(13) A smooth circular ring is fixed in a horizontal position, and a small ring P sliding upon it is in equilibrium when acted on by two strings in the direction of the chords PA, PB; shew that, if PC be a diameter of the circle, the tensions of the strings are in the ratio of BC to AC. SECT. 3. Each force equal and opposite to the resultant of the other two. If three forces act on a point, it is sufficient and necessary for equilibrium, that any one of them be equal and opposite to the resultant of the other two. If P, Q, R, be the three forces, these conditions are equivalent to the equations P2 = Q2 + R2 + 2QR cos Q, R, Q2 = R2 + P2 + 2RP cos R, P, R2 = P2 + Q2 + 2PQ cos P, Q. (1) If a point O, fig. (17), be acted on by three forces represented in magnitude and direction by OA, OB, OC, O being the point of intersection of lines drawn from the angles of a triangle ABC to bisect opposite sides, to prove that the point will be in equilibrium. Let P, Q, R, be the points of bisection of the sides: then the resultant of the forces OA, OB, is equal to a force 20R along OR. But, by a known property of triangles, the line OC is double the line OR: hence the forces OA, OB, are equivalent to CO, and therefore balance OC. (2) A force P, acting at an angle of 30° to an inclined plane, supports a weight W: supposing R to be the pressure on the plane, to prove that W2 = P2+PR+ R2. The reaction of .the inclined plane against the weight is equal to R, and the angle between this reaction and P's direction is equal to 60°: hence W2 = P2 + R2 + 2PR cos 60° = P2+PR+R2. (3) Two weights P and Q hang at the ends of a string which passes over two smooth pegs A and B: a weight Wis suspended from a point 0 at a point of the string between the pegs to find W in terms of P and Q in order that the angle AOB may be a right angle. The weight W is given by the relation W2 = P2 + Q2. (4) Three forces act at the three angles of a triangle towards a point within the triangle and are proportional respectively to the distances of the point from the angles: give a geometrical construction for the position of the point in order that equilibrium may subsist. CHAPTER III. RESULTANT OF ANY NUMBER OF FORCES ACTING ON A POINT. (1) D is the middle point of the side BC of a triangle ABC, fig. (18): three forces, represented by AB, AC, DA, act upon the point A: to find the magnitude and direction of their resultant. Complete the parallelogram ABEC and join AE: the intersection of AE, BC, will coincide with the point D. The resultant of AB, AC, is AE: hence the three forces AB, AC, DA, are equivalent to the two forces AE, DA, and therefore their resultant is the force DE or AD. (2) ABC, fig. (19), being a triangle, the point A is acted upon by three forces, represented in magnitude and direction by the lines AB, AC, BC: to draw the line which shall represent their resultant in magnitude and direction. Complete the parallelogram BADC. Then the two forces AB, BC, are equivalent to the two AB, AD, and therefore to the force AC: thus the three forces AB, AC, BC, are equivalent to double the force AC. Produce AC to E, making CE equal to AC. Then AE represents the resultant of the three forces AB, AC, BC, in magnitude and direction. (3) A quadrilateral ABCD, fig. (20), is acted on by forces, which are represented in magnitude and direction by AB, AC, DB, DC, respectively: to prove that their resultant is represented in magnitude and direction by four times the distance between the middle points of AD, BC. Let F, G, be the middle points of BC, AD, respectively. Join AF, DF, FG. Then the forces AB, AC, are equivalent to 24F; and the forces DB, DC, to 2DF. Again, the forces AF, DF, are equivalent to 2GF. Hence the forces AB, AC, DB, DC, are equivalent to 4 GF. (4) Three equal forces are represented in magnitude and direction by the lines drawn from the angles of a triangle to the centre of the circumscribing circle to prove that the resultant is represented in magnitude and direction by the line joining that centre with the intersection of the three perpendiculars drawn from the angles of the triangle to the sides respectively opposite. Let ABC, fig. (21), be the triangle, and O the centre of the circumscribed circle. Produce CO to meet the circle in F and join BF. Complete the parallelogram BFOK. Then the force KO is equivalent to forces BO, KB, applied at O, that is, to forces BO, OF, that is, to the two forces BO, CO. Let E be the intersection of the perpendiculars AP, BQ, CR, drawn from A, B, C, to BC, CA, AB, respectively. Then BE, FA, are parallel, for FAC, being an angle in a semicircle, is a right angle, and therefore equal to BQC. Also BF, EA, are parallel, since FBC, being an angle in a semicircle, is a right angle, and therefore equal to ‹ APC. Hence EA = BF = KO: but EA, KO, are parallel: hence OAEK is a parallelogram: hence the forces AO, KO, are equivalent to the force EO, that is, the forces AO, BO, CO, are equivalent to the force EO. (5) AB, CD, are any two equal and parallel chords in a circle, and Pis a point on the circumference, half way between A and B: prove that if forces, represented by the lines PA, PB, PC, PD, act at the point P, their resultant is constant. (6) Forces are applied at one of the corners of a regular hexagon, acting towards the other corners, and are proportional in magnitude to the distances of those corners from the point of application to find the ratio which the magnitude of the resultant bears to that of one of the forces acting along a side. The required ratio is that of 6 to 1. (7) If O be the centre of the circle circumscribing a triangle ABC, and D, E, F, be the middle points of the sides; the system of forces represented by OA, OB, OC, will be equivalent to those represented by OD, OE, OF. (8) A quadrilateral ABCD is acted on by forces represented in magnitude and direction by AB, AD, CB, CD, respectively : prove that their resultant is represented in magnitude and direction by four times the line joining the middle points of the diagonals. SECT. 2. Trigonometrical method. (1) A force of two pounds and a force of three pounds act upon a point; the direction of the former force being inclined at an angle of 60°, and that of the latter at an angle of 45° to a given straight line passing through the point: to find the magnitude and direction of the resultant. Let R= the magnitude of the resultant and 0 = its inclination to the given straight line. Then (2) Three forces P, Q, R, acting on a point O, are inclined at angles a, ẞ, y, to a given line passing through 0: to find the magnitude and direction of the resultant of these forces. |