When bones are healing it is critical that they are held in
Solution
Here, it needs to be understood that the net forces along the horizontal and the vertical direction will remain balanced as there is not net motion for the leg and the cast.
We will assume that the angle at which the knee cable is held is . Now we have the ankle cable acting at an angle of 60 degrees from horizontal and is of 150 N magnitude.
Now we will balance the forces along the horizontal and the vertical direction to get the required values:
Hence, along the horizontal we get: TkCos = 150 cos60
or, Tk Cos = 75 [Equation 1]
Along the vertical we have: TkSin + 150 Sin60 = 150
or, TkSin = 20.0962 [Equation 2]
We divide the equation 2 by equation 1 to get:
Tan = 0.26795
or, = 15.0
Using this in equation 1 we get: Tk = 75 / cos15 = 77.646 Newtons
Therefore the tension in the knee cable is 77.646 Newtons while it is held at angle of 15 degrees with the horizontal.
