Since we were able to find the shoulder force using the arm system, we can now turn our focus back to the spinal system so that we can continue on our route in examining the force the pelvis and erector spinae muscles exert on the spine. In order to find the force of the muscle exerts on the spine we must now refer back to Table 2. As mentioned earlier the net torque of the spinal system is 0, so the counterclockwise torques must be equal to the clockwise torque because there is no rotation taking place. In knowing this, we were able to create the equation ⅔ FM (sin 12°) + (FS)(sin) = ½ (mg) to find the muscle force at the three angles. After calculating each component of the force, we are now able to use the Pythagorean Theorem to solve for …show more content…
These values are then plugged into the Pythagorean Theorem,-502+562=c2 , and c2 = 5636 N. The square root of 5636 N is found to be 75 N, and this value is the sum of the two shoulder forces. To findwe did the inverse tangent of the triangle created in Figure 4, tan-1 (56/50), and found that it is 48°. So as the person pushes the box at an angle of 60°, the shoulder force exerted on the spine is 75 N at a 48° angle. We then plugged these values into our torque equation to find the muscle force if the box is being pushed at a 60° angle. This is what the equation looks like once all of the values have been plugged in, ⅔ FM (sin 12°) + 75 N (sin 48°) = ½ (123 N). The answer to the equation is 42 N, meaning that as the box is being pushed at a 60° angle the erector spinae muscle is exerting a force of 42 N on the spine. This series of steps used to find the force the muscle exerts on the spine at a 60° angle are the same steps that we used to calculate the muscle force when the box is being pushed at the other two …show more content…
First we found the force of the pelvis acting in the “x” direction using (FP)x - FM (cos 12°) + (FS)x = 0. We then plugged in the values for the angle 60° angle, (FP)x - 42 (cos 12°) N- 50 N = 0, and found that the pelvis force in the “x” direction is 91 N. The same was done for the “y” direction using (FP)y + FM (sin 12°) - 123 N + (FS)y = 0, and after the values are plugged in the equation looks like this, (FP)y + 42 (sin 12°) - 123 N + 56 N = 0, so the pelvis force in this direction is 58N. We then used another pictorial representation (Figure 5)of the sum of the pelvic forces in the two directions to find its total sum acting on the