Russel Traction apparatus

Now lets look at an example from the orthopedic ward. (Show the Russel Traction apparatus). The first question to ask is how large should the mass of the counter-balance marked W be? Here we recall that here is just a single tension in the cord. That same tension acts on the leg and on the mass. In order for the tension to cancel the gravitational force on both these we need to choose the mass of W to be the same as the mass of the leg which we denote M. With this condition we have a static situation and the tension in the cord is

Now we ask what is the horizontal force on the leg because the foot is attached to the pulley. The two strings in tension produce forces on the pulley of magnitude the tension in the cord and directed as a pull! Thus

I wrote the equations slightly more general than required for this problem where the angles . In that special case the force in the vertical direction cancels out and if the weight of the leg Mg=8 lb we are left with the horizontal component with magnitude

We can also make the following observations:

1. If I move the points of attachement of the pulleys on the wall closer together I get a large horizontal force.
2. If the leg does not lie centered between the pulleys on the wall then the force on the foot will not be exactly horizontal; the vertical component will no longer cancel out. This will in turn bring the system out of balance and require a new choice of M.
3. There is an inherent stability in the system. If we choose a too small mass M, the leg will accelerate down-wards which will bring in a vertical component of the cord tension which in turn will bring the acceleration to a halt and stop the leg in a new equilibrium. Thus no disaster occurs if we do not get the mass right the first time around. We can adjust the mass until we have the leg centered between the pulleys on the wall.