Calculating Traction Force for Leg Fracture Treatment

[TyroneTheDino]

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?

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

 

[CWatters]

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]

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.

 

[theodoros.mihos]

I solve: tan(35)=oppostie/25.

What is this?