Question

The split Russell traction mechanism is an orthopedic device that combines suspension and traction to immobilize, position, and align the lower extremities in the correction of orthopedic deformities and in the treatment of congenial hip dislocation and hip and knee contractures. Split Russell traction uses a sling to relieve the weight of the lower extremities. The traction weights are suspended from pulley-and-rope systems at the foot and head of the patient's bed. A leg in a cast is stabilized by such a system of weights and pulleys as shown in the figure below.

The leg is held in the position shown by two weights that are connected to the leg via two cables. The combined weight of the leg and the cast is 340 N. Cable 2 has a weight W₂ of 287 N hung above the patient while the other end of this cable makes an angle β of 45 degrees with the horizontal.

What should the angle α be so that the leg is in equilibrium with no force on the leg from the rest of the body

Answer 1

free body diagram 4 W sing5 twisoid > W_C0545° W, cost 340N

As the leg is in equilibrium,

Equating vertical forces

W₂sin45° + W₁sin$\alpha$ = 340

W₁sin$\alpha$ = 340 - W₂​​​​​​sin45°

W₁sin$\alpha$ = 340 - 287sin45°

W₁sin$\alpha$ = 340 - 202.97

W₁​​​​​​sin$\alpha$ = 137.03 N ..............eq(1)

Equating horizontal forces

W₁cos$\alpha$ = W₂​​​cos45°

W₁cos$\alpha$ = 287cos45°

W₁cos$\alpha$ = 202.97 ...............eq(2)

Dividing eq(1) by eq(2)

tan$\alpha$ = 137.03 / 202.97

tan$\alpha$ = 0.675

$\alpha$ = 34.02°