# Homework Help: Tension in ropes at rest

1. Aug 16, 2011

### bulbasaur88

Problem XIV, page 18 ------>http://www.physics.princeton.edu/~mcdonald/examples/ph101_2006/learning_guide_ph101_2006.pdf
A patient's leg is held in traction as shown in the diagram. The top and bottom pulleys are fixed, while the central pulley is not, but has come to rest at a position determined by the various forces acting on it. Note that one single rope runs between the mass and the pulleys and is fixed to the middle pulley, while a separate strap connects the foot to the center pulley. What is the magnitude and direction of the force on the patient's foot produced by the traction apparatus shown int he diagram?

Also express this force as a vector (using the coordinate system shown). Assume that all three pulleys are frictionless.

What I Understand About the Problem
The middle pulley has NO acceleration in any direction because the foot has come to a rest position.

Forces Acting on the Center Pulley
1. Tension in the direction 40 degrees south of east, T = (2.5)(9.8) = 24.5 N, due to the weight of the box

2. The two tensions, T', in the northeasterly direction

3. The tension caused by the foot in the southwesterly direction, Tf

I realize that in order for the middle pulley to stay in a rest position, all the of the y-components of force must cancel out. Same for the x-components of force.

I also think, that since it is the same single rope that holds the weight of the block, the tension in the whole rope remains the same magnitude; however, since the tension in the two rope segments at the very top share the weight of the block, T' = T/2. Is this a correct assumption?

I honestly do not know how to continue if all of my assumptions are correct up to here, butI would like to be assured if I am going in the right direction before continuing.

Thank you for any help!!!

Answer: 56.7 N, 19.5 degrees above horizontal

2. Aug 16, 2011

### PeterO

There are 118 pages to the guide you included - would you like to help out by mentioning which page the problem is?

3. Aug 16, 2011

### bulbasaur88

Hello :) I did...It is on page 18...Thank you for looking :D

4. Aug 16, 2011

### PeterO

The tension in the rope from the block to the middle pulley, via the top pulley is the same throughout, so your T' and T have the same value.

5. Aug 16, 2011

### bulbasaur88

Ok, thank you. That simplifies things a little bit for me..

So in the y-direction the forces acting on the middle pulley are:
ΣFy = Tsin45 + TsinƟ - Tsin40 - (Tf)sinβ= 0, where β is the angle the string attached to the foot makes with the horizontal, and Ɵ is the angle between the highest reaching rope and the horizontal.

And the forces in the x-direction are:
ΣFx = Tcos45 + TcosƟ + Tcos 45 - (Tf)cosβ = 0

----------------------------------------------------------------------------------------

We know that T = weight of block = 24.5 N

So we have
ΣFy = Tsin45 + TsinƟ - Tsin40 - (Tf)sinβ= 0
ΣFy = 24.5sin45 + 24.5sinƟ - 24.5sin40 - Tfsinβ = 0

and

ΣFx = Tcos45 + TcosƟ + Tcos 45 - (Tf)cosβ = 0
24.5cos45 + 24.5cosƟ + 24.5cos45 - Tfcosβ = 0

This leaves me with 3 unknown variables: β, Ɵ, and Tf (tension in the string attached to the foot)...Am I missing something?

6. Aug 16, 2011

### PeterO

I think you are supposed to assume that the ropes going "North East" are parallel.

I would do the vector sum of the "three tensions" to find the net force they apply. That is the force they apply to the foot/cast.

Note: if the "North East" ropes were not to be considered parallel, there would have been some measurements there to enable you to calculate what the angle actually was.

The diagram might actually be showing the pulleys much bigger i diameter that those used in hospitals; when compared to the size of a foot.

7. Aug 16, 2011

### bulbasaur88

THANK YOU ! Gosh, the picture is so terrible....I would've never thought to look at the ropes as parallel. I will get back to you! :)

8. Aug 16, 2011

### bulbasaur88

GOT IT!!! You are awesome!!!!!!!!!!!! Thanks a million!!!!