Question
gPlysiconyhycrrenlewfassignmentProblemlD 84104210 CH3 1Problem 8.47 austin previous | 10 of 13 | next aterials When you bend over, a series of large muscles, the erector spinae, pull on your spine to hold you up Figure 1) shows a simplified model of the spine as a rod of length L that pivots at its lower end. In this model, the center of gravity of the 330 N weight of the upper torso is at the center of the spine. The 150 N weight of the head and arms acts at the top of the spine. The erector spinae muscles are modeled as a single muscle that acts at an 12° angle to the spine Suppose the person in (Fiqure 1) bends over to an Part A What is the tension in the erector muscle? Hint: Align your z-axis with the axis of the spine. Express your answer to two signiticant figures and include the appropriate units. TValue Units Submit My Answers Give Up Figure! _VI of 1 Part B A force from the pelvic girdle acts on the base of the spine. What is the component of this force in the direction of the spine? (This large force is the cause of many back injuries). Effective location of erectorrtnar muscles 30 Express your answer to two significant figures and include the appropriate units Center of gravity of head a .. Center of gravity ot upper torso Pivot 21Es PR Submit My Answers Give Up Print Assig Continue
Explanation / Answer
Torque summation around pivot:
150*L*cos30º 330*(1/2L)*cos30º - F*(2/3L)*sin12º = 0
F*(2/3)*sin12º = [150 + 330*(1/2)]*cos30º
F = 315*cos30º / [(2/3)*sin12º]
F = 1968 N
b)
The force along the length of the spine is the sum of force components along the spine from the three forces
Fp,g = 330*sin30º + 150*sin30º + 1968*cos12º
Fp,g = 165 + 75 + 1925
Fp,g = 2165 N
