proportional to these perpendiculars: to prove that the three forces are in equilibrium. Let ABC, fig. (13), be the triangle; AP, BQ, CR, the three perpendiculars; P, Q, R, the respective forces. Then, the two triangles BAQ, CAR, being similar, BQ CR: AB : CA, and therefore, the magnitudes of the forces being inversely as the perpendiculars, This proportion shews that the forces are in equilibrium. (6) If three forces, the magnitudes of which are 3P, 4P, and 5P, act at one point and are in equilibrium, shew that the forces 3P and 4P are at right angles to each other. (7) Two weights are attached to given points of a fine string, the ends of which are tied to two fixed points: prove that the tensions of the three portions of the string cannot be all equal. (8) Two small rings slide on the arc of a smooth vertical circle; a string passes through both rings and hangs below the arc, three equal weights being attached to it, one at each end and one on the portion between the rings; to find the position of the rings when they are in equilibrium. If A, B, be the two rings, and C the position of the middle weight, ABC must be an equilateral triangle, of which the side AB is horizontal. (9) A circular disc is kept at rest by three forces acting all outwards or all inwards along normals at three points of the circumference: shew that the forces are as the sides of the circumscribing triangle which touches the disc at those points. SECT. 2. P Q R :: sin Q, R : sin R, P: sin P, Q. If three forces P, Q, R, act on a point, and Q, R, R, P, P, Q, denote the respective angles between Q, R; R, P; P, Q; the sufficient and necessary conditions of equilibrium are expressed by the following proportion P: Q R sin Q, R : sin R, P: sin P, Q. (1) From a fixed point 0 in a string AOB, fig. (14), which is fixed to the two points A, B, in a horizontal line, a weight W is suspended: to compare the tensions of the two strings AO, BO, a and ẞ being their inclinations to the horizon. Since < OAB = a and ‹ OBA = B, it follows that hence and COR. Since AOB π-a-B, we see also that = S: W:: cosẞ : sin (a + B), T: W:: cos a: sin (a+B). (2) Three forces, the directions of which bisect the angles of a triangle, are in equilibrium; to shew that they are proportional to the cosines of the respective half-angles. Let the directions of the three forces P, Q, R, intersect in the point 0, fig. (15). Then, the forces being in equilibrium, P Q R sin Q, R sin R, P: sin P, Q. (3) A hemispherical bowl ACB, fig. (16), contains a weight W which is attached to a weight P by means of a string passing over the rim of the bowl at a point A to determine W's position of equilibrium. Let M be the position of the weight's equilibrium, O being the centre of the sphere: let AOM=0: then, the triangle AOM being isosceles, OMA== (π-0). The weight is kept at rest by the reaction R of the bowl, acting along MO, the force P, acting along MA, and the force W, acting vertically downwards: hence follows that the negative sign in the numerator must be rejected: therefore the position of equilibrium is defined by the equation A If P> W, then cos> 1, and the weight could not rest in the bowl. (4) Three forces P, Q, R, acting upon a particle, produce equilibrium supposing the angle between P's direction and R's to be π, and that between Q's and R's to be, to compare the magnitudes of the forces. P, Q, R, are to each other respectively as 2, √3, 1. (5) A sphere rests upon two inclined planes: to find the pressure which it exerts upon each. Let W be the weight of the sphere; a, ß, the inclinations of the two planes, and R, S, the respective pressures exerted upon them. Then (6) A boat B experiences a force F from the stream, parallel to the bank OA, and a force W from the wind, at right angles to OA towards the opposite bank: to find the tension of the rope OB by which the boat is tied to a point O in the bank, and to determine the inclination of OB to OA. (7) A weight is supported on an inclined plane, the inclination of which to the horizon is 30°, first, by a power parallel to the plane, and, secondly, by a power parallel to the base: to compare the pressures on the plane in the two cases. The pressure in the former case is to that in the latter as 3 to 4. (8) One end of a string AOP is attached to a fixed point A, and to the other end is fixed a weight P: a string BO, shorter than AB, is attached to a point B in the same horizontal line with A, and is connected with the string AOP by a moveable loop O, which is capable of sliding along AOP: to find the inclination of BO to the horizon, when there is equilibrium. W. S. 3 (9) A string of length c is tied to two points A, B; the inclination of AB to the horizon is a; a weight, sliding freely on the string, hangs in equilibrium: shew that, if AB= a, the weight divides the string into two parts equal to (10) A string AHKB is attached to two fixed points A, B, in a horizontal line: from fixed points H, K, in the string are suspended equal weights P, P: to find the tensions of the several portions of the string, supposing AH, HK, KB, to be all equal. If AB=c, and 7= the whole length of the string, the tension of HK is equal to (11) Two forces P and P', acting in the diagonals of a parallelogram, keep it at rest in such a position that one of its edges is horizontal: shew that P sec a' = P' sec a = W cosec (a + a'), where W is the weight of the parallelogram, and a, a', are the angles at which its diagonals are inclined to the horizontal edge. (12) A particle is placed in the centre of a circle and is acted on by three forces P, Q, R, tending towards the angular points A, B, C, of the circumscribed triangle: prove that, when there is equilibrium, where A, B, C, are the angles of the triangle. |