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°) ×
= ![F_{Mx}](https://tex.z-dn.net/?f=F_%7BMx%7D)
sin(18°) ×
= ![F_{My}](https://tex.z-dn.net/?f=F_%7BMy%7D)
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
![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})) }](https://tex.z-dn.net/?f=F_M%20%3D%20%5Cdfrac%7B0.72%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%20w_1%20%2B%200.48%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%20w_2%20%2B%200.36%20%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%20w_3%7D%7B%280.48%20%5Ctimes%20sin%2830%5E%7B%5Ccirc%7D%29%5Ctimes%20cos%2818%5E%7B%5Ccirc%7D%29%20-%200.48%5Ctimes%20cos%2830%5E%7B%5Ccirc%7D%29%5Ctimes%20sin%2818%5E%7B%5Ccirc%7D%29%29%20%7D)
Therefore;
![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})) }](https://tex.z-dn.net/?f=F_M%20%3D%20%5Cdfrac%7B0.72%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes0.07%5Ccdot%20w%20%2B%200.48%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%200.12%20%5Ccdot%20w%20%2B%200.36%20%20%5Ctimes%20sin%2860%5E%7B%5Ccirc%7D%29%20%5Ctimes%200.46%5Ccdot%20w%7D%7B%280.48%20%5Ctimes%20sin%2830%5E%7B%5Ccirc%7D%29%5Ctimes%20cos%2818%5E%7B%5Ccirc%7D%29%20-%200.48%5Ctimes%20cos%2830%5E%7B%5Ccirc%7D%29%5Ctimes%20sin%2818%5E%7B%5Ccirc%7D%29%29%20%7D)
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
![Force \ on \ vertebra, \ F_v = \sqrt{\left(\sum F_x \right)^2 + \left(\sum F_y \right)^2}](https://tex.z-dn.net/?f=Force%20%5C%20on%20%5C%20vertebra%2C%20%5C%20F_v%20%3D%20%5Csqrt%7B%5Cleft%28%5Csum%20F_x%20%5Cright%29%5E2%20%2B%20%5Cleft%28%5Csum%20F_y%20%5Cright%29%5E2%7D)
Therefore;
![Force \ on \ vertebra, \ F_v = \sqrt{\left(2.258\right)^2 + \left(1.384 \right)^2} \approx 2.648](https://tex.z-dn.net/?f=Force%20%5C%20on%20%5C%20vertebra%2C%20%5C%20F_v%20%3D%20%5Csqrt%7B%5Cleft%282.258%5Cright%29%5E2%20%2B%20%5Cleft%281.384%20%5Cright%29%5E2%7D%20%5Capprox%202.648)
The force acting on the fifth vertebra,
≈ <u>2.648·w</u>
![The \ direction \ of \ the \ force,\, \theta = tan^{-1} \left(\dfrac{F_{My}}{F_{Mx}} \right)](https://tex.z-dn.net/?f=The%20%5C%20direction%20%5C%20%20of%20%20%5C%20the%20%5C%20%20force%2C%5C%2C%20%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5Cleft%28%5Cdfrac%7BF_%7BMy%7D%7D%7BF_%7BMx%7D%7D%20%5Cright%29)
![\theta = tan^{-1} \left(\dfrac{1.384}{2.258} \right) \approx 31.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5Cleft%28%5Cdfrac%7B1.384%7D%7B2.258%7D%20%5Cright%29%20%5Capprox%2031.5%5E%7B%5Ccirc%7D)
The direction of the force, θ ≈ <u>31.5°</u>
Learn more here:
brainly.com/question/1858958