Algorithms for Constrained Minimization of Smooth Nonlinear by A.G. Buckley, J-.L. Goffin

By A.G. Buckley, J-.L. Goffin

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If W is the weight of the segment, the potential energy, V, is given by y r, 21a ^ — = 2a cos 6 cos 26 W 40 1 dV . n 21a . ^n 21a . / n = - 2a sin 6 H sin 26 = sin 6\ cos 6 27 ) W dö 20 10 V Id 2 V ^ n 21a => r- = — 2a cos 6H cos 26. Wd62 10 260 Ch. 25. FIG. Hence dV/dO vanishes for 0 = 0 and cos0 = 20/27. When 0 = 0, d2V/d62 = W\-2a + 21a )>a Also, when cos 0 = 20/27, d2V/d02= , 40 27 400 27\ Wa[1 - — + — . — -2 - — < 0 . 27 5 (27) 10/ Therefore the position 0 = 0 is a position of stable equilibrium ( V has a minimum there) and the oblique position, cos 0 = 20/27, is a position of unstable equilibrium ( V has a maximum there).

Prove that the least couple required to rotate the cylinder about its axis has moment \iWa (1 +/* 2 )cosa' where \i is the coefficient of friction between the cylinder and each plane. ) 7. A chain of four equal uniform rods, AB, BC, CD, DE, each of weight Wand freely jointed together at B, C, D, hangs symmetrically from two fixed points A and E in the same horizontal line. The angles of inclination of AB and BC to the vertical are 6 and respectively. Prove that tan = 3 tan 6. Show also that the resultant reaction at B is WV(l+itan2tf>).

13), s = yJ(y 2 -c2)= V[c2cosh2(x/c)-c2] => s = c sinh (x/c). 12) in the form from which sec \j/ + tan \j/ — e x/c sec i// — tan ij/ = e " x/c. ] 264 From we have FURTHER STATICS Ch. 15) > T = wy. This equation shows that the tension at any point of the string is proportional to the ordinate at that point. Example 1. A uniform heavy chain of length 32 m hangs symmetrically over two smooth pegs at the same level so that the lowest point of the portion of the chain between the pegs is 2 m below the level of the pegs.

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