Physics dynamics problem

Question

The Russell traction apparatus is used for a fractured femur. When the femur is healing, the muscles of the upper leg tend to pull the two sections of the femur together, which could result in a shortened leg after healing. The Russell apparatus provides a force parallel to the femur that prevents contraction of the femur. The sling supports the weight of the lower leg.

(a) Find the magnitude and direction of the horizontal force on the leg (i.e., the sole of the foot).

(b) To adjust the size of this force, the angles at the pulley on the foot can be changed. Would the angles be made larger or smaller, to decrease the force on the leg? Explain your reasoning.

(c) Why are the angles kept equal, as shown?

Link to Image: https://docs.google.com/document/d/18KwCHsVk5ah4cE6B7w2ymyYvtQ95QjRlq2-3DtO4Qbw/edit?usp=sharing

Answer 1

Let

m₁ = 4kg be the mass of the lower leg,

m₂ = 4kg be the mass of the weight,

α₁ = 30° be the upper angle at the pulley,

α₂ = 30° be the lower angle at the pulley,

g = 9.81 m/s² be gravitational acceleration.

The force on the leg is the vector sum of two forces:

the tension in the upper string with magnitude m₁g, and

the tension in the lower string with magnitude m₂g.

Each of the two forces can be decomposed into a sum of a horizontal and vertical force.

Let's consider upward direction as positive and downward direction as negative.

The horizontal component of the upper force is m₁gcos(α₁), and

the vertical component of the upper force is m₁gsin(α₁).

The horizontal component of the lower force is m₂gcos(α₂ ), and

the vertical component of the upper force is -m₂gsin(α₂ ).

(a)

The net horizontal force on the leg is the sum of the two horizontal components:

m₁gcos(α₁) + m₂gcos(α₂ ) =

4×9.81×cos(30°) + 4×9.81×cos(30°) ≈ 68 N

(b)

Increasing the angles would decrease their cosine, which would decrease the

horizontal force.

(c)

It is important that the vertical force on the leg be 0.

Otherwise the leg would be constantly pulled up or down.

The vertical force on the leg is the sum of the vertical components:

m₁gsin(α₁) - m₂ gsin(α₂ )

So the angles are kept equal because the masses m₁ and m₂ are kept equal.

The angles would not be made equal if the masses were not equal.

The only equality that needs to be maintained is

m₁gsin(α₁) = m₂gsin(α₂ )