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Hi, I have a question for Physics Lab. I'm not sure how to proceed in answering this question. Any help would be much appreciated, thank you.

In order to treat certain broken bones and other injuries, it is sometimes necessary to immobilize the affected regions and to remove the usual forces from the fracture while it heals. Various systems of traction using weights, ropes, and pulleys are found in current practice [6]. Figure 5.5 (I've attached this figure below) shows one way to maintain a force on a patient's leg; this configuration is physically similar to the one studied in the experiment.

Figure 5.6 (I've also attached this picture below) shows the more elaborate Russell traction system for immobilizing a fractured femur [6].

The pulley positions are adjusted so that the vector sum of the forces pulls the thigh at the proper angle with respect to the horizontal. The forces vector T1 and vector T2 are transmitted through the lower leg. [6] Jerry B. Marion, William F. Hornyak, GENERAL PHYSICS WITH BIOSCIENCE ESSAYS, 2nd Ed. (New York, John Wiley & Sons, 1985). If m = 5 kg, compute the magnitude of the force on the femur and the angle of the thigh in the Russell traction system. (Hint: T = mg)

_______ Newtons

________ degrees

Essentially, this is a vector addition problem! All you have to do is to add together the three forces and find the resultant.

The tension T in the cable is:

T=m*g=5*9.8=49N

First break each force vector into its respective components:

F1: 49N at 12 degrees NW becomes:

Fx1=-49*cos(12)=-47.9N and Fy1=49*sin(12)=10.2N

F2: 49N at 12 degrees SW becomes:

Fx2=-49*cos(12)=-49.7N and Fy2=-49*sin(12)=-10.2N

F3: 49N at 26 degrees WN becomes:

Fx3=-49*sin(26)=-21.5N and Fy3=49*cos(26)=44.0N

Adding all the x components together:

Fx1+Fx2+Fx3=-47.9-47.9-21.5=-117N

Adding all the y components together:

Fy1+Fy2+Fy3=10.2-10.2+44.0=44.0N

Add these two vectors together using the Pythagorean Theorem the resulting vector magnitude will be:

F=sqrt(117^2+44^2)=125N

And the direction of the force can be found using the inverse tangent function:

Alpha=inversetan(44/117)=20.6 degrees.

So the the force pulling on the leg will be 125N at an angle of 20.6N above th horizontal.

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