anonymous
  • anonymous
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. (See the figure.)
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
i hope theres a picture :)
anonymous
  • anonymous
If the physical therapist specifies that the traction force directed along the leg must be 20.0 N, what must Wbe? yah one sec
anonymous
  • anonymous
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amistre64
  • amistre64
im not sure, but the top angle seems redundant to me
amistre64
  • amistre64
Wcos(35) = 20 W = 20sec(35) would be my best guess
anonymous
  • anonymous
is w the same thing as T
amistre64
  • amistre64
yes
anonymous
  • anonymous
i thought the goal was to find (W) which is the mass
amistre64
  • amistre64
find w that produces the tension in the system
amistre64
  • amistre64
of course like i said before, i aint to sure about my physics applications tho
amistre64
  • amistre64
so, if you have anything to offer other than me taking a blind stab at it, feel free to contribute it
anonymous
  • anonymous
so w is what cause the tension
anonymous
  • anonymous
The leg is seeing two tensions. Therefore, \[2T \cos(35) = W\]
amistre64
  • amistre64
thats where my doubts were :)
anonymous
  • anonymous
so is w the total force ?
anonymous
  • anonymous
These one always make me stop and think twice. Remember if the system is stationary, the forces must cancel. The first time the wire is diverted towards the leg, we created a horizontal component of the tension force. This horizontal component is created again when we turn away from the leg. Therefore, the force on the leg must be twice the horizontal component such that they cancel. @nath We have two expressions here.
anonymous
  • anonymous
The first one is\[2T \cos(35) = 20\]and the second\[T = W\]Solve for T, then we know W
anonymous
  • anonymous
why am i able to replace t with w
anonymous
  • anonymous
What forces act in the vertical direction? Tension and weight, right? Note that because the angles about the pulley attached to the leg are symmetric, their vertical components cancel out.
anonymous
  • anonymous
dont the x force cancel each other out?
anonymous
  • anonymous
The x forces cancel out with the tractive force on the leg. Let's write out our expressions for summation of forces in the x and y-directions. \[F_x = 0 \rightarrow 2 T \cos(35) = 20\]\[F_y = 0 \rightarrow T \sin(35) - T \sin(35) + T_{ceiling} - W = 0\]The first two terms are at the pulley attached at the leg, the \(T_{ceiling}\) is the tension at the ceiling.
anonymous
  • anonymous
both the net forces of the x and y component are equal to zero
anonymous
  • anonymous
Yes. The leg isn't moving nor is the weight.
anonymous
  • anonymous
so Tceiling =W?

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