# 10 - 2/8/2010 Problem (text p193): Three bungee jumpers are...

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2/8/2010 1 Problem (text p193): Three bungee jumpers are connected as shown in the diagram. How far will each jumper fall? x 1 Initial position (cords unstretched) x 2 x 3 Final position (cords stretched) The three bungee cord segments have spring constants k 1 , k 2 , and k 3 . The three jumpers have masses m 1 , m 2 , and m 3 . The forces acting on each jumper must add up to zero. Convention: downwards forces positive. 0 x k - ) x - (x k g m : 1 jumper for 1 1 1 2 2 1 Collecting terms and rearranging gives: 0 ) x - (x k g m : 3 jumper for 0 ) x - (x k - ) x - (x k g m : 2 jumper for 2 3 3 3 1 2 2 2 3 3 2 g m x k - )x k (k 1 2 2 1 2 1 g m x k x k - g m x k - )x k (k x k - 3 3 3 2 3 2 3 3 2 3 2 1 2

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2/8/2010 2 These equations can be expressed in matrix format: g m x k - )x k (k x k - g m x k - )x k (k 2 3 3 2 3 2 1 2 1 2 2 1 2 1 g m g m g m x x x k k - 0 k - k k k - 0 k - k k g m x k x k - 3 2 1 3 2 1 3 3 3 3 2 2 2 2 1 3 3 3 2 3 b Ax One way of solving an equation of the form Ax=b is to premultiply both sides by A 1 : b A Ix b A Ax A b Ax 1 1 1 Ways of obtaining the inverse of a square matrix in Matlab: >> A = [1 2; 3 8] A = 1 2 3 8 >> inv(A) b A x 1 ans = 0.5714 0.1429 0.2143 0.0714 >> A^ 1 ans = 0.5714 0.1429 0.2143 0.0714
2/8/2010 3 Matlab Solution (using inverse matrix): >> k1 = 50; k2 = 100; k3 = 50; % in N/m >> m1 = 60; m2 = 70; m3 = 80; % in kg >> g = 9.81; >> A = [ k1+k2 -k2 0 -k2 k2+k3 -k3 0 -k3 k3 ]; >> b = [ m1*g; m2*g; m3*g ]; >> x = inv(A) * b x = 41.2020 55.9170 71.6130 Very Important Note: This is not a very good way of solving systems of equations This is not a very good way of solving systems of equations. “Left division” is much better. Left Division: Normal “right division” can be thought of as post multiplying by the inverse: x / y is equivalent to x * y 1 “Left division “is logically equivalent to pre multiplying by the inverse: y \ x is equivalent to

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## 10 - 2/8/2010 Problem (text p193): Three bungee jumpers are...

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