Solving Horizontal Force for Femur Injury Treatment

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Homework Help Overview

The discussion revolves around a problem related to the treatment of a femur injury, specifically focusing on calculating the horizontal force acting on the leg using a pulley system setup. Participants are exploring the mechanics involved in this scenario, including the roles of tension and gravitational forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are considering modeling the situation as a pulley system, questioning the implications of uniform tension and the relationship between the masses involved. There are discussions about the role of the fixed board and how it affects the forces acting on the leg.

Discussion Status

The conversation is ongoing, with participants raising questions about the setup and the assumptions being made. Some guidance has been offered regarding the treatment of the pulley system, but there is no explicit consensus on the approach to take or the values of the tensions involved.

Contextual Notes

Participants have noted specific numerical values for the masses involved, but there are still uncertainties regarding the configuration of the system and the forces at play, particularly concerning the fixed board and its interaction with the leg.

stunner5000pt
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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)
[tex]T_{1} - mg = T_{2} - m_{L}g[/tex]
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?

Please help! Thank you very much!
 

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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 [tex]m = m_L[/tex], since the pulley system provides no mechanical advantage. (Not a particularly clever design!)
 
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...
 
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.
 

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