When you bend over, a series of large muscles, the erector spinae, pull on your

Question

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 330 N weight of the upper torso is at the center of the spine. The 160 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 the figure bends over to an angle of 30 from the horizontal.

Part A

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.

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).

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

Explanation / Answer

net toprque about the pivot = 0

torque due to erector muscle

F*cos18*(2/3)*L*sin30 + F*sin18*(2/3)*L*cos30

torque due to weight of uppertorso

W1*L/2*cos30

torque due to weight of head and arms

W2*L*cos30

F*cos12*(2/3)*L*sin30 + F*sin12*(2/3)*L*cos30 = W1*L/2*cos30 + W2*L*cos30

(F*(2/3)*sin(30+18) = (330*(1/2)*cos30)+(160*cos30)

F = 568.1 N

++++++++++++

part(B)

along horizantal

Px = F*cos18

Px = 540.3 N

along vertical

Py - F*sin12 - W1 - W2 = 0

Py = (568.1*sin12)+(330)+(160)

Py = 608.11 N

P = sqrt(Fx^2+Fy^2)

P = sqrt(540.3^2+608.11^2)

P = 813.46 N

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