we will model the spine and upper body as a horizontal rigid rod of uniform density with a length of 50.0 cm and a mass of 40.0 kg. assume that the person attempts to lift an object with their arms, which we will model as attached at the far end of the rod. support of the back in this position is provided primarily by the erector spinalis muscle which we will model as being attached at one end to the spine at a point 33.0 cm from the hip at an angle of 10

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09-12-2022
Physics

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we will model the spine and upper body as a horizontal rigid rod of uniform density with a length of 50.0 cm and a mass of 40.0 kg. assume that the person attempts to lift an object with their arms, which we will model as attached at the far end of the rod. support of the back in this position is provided primarily by the erector spinalis muscle which we will model as being attached at one end to the spine at a point 33.0 cm from the hip at an angle of 10

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kjuli1766 kjuli1766 · Answer 1 · 2022-12-18T03:27:17-05:00

Tension in the back muscle for this scenario is 3095.4 N and the compressive reaction force is 3048 N, and the ratio is 5.53.

Two forces are acting on the compartment, Gravity acting downward, Tension T in cable acting upward.

Force is an influence which could change the movement of an object. A pressure can reason an object with mass to change its pace (e.g. moving from a kingdom of relaxation), i.e., to boost up. pressure can also be described intuitively as a push or a pull. A pressure has both importance and course, making it a vector quantity. it's far measured within the SI unit of newton (N). it's far a wellknown mechanical quantity. pressure is represented with the aid of the symbol F (previously P).

A) In equilibrium net torque about the hip = 0

T*sintheta*x - Fgup*L/2 - Fgob*L = 0

T*sin10*0.33 - 40*9.8*0.5/2 - 16.2*9.8*0.5 = 0

Tension(T) = 3095.4 N

B) Fv = FRx = T*costheta

Fv = 3095.4*cos10

= 3048 N

C) Fv/(wub + wobj) = 3048/((40+16.2)*9.8) = 5.53

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