Question

Application to the Russell traction system Any traction system can be analyzed by keeping in mind two things. First, the force applied is in the direction of the wire at the point where it is attached to the patient. Second, the force is equal to the weight hung on the wire. When a thigh bone of femur becomes fractured, a potential problem arise in the healing process. There is a tendency for the major leg muscles to pull together and misalign the two segments of the broken bone at the fracture point, thus sometimes causing shortening or ultimately a limp. To prevent this, traction is used to maintain tension and hold the adjoining structure in alignment. Figure 4 shows the Russell traction system which generates a force along the direction of the femur. The Russell traction apparatus (Figure 5.4) is used for a fracture of a femur. This system is designed to elevate the upper leg at an angle of roughly 53 degrees above the horizontal. The mass of the leg is 4.00 kg, and the horizontal tension required to keep the leg stretched is 60.0 N. Find the values of masses m_1 and m_2. What is the angle, relative to the horizontal, of the art force on the kg due to the traction system? How does this compare with the angle of the femur?

Answer 1

ANSWER

Part (a) Let the tension in the horizontal rope beT and this tension force equally splits into tweo parts and they acts along two horizontal ropes. So, the tension force in the each of the parallel ropes is -. This force balances the weight of the block of mass m2- Thus =m,g 60 N 2g 2(9.8 m/s' m- 3.06 kg ANS. Let the tension in the rope that is connected to the block of mass m, beT'. Now, it is known that the weights of both the blocks together balance the weight of the leg. So, m g+m,g = mg (Here, m is the mass of the leg.) m,-m-m,-4.0 kg-3.06 kg m2-0.94 kg ANS Part (b) The tension force in the horizontal rope and the tension force T' are perpendicular to each other. So, the magnitude of the resultant force is, F, =JF. + r.2-F2 + (ng). (60 N), ((0.94 kg)(9.8 m/s*))2 F 60.7 N The direction of the resultant relative x-axis is, (0.94 kg)(9.8 m/s) 60 N -1 θ= 8.730 ANS

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