# Homework Help: Center of Mass and Equilibrium

1. Nov 11, 2008

### TJDF

1. The problem statement, all variables and given/known data

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In class you learned that when a person walks, their centre of mass must be positioned directly above the foot their weight is resting on. The figure shows the anatomy of the right leg of length L = 0.96 m for a person weighing Wb = 519 N who is walking with a cane on their left side. The figure shows the instant during the stride when the person has one sixth of their body weight resting on the cane. The cane is located a distance D = 34.00 cm from their vertical axis. By using the cane, the person is able to position their vertical axis a distance x, 6.80cm, from their right foot. The weight of the leg is WL = Wb/7 and has a centre of mass that is located a distance 0.43·L from the ankle. The socket of the pelvis (acetabulum) exerts a force FA on the head of the femur. The net force FM of a series of abductor muscles is applied to the top of the femur (greater trochanter) at an angle of 70° with respect to the horizontal. The distance from the greater trochanter to the acetabulum is dA = 7.8 cm, and to the sacrum is dS = 20.1 cm. Calculate the magnitude of FM.

2. Relevant equations

Sum of all Torque = 0

3. The attempt at a solution

0 = -(0.078m x Fm sin70) + Torque of Leg (?) + (5/6 x 519 N x 0.055 m)

Solve for Fm, right?
but I can't figure out the Torque of the leg.

Last edited by a moderator: May 3, 2017
2. Nov 11, 2008

### TJDF

I've recently tried this...

Thinking of similar triangles
rleg*moment arm of leg* = ([(ds-da)-x]/L) x ( L x 0.43)

then, I did this
0 = -T(leg) + T(n) - T(m)
0 = -(rleg x [Wb/7]) + ([(ds-da)-x] x [5/6 x Wb]) (sin theta x da x Fm)

Solve for Fm

but I have no luck.