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Tension/Traction problem

  • Thread starter Minusu
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  • #1
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Homework Statement



An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a m = 6.9kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed. Find the angle θ.

Fgw(force of gravity on weight)= 68N
m1(mass of foot and boot=4kg
∅= 17 degrees

Homework Equations


Fnet=ma
Fnety=Tpicos∅+Tp2cosθ-Fgw

The Attempt at a Solution



I tried to find for the angle between the top rope and horizontal by doing the following;
Fnety=Tp1cos∅+Tp2cos-Fgw
Fnety=0
68/68cos17=Tp2cosθ
Fnetx=0
Fnetx=Tp1sin∅+Tp2sinθ=0
Tp2=Tp1sin∅/sinθ
Plug tp2 into first equation
68/68cos17=(68sin17/sinθ)*cosθ
θ=36.501 degrees
I'm not sure if this makes sense though and then a later part of the question asks, "what is the traction force?" and it states that the traction force must be pulling outward. Help please! http://puu.sh/4F63M.png [Broken]
 
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Answers and Replies

  • #2
tiny-tim
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Hi Minusu! :smile:
Fnety=Tp1cos∅+Tp2cos-Fgw
Fnety=0
No: you have to decide which body you're adding the forces on.

In this case, it's the forces on the boot (whose y components have to add to zero) …

there are only two forces on the boot! :wink:
 
  • #3
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Would these two forces be the Fg on the boot and the tension of the ropes?
So Fnety=-Fgboot+tp1cos∅+tp2cosθ=0
39.2-68cos17/tp2=cosθ

And then solve for the x direction so
Fnetx=Tp1sin17+tp2sinθ=0
68sin17/sinθ=tp2

Plug into first

39.2-68cos17/68sin17=tanθ
θ=52.4
 
  • #4
tiny-tim
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Hi Minusu! :smile:

(just got up :zzz:)
… a later part of the question asks, "what is the traction force?" and it states that the traction force must be pulling outward. … http://puu.sh/4F63M.png [Broken]
hmm … i've read the question again, more carefully, and i don't understand how that is compatible with "The boot is held suspended by the ropes and does not touch the bed." :confused:

if the net traction force (ie the resultant of the two tension forces) is horizontal, then there's nothing to hold the boot up

i suggest you try ignoring "does not touch the bed", and see if the computer accepts the answer that gives you

so just solve for the angles needed for the resultant of the two tensions to be horizontal :smile:
 
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  • #5
haruspex
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So Fnety=-Fgboot+tp1cos∅+tp2cosθ=0
Think again about the vertical components of the tensions.
 
  • #6
haruspex
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, and i don't understand how that is compatible with "The boot is held suspended by the ropes and does not touch the bed." :confused:
I don't see the difficulty. The upper part of the leg is to be taken as a horizontal weightless string. The patient touches the bed, of course.
 
  • #7
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Thanks for all the help. I just solved for the angle. The tension in Tp1 and Tp2 is the same, so Fnety=Tp1sin∅+Tp2sinθ-Fg
Fnety=0
0=68sin17+68sinθ-39.2
(39.2+68sin17)/68=sinθ
.868=sinθ
θ=sin^-1(.868)
θ=60.3 degrees.
 
  • #8
haruspex
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Thanks for all the help. I just solved for the angle. The tension in Tp1 and Tp2 is the same, so Fnety=Tp1sin∅+Tp2sinθ-Fg
Fnety=0
0=68sin17+68sinθ-39.2
(39.2+68sin17)/68=sinθ
.868=sinθ
θ=sin^-1(.868)
θ=60.3 degrees.
Looks good. Of course, strictly speaking, the leg is a rod with weight distributed along it. There will be a vertical force from the bed at the hip, so not all of the weight of the leg will be taken by the ropes. However, I don't think you have enough information to take that into account.
 

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