Russell Traction Apparatus: Find W & Traction Force

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The discussion focuses on the Russell traction apparatus, which is designed to support an injured leg while maintaining traction to prevent misalignment during healing. The key problems involve calculating the weight W necessary to balance a leg weighing 47.0 N and determining the traction force produced by the system. Participants emphasize the importance of using free body diagrams to analyze forces and maintain equilibrium, noting that the net force is zero due to the lack of movement. There is uncertainty about how to apply the equations of motion and balance forces effectively to find W and the traction force. Understanding the forces acting on both the weight and the pulley is crucial for solving the problem accurately.
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Homework Statement


In order to prevent muscle contraction from misaligning bones during healing (which can cause a permanent limp), injured or broken legs must be supported horizontally and at the same time kept under tension (traction) directed along the leg. One version of a device to accomplish this aim, the Russell traction apparatus, is shown in the figure. This system allows the apparatus to support the full weight of the injured leg and at the same time provide the traction along the leg.

A.If the leg to be supported weighs 47.0 N , what must be the weight of W?
B. What traction force does this system produce along the leg?
upload_2017-3-8_17-46-52.png

Homework Equations


F=MA
W=MG

The Attempt at a Solution


I know that the F(nety)=0 because there is no movement.
I am unsure on how to relate the equations to finding W.
I tried using the weight given to find W by using 47Ntan(50).
 
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The standard technique is to draw a free body diagram for each component and consider the balance of forces on each.
What are the forces acting on the weight? What does that tell you about the tension?
What are the forces acting on the pulley next to the foot? What equation(s) for those?
Etc.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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