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Bungee cords designed to U.S. Military specifications (DoD standard MIL-C-5651D, available at http://dodssp.daps.dla.mil) are characterized by a force constant times unstretched length in the range kL ~ 800-1500 N. Jumpers using these cords intertwine three to five cords to make a thick rope that is strong enough to withstand the forces of the jump. Suppose that you are designing a bungee jump off of a bridge that is 50.0 m above the surface of a river running below. You have read that you should expect the cord to stretch (at peak extension) to about 210% of its natural length. You have also read that you should use 3 cords together for jumpers with weights in the range 100-150 lbs, 4 cords for 150-200 lbs, and 5 cords for 200-250 lbs. Suppose you decide to use cords of length 20 m, which would seem to offer a safety zone of about 8 m (or really about 6 m if the cord is attached at the ankles).

Part A

Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that kL = (4)(800 N) = 3200 N. Since you are evaluating the safety factor, ignore any losses due to air resistance or dissipation in the cord. Ignore the mass of the rope.

Solution

We begin with an initial-state final-state diagram and the corresponding energy bar diagrams:

As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model:

\begin{large} \[ E_{\rm i} = mgh_{\rm i} = E_{\rm f} = mgh_{\rm f} + \frac{1}{2}k x_{f}^{2} \] \end{large}

It seems that we have a problem here, because we do not know h_f or x_f. The easiest way to deal with this problem is to utilize our freedom to choose the zero point of the height axis. If we restructure our coordinate system to place h = 0 m at the point where the cord is stretched to its natural length L (as shown below).

This redefinition of the origin of the height in our coordinate system greatly simplifies the equation describing the evolution of the jumper's mechanical energy:

\begin{large} \[ mgL = - mgx_{f} + \frac{1}{2} k x_{f}^{2} \] \end{large}

The discussion around the diagrammatic representation has left us with a quadratic equation in x_f, which is solved to obtain:

\begin{large} \[ x_{f} = \frac{mg \pm \sqrt{(mg)^{2} + 2 k mg L}}{k} \]\end{large}

It is not sensible that we should find a negative value for x_f, so we must select the plus sign, giving:

\begin{large} \[ x_{f} = \frac{mg}{k}\left(1+\sqrt{1+\frac{2kL}{mg}}\right) = 21.5\:{\rm m} \] \end{large}

Part B

Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords?