Ex. 5. Ifl be the mechanical advantage of a lever, and s the mechanical advantage of a screw, shew that the mechanical advantage of the common vice is l. s. Ex. 6. Apply the principle of virtual velocities to find the relation of the power to the weight in the endless screw, as found in Article 56. Ex. 7. Shew that, in the annexed system of pullies, W=5P. Apply the principle of virtual velocities to find the same result. Ex. 8. In the annexed system of three moveable pullies, with the cords at each pulley inclined 60° to each other, shew that CHAPTER IX. ON THE EQUILIBRIUM OF A SYSTEM OF WEIGHTS SUSTAINED BY CORDS OR BEAMS. 74. If a weight W be hung from a knot C at which the cords AC, BC (supposed without weight) meet, and the lengths of the cords and the position of A and B are given; to find the tensions in the cords. The geometrical data give the angles which AC and BC make with the vertical direction; let them be a and ẞ respectively. B If t and to be the tensions in the two cords AC and BC respectively, we may find their values by the properties of the parallelogram of forces or otherwise analytically as in Article 23, as follows: Resolving horizontally and vertically, t1 sin. a-t2 sin. B=0 ticos. a+to cos. B-W= 0 W C (1) Multiply (1) by cos. ß, (2) by sin. ß, and add, we find, t1 (sin. a. cos. ẞ+cos. a. sin. ß) — W. sin. ß=0 A We should find these results at once from Article 6, but the equation (1) shews us a property in addition, which we shall find to hold for systems of any number of cords and beams, namely, that the resolved parts of the tensions horizontally are the same in each. 75. PROP. To investigate the form and properties of the funicular polygon, or a system of cords connected by knots from which given weights are hung. Let t1 tension in AB, t=tension in BC, t=tension in CD, &c. Let a1, B1 be angles of the cords with the vertical direction at B. If n be the number of cords, we have n tensions, and 2 (n-1) angles to determine, from the following equations: AF-1 sin. a-l sin. a-&c. . . . -In-1 sin. a—1— Resolving horizontally and vertically at each knot, we have We have also B1 + a2 =πT, B2+az=π, &c. . . ẞn-2+an-1=π, which are n-2 equations. We have, consequently, the equations (a) and (b), 2(n−1) and n-2; or, in all, 3n-2 equations, to find the n tensions and 2n-2 angles as required. Since B1+=π, we have sin. ß1=sin. a, and so onwards, sin. ẞ=sin. α, &c. . . sin. ẞn-2=sin. an-1; therefore we see that the first of each pair of the equations (1), (2), . . (n−1) shews the horizontal component of the tension in each cord to be the same; for we have t1 t1 sin. aq=t, sin..=t, sin. a,=tgsin.B,=&c. ...=tsin.ßn but t1 sin. a1=t1 cos. A=t„ sin. ẞn-1=t2 cos. F n-1 or the resolved parts of the extreme tensions in the line AF are equal and opposite. From the equations (1), (2), (n-1), we find by eliminating alternately one of the tensions in each pair, 4 When the weight of a chain supplying the place of the cords of the polygon is taken into account, the problem becomes one of considerable difficulty, and is connected with the construction of chain bridges. 76. PROP. Three uniform beams connected together form a triangle, ABC, with AB horizontal, and have a weight, W, hung from C; to find the reaction in each of the beams, and to shew that the horizontal part is the same in each beam. the beam AC, R2 that in BC; and let b1 be the weight of AC, b2 that of BC. Since the beams are uniform, the center of gravity of each will be at its middle point, and we may consider half the weight of each to be at each end, so that the whole weight hanging The equation (1) shews that the horizontal parts of the reactions are equal. |