Calculating Traction Force for Leg Fracture Treatment

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SUMMARY

The discussion focuses on calculating the traction force required for leg fracture treatment using a Russell traction apparatus. The specified traction force is 25N, leading to the equation W=mg, where W represents the weight. The participant initially calculates the vertical component of tension but questions the accuracy of their solution, concluding that the vertical component does not equal W. The analysis reveals that the tension in the rope remains consistent despite changes in angle, emphasizing the importance of understanding tension dynamics in traction systems.

PREREQUISITES
  • Understanding of basic physics concepts, specifically forces and tension.
  • Familiarity with trigonometric functions, particularly tangent.
  • Knowledge of the Russell traction apparatus and its application in physical therapy.
  • Basic principles of mechanics, including the relationship between weight and mass.
NEXT STEPS
  • Study the mechanics of tension in pulley systems.
  • Learn about the Russell traction apparatus and its variations.
  • Explore the application of trigonometry in force calculations.
  • Investigate the effects of angle changes on tension in physical therapy settings.
USEFUL FOR

Physical therapists, medical students, and professionals involved in orthopedic treatment and rehabilitation, particularly those focusing on traction methods for leg fractures.

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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