- #1

TyroneTheDino

- 46

- 1

## 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