Question: Consider five jumpers (m, m2, m3, m4, and ms) that are connected by 10 m long bungee cords and are being held in place

Consider five jumpers (m, m2, m3, m4, and ms) that are connected 

Consider five jumpers (m, m2, m3, m4, and ms) that are connected by 10 m long bungee cords and are being held in place vertically so that each cord is fully extended but unstretched (m1 is at the top, followed by m2, etc.). We can define five distances, X1, X2, X3, X4, and xs, as measured downward from each of their unstretched positions. After they are released, gravity takes hold and the jumpers eventually come to rest at their equilibrium positions. Assume that each cord behaves as a linear spring and follows Hooke's law (F= kx). Using Newton's second law, force balances can be written for each jumper. Given: m = 55 kg, m2 = 75 kg, m3 = 60 kg, m4 = 75 kg, and m5 = 90 kg k = 80 N/m, k= 70 N/m, k3 = 70 N/m, k4 = 100 N/m, and ks = 20 N/m a) Draw free body diagrams for each of the five jumpers and b) define the force balance equations. c) Express the system equations as a matrix equation and d) use MATLAB to solve for the values of X1, X2, X3, X4, and xs at equilibrium.

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