Calculating Traction Force for Leg Fracture Treatment

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Homework Help Overview

The discussion revolves around calculating the traction force required for leg fracture treatment using a Russell traction apparatus. The original poster presents a scenario where a specified traction force of 25N is directed along the leg, and they seek to determine the weight (W) involved in the setup.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to relate the horizontal traction force to the vertical components of tension and expresses uncertainty about their calculations. Some participants question the relationship between the vertical component of tension and weight, while others explore the implications of changing the angle of the pulleys on the system's functionality.

Discussion Status

The discussion is active, with participants raising questions about the assumptions made regarding tension and weight. There is a focus on clarifying the relationships between the forces involved, and some guidance has been offered regarding the implications of friction and pulley angles.

Contextual Notes

Participants are considering the effects of angles and friction in the pulley system, which may influence the calculations and assumptions about tension and weight. The original poster's calculations appear to be under scrutiny, indicating a need for further exploration of the problem setup.

TyroneTheDino
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Homework Statement


When the thigh is fractured, the patient's leg must be kept under traction. One method of doing so is a variation on the Russell traction apparatus. If the physical therapist specifies that the traction force directed along the leg must be 25N, what must W be?
8b998626-cd0d-4df1-b3c6-7e95154208c7-original.png


Homework Equations



W=mg
T=mg

The Attempt at a Solution


I begin by saying that the horizontal dotted line has a force of 25N in each direction. Because that is where the traction force is. I believe that the horizontal component of tension in the rope is going to equal 25N as well. I use that to find my vertical components of tension. I solve: tan(35)=oppostie/25. I get the opposite to side to equal .028N. Because there are two of these tension in the horizontal direction. I get that the total tension that should be equal to W= .056N

I am wondering if somehow my solving is flawed
 
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The vertical component of tension is not equal to W.

There is no friction in the pulleys so what does that mean for the tension in the rope at all points?
 
CWatters said:
The vertical component of tension is not equal to W.

Just to clarify.. Suppose the 35 degree angle was reduced to zero by moving the two pulleys closer together. The vertical component would be zero but the system would still work wouldn't it.
 
TyroneTheDino said:
I solve: tan(35)=oppostie/25.

What is this?
 

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