A person bending forward to lift a load "with his back" rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. For a person bending forward to lift a 200-N object, the spine and upper body are represented as a uniform horizontal rod of weight 350 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12degrees. Find the tension in the back muscle and the compressional force in the spine.
Please show steps!
The answer is T=2.71 kN, Rx=2.65kN.
2 Answers
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Sorry, I just imagine the forces and flexural moment diagram on my mind.....
∑τ = 0
(W_object) L + (W_spine) (½ L) - T (⅔ L) sin θ = 0
(200) + (350) (½) - T (⅔) sin 12° = 0
T = 2705.476 N
∑Fx = 0
Rx - T cos θ = 0
Rx - 2705.476 cos 12° = 0
Rx = 2646.355 N
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you need to draw a free body diagram for this guy. Probably didn't want to hear that. this would be very difficult and lengthy to answer in words. basically what they are getting at is that given the relatively small angle of 12 degrees the components of the force acting perpendicular to the load must be much larger to achieve the required result. All you need to do is draw your force vectors, split them into triangles and do a little SOH CAH TOA and you'll be golden. If you are unfamiliar or just not practiced with this method go ask a TA, that's what they are there for. Good luck, we've all been there. It ꜱᴜcκs
