The moment arm for the force FM produced by the gluteus scle is d = 9 cm relativ

Question

The moment arm for the force FM produced by the gluteus scle is d = 9 cm relative to the center of rotation of the hip. makes an angle = 20° with the vertical axis. Given the PRoromasions from the mid-sagittal plane to the: center of rotation of the hip-a = 18.0 cm center of gravity of the leg = b = 19.5 cm normal force on the foot = c = 8.0 cm he line of action for R runs through the hip's center of rotation. unknouns F (10 points) Write the equationg of equilibrium needed to solve for the unknowns. SortCAHto x = FME5% oo)-R.laz 2 FM#0AR- o.la5 w/o-O/9 2. t W/6 (10 points) If the patient weighs W 690 N, compute the muscle force FsM and the reaction force components Rx and Ry acting on the femoral head.

Explanation / Answer

a) There will be three equilibrium conditions

1) moment equilibrium

2) net horizontal force

3) net vertical force

For moment equilibrium taking moment about centre of rotation of hip

Fm × d = (W/6)×(a-b) + (W/2) × (a-c)

Substituting the values of a,b,c,d in above equation

Fm × 9 = (W/6) ×1.5 + (W/2) ×10........(1)

for horizontal equilibrium

Fm × SinA = Rx.......(2)

For vertical equilibrium

Fm × CosA + W/2 = Ry + W/6.......(3)

b) when W= 690 N then substituting in equation (1)

We get Fm × 9 = 3622.5

Fm= 402.5 N

From equation (2)

Fm × sin20 = Rx ( A= 20° given)

Rx = 137.66 N

From equation (3)

Fm× CosA + W/2 = W/6 + Ry

Substituting the values

Ry = 608.226 N

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