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 4.0kg, and the doctor has decided to hang a 6.0kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed.
a) The net traction force needs to pull straight out on the leg. What is the proper angle theta for the upper rope?
b) What is the net traction force pulling on the leg?
c) what is the net traction force pulling on the leg?
Heres a link to the picture http://session.masteringphysics.com/problemAsset/1...
2 Answers
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The tension in the rope T = Mwg = 6.09.8 = 58.8 N
For the leg to be pulled straight out, the vertical components of forces on the leg must sum to zero.
The vertical component from the lower part of rope is -T*sin15º
The vertical component from the upper part of the rope is T*sinθ;
The vertical force from the weight of the foot and boot is -4.0*9.8 = -39.2 N
Then
-Tsin15º + Tsinθ - 39.2 = 0
sinθ = (39.2 + 58.8*sin15º)/58.8 = 0.925
θ = 68º
The net traction force is the sum of the horizontal components:
T*cosθ + T*cos15º = 79 N
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Broken Leg Traction
Source(s): https://shrinkurl.im/a8pDX