Shows the more elaborate Russell traction system for immobilizing a fractured fe

Question

Shows the more elaborate Russell traction system for immobilizing a fractured femur [6]. The Russell traction system used to immobilize a fractured femur. The system is adjusted so that the angle of elevation 0 of the thigh is approximately 20 degree (adapted from Williams and Lissner). The lower leg is isolated as a free body. The force F' is the force exerted by the femur on the lower leg. The reaction force F = -F is exerted on the femur. The pulley positions arc adjusted so that the vector sum of the forces pulls the thigh at the proper angle with respect to the horizontal. The forces h and T_2 are transmitted through the lower leg. [61 Jerry B. Marion, William F. Hornyak, GENERAL PHYSICS WITH BIOSCIENCE ESSAYS, 2nd Ed. (New York, John Wiley & Sons, 1985). If m = 3 kg, compute the magnitude of the force on the femur and the angle of the thigh 9 in the Russell traction system.

Explanation / Answer

let,

mass, m=3kg

magnitude of the forece,

T1=T2=T3=m*g =3*9.8=29.4 N

here,

T1=T1*cos(theta)i + T1*sin(theta)j

T1=29.4*cos(theta)i + 29.4*sin(theta)j

and

T2=T2*cos(12)i + T2*sin(12)j

T2=29.4*cos(12)i + 29.4*sin(12)j

T2=28.76i+6.11j

and

T3=T3*cos(12)i - T3*sin(12)j

T3=29.4*cos(12)i - 29.4*sin(12)j

T3=28.76i-6.11j

total force, F=T1+T2+T3

F=(29.4*cos(theta)+28.76+28.76)i + (29.4*sin(theta)+6.11 -6.11)j

F=(29.4*cos(theta)+57.52)i + (29.4*sin(theta))j

use,

tan(20)=(29.4*sin(theta))/(29.4*cos(theta)+57.52)

===> theta=62 degress

therefore,

magnitude of force, F=(29.4*cos(theta)+57.52)i + (29.4*sin(theta))j

magnitude of force, F=(29.4*cos(62)+57.52)i + (29.4*sin(62))j

F=(71.32)i+(25.96)j

F=sqrt(71.32^2+ 25.96^2)

F=75.9 N --------------------------->

and

theta=62 dgrees --------------->

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