Tension of cable at the ankle = ? N Tension of cable at the knee = ? N Degrees f
ID: 1492635 • Letter: T
Question
Tension of cable at the ankle = ? N
Tension of cable at the knee = ? N
Degrees from the horizontal that the cable at the knee must be directed = ? degrees
6.10 When bones are healing, it is critical that they are held in a fixed position. This is a primary purpose of casts. It is also sometimes necessary to suspend the limb and cast in a fixed position. Consider the two cables suspending a leg with a cast in Figure 6.40. If the leg and cast weigh 150 N and the angle of the ankle cable is 60 degrees above the horizontalwhat are the cable ten sions and the angle of the knee cable from the horizontal such that the leg is supported? Cable 25 cm 15 cm , Cast w = 150N Figure 6.40 Two cables suspending a leg cast.Explanation / Answer
Net torque about an axis passing through knee is zero
Torque = rxF
150*0.25 = (0.25+0.15)sin(60)*TA
TA = 108.25 N
Net force along horizontal direction is zero
TA*cos(60) = Tk*cos(x)
Tkcos(x) = 10.8.25*0.5 =54.125 N ... (2)
Net force along vertical direction is zero
TA sin(60)+Tksin(x) = W
108.25*sin(60) +Tksin(x) = 150
Tk sin(x) = 56.253 ..(3)
(3)/(2)
tan(x) =56.253/54.125
from(2) and (3) angle x = 46.1 degrees
Tk = 78.06 N
Tension of cable at the ankle = 108.25 N
Tension of cable at the knee = 78.06 N
the horizontal that the cable at the knee must be directed = 46.01 degrees
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