By Brian H. Chirgwin
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Additional info for A Course of Mathematics for Engineers and Scientists. Volume 3: Theoretical Mechanics
Since this tension supports the vertical portion HB, TR = wHB; but T H = wyH. •. yH = HB. Similarly yK =KC. Hence the free ends B, C both lie on the directrix (y = 0) of the catenary as shown in Fig. 14. By symmetry yH =yK . 8) = C2 + (1— YH)2. (1) But since the sag (the depth of A below KH) is 1/18, yH = c 08. Substitution in (1) gives 9c2 — 18c/ + 8/2 -= (3c — 4/) (3c — 2/) = 0. ) HB = yH 13//18. •. 2xH = 4/ yH c 13 12 • tampH) = / log (1-: + 15i). log (3\ 2 / Since TH = wyH =13w1/18 and cos /pH = lt, the force exerted by the chain on either peg is the resultant of two forces, each of magnitude 13 w//18, inclined at an angle cos-1(A) to each other.
Tl = W V10. From the coefficient of Sip we obtain 4 T2 siny, = 8 W sin(0 21,v — 99) whence 772— yio w \ 5) Exercises 2:4 1. A uniform rod AB of weight W and length 2/ has one end A in contact with a smooth horizontal plane and it rests against a small smooth peg C at a distance n l from the plane (n < 2). A horizontal force P is applied at the end B in a vertical plane through AB. Find the work done by the forces acting on the rod when the angle 0 between AB and the horizontal is increased by a small amount 60.
32 W. 10. Two equal uniform heavy rods AB, BC are freely pivoted at B, and stand in a vertical plane with the ends A and C resting on a rough horizontal board and the angle ABC equal to 2 a. The board is gradually tilted about a horizontal axis perpendicular to the plane of the rods. Show that, if motion is confined to the vertical plane of the rods, equilibrium can never be broken by slipping up the plane, but that slipping down the plane will occur first at the upper or lower of the two points of contact according as the coefficient of friction is greater than or less than tan a.
A Course of Mathematics for Engineers and Scientists. Volume 3: Theoretical Mechanics by Brian H. Chirgwin