w + 0.46·w ≈ 1.384·w
Therefore;
The force acting on the fifth vertebra, ≈ <u>2.648·w</u>
The direction of the force, θ ≈ <u>31.5°</u>
Learn more here:
brainly.com/question/1858958