Question

When you bend over, a series of large muscles, the erector spine, 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 360N weight of the upper torso is at the center of the spine. The 170N weight of the head and arms acts at the top of the spine. The erector spine muscles are modeled as a single muscle that acts at an 12 angle to the spine. Suppose the person in (Figure 1) bends over to an angle of 30 from the horizontal.

What is the tension in the erector muscle? Hint: Align your x-axis with the axis of the spine.

Express your answer to two significant figures and include the appropriate units.

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).

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer 2

Torque summation around pivot:

170*L*cos30

Answer 3

When you bend over, a series of large muscles, the erector spinae, pull on your spine to hold you up. The figure 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 310 N weight of the upper torso is at the center of the spine. The 155 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 degree angle to the spine. Suppose the person in the figure bends over to an angle of 30 degree from the horizontal.

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What is the tension in the erector muscle? Hint: Align your x-axis with the axis of the spine. ANSWER: 1900N

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).

Best Answer

Torque summation around pivot:

155*L*cos30º 310*(½L)*cos30º - F*(⅔L)*sin12º = 0

F*(⅔)*sin12º = [155 + 310*(½)]*cos30º

F = 310*cos30º/[(⅔)*sin12º] = 1.9*10^3 N

The force along the length of the spine is the sum of force components along the spine from the three forces

310*sin30º + 155*sin30º + 1900*cos12º = 2.1*10^3 N