# How do I find the horizontal force/traction force of this?

1. Oct 1, 2012

### riseofphoenix

8. A setup similar to the one shown in the figure below is often used in hospitals to support and apply a traction force to an injured leg.

(a) Determine the force of tension in the rope supporting the leg.

F = mg
F = (8.00)(9.81)
F = 78.48
F = 78.5 N

Is that all I had to do? It says I got the answer for this part right.

(b) What is the traction force exerted on the leg? Assume the traction force is horizontal.

This is where I'm completely lost and confused.
First of all, the horizontal traction force they are referring to is the longest one on top right?
If so, then do I have to find the value of that horizontal line?
How???
They say the answer is supposed to be 105 N

Help!

2. Oct 2, 2012

### collinsmark

'Looks right to me!

If you still have questions why this is, you should assume that the system is in static equilibrium (i.e. nothing is moving). Second, assume that the pulleys are frictionless.

For a frictionless, simple pulley (with just one rope hanging off of it) in static equilibrium, the tension in the rope on one side is the same as the other side. [Edit, and assume the rope is ideal, mass-less, too.]
What is meant by "traction force" is the total horizontal component of force on the foot, from the pulley attached to the foot. Notice that the pulley attached to the foot is being "pulled" to the right a little by the rope. (Well, technically it is still in static equilibrium due to the friction of the patient on the attached bed. But that's not the important thing. The point is that there is still a horizontal component of force acting on the foot-pulley ultimately from the rope.) This force acts to pull the foot away from the rest of the leg, in a net horizontal direction. Find this horizontal component of that force.

Last edited: Oct 2, 2012