# Pulley system

1. Oct 26, 2005

### stunner5000pt

ok have a look at the diagram posted. When someone broke their femur (thigh bone) the muscles around the femur would compress tightly and thereby shorten the bone once it heals back. SO the apparatus described in the diagram is what they used to do to prevent the femur from growing back shorter. The question is to find the horizontal force - in terms of the mass,m and the mass of the leg(ML).

What i was thinking of doing was to assume that htis was a simply pulley system where there were two masses on either side of the one pulley... but that wouldn't yeild an answer since it would get (since there is no acceleration)
$$T_{1} - mg = T_{2} - m_{L}g$$
but then i cnanot solve for T1 or T2 (can i?)

Now lookin at my diagram and my various labelled tensions...
T1 = mg
T2 T2cos 50 = mg and T2 = mg / cos 50
vertical component is t2 cos 50\
horizontal t2 sin 50

is this the same value for T3 ? When does trhe value of the other force (gravity on the leg) come into play?

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2. Oct 26, 2005

### Staff: Mentor

I'm not sure I understand the setup. Is the cable supposed to totally support the weight of the leg? (I don't know what the board-like thing attached to the leg is. Is it fixed? Or does it move with the leg?)

In any case, I see no reason why you can't model this as a simple pulley system with uniform tension throughout. Of course, that implies that $$m = m_L$$, since the pulley system provides no mechanical advantage. (Not a particularly clever design!)

3. Oct 27, 2005

### stunner5000pt

that board attafcehd to the leg is not supposed to move. The leg just rests on that board.

Actually there are numerical values for m=5kg and ML = 3.75 kg

the question is to find the horizontal force on the leg...
So the concern is what is the tension T2 and T3...

4. Oct 28, 2005

### Staff: Mentor

Since the board is fixed, other forces must act on it. I see no reason not to treat the pulley system as frictionless & massless, and thus the tension as uniform throughout. The tension will equal the weight of the hanging mass.