between CA and CB. by the triangle ABC. and thence OC. Another relation between them is given In (B) we have given that LOAC = 30°, LOBC = 18°, LCAB = 90°, and AB Let then and therefore y' tan 18° y tan 30°.. Also from the right-angled triangle ABC, 1848. (4). Shew how to determine the height of a mountain by observations at two stations in the same horizontal plane, the distance between the stations being known. (B). If the stations are in the same vertical plane passing through the summit, and the summit (S) is observed from the further station, but a lower point (S) is observed by mistake from the nearer, shew that the height determined by the process lies between the heights of S and S'. The process of finding the height of the mountain would in this case be to observe the angle NAS (fig. 31) from the further station A, and the N'BS' from the nearer station, and, knowing the distance AB, to determine the height (x) from the triangles ASN, BS'N' on the supposition that SN, S'N' were each; the result obtained would correspond to the height S"N" of the point where BS' produced meets AS; and since S is supposed to be invisible at B, S" must be lower than S but higher than S'. 1848. STATICS. (A). If two forces, acting on a particle, be represented by two adjacent sides of a parallelogram, prove that their resultant will act in the direction of the corresponding diagonal. (B). Explain how the force of the current may be taken advantage of to urge a ferry-boat across a river, the centre of the boat being attached, by means of a long rope, to a mooring in the middle of a stream. Let the boat be kept in a position inclined to the direction of the stream at an angle of about 45°, its foremost end pointing up the stream. The current will produce upon the boat a pressure consisting of two parts, one perpendicular to the side of the boat, the other in the direction of its length. The latter is inconsiderable compared with the former, because the surface offered to the perpendicular resistance is much greater than that which the longitudinal pressure acts upon. Considering the rope (since it is long) as nearly parallel to the direction of the stream, we shall have two forces, viz. the tension of the rope and the pressure of the water perpendicular to the side of the boat, the resultant of which, by the proposition, acts between them, and therefore tends to propel the boat towards the bank, to which the foremost end of the boat is directed. 1849. (A). Assuming the parallelogram of forces, so far as the direction of the resultant is concerned, shew that the diagonal of the parallelogram represents the magnitude of the resultant. (B). The resultant of two forces is 10 lbs., one of them is equal to 8 lbs., and the direction of the other is in clined to the resultant at an angle of 36°: find the other force, and the angle between the two. Let OA, OB, (fig. 32) represent the two forces; then, by (A), OC, the diagonal of the parallelogram OACB, represents their resultant. We have given 04 proportional to 8, OC to 10, and OB to x (the number of lbs. in the required force); and LBOC = 36°. From the triangle OAC, therefore OA2 = 0С2 + AC2-20C.AC cos OCA; 82 = 10" + x2-2.10 x cos 36°, 1850. (A). If three forces which act in a plane keep a rigid body at rest, prove that their lines of action are either parallel or pass through a point; and in both cases, shew that any two of the forces are inversely proportional to the perpendiculars drawn on their respective lines of action from any point in the line of action of the third. (B). An uniform heavy rod of given length is to be supported in a given position with its upper end resting at a given point against a smooth vertical wall, by means of a fine string attached to the lower end of the rod and to a point in the wall. Find by geometrical construction the point in the wall to which the string must be attached. Let AB (fig. 33) be the rod, G (its middle point) its centre of gravity; the forces which keep it at rest are and its weight acting along GW, tension of string along BP, reaction of wall along ARI NP. Now the principle of which (4) is the enunciation tells us that these three forces must meet in a point (because they are not parallel) in order that there may be equilibrium. We must therefore suppose the string BP placed in such a position that it shall pass through O the intersection of AR and WG produced. When this is done, all the statical conditions have been brought in, and the problem of finding the position of equilibrium is solely a geometrical one. We have then from the geometry of the figure, But therefore AP: AN :: PO : OB :: AG: GB. AG = GB; AP AN Since the position and length of AB is given, AN is known, and therefore the point P where the string is to be tied, is determined by taking AP = AN. 1848. (A). When a body is kept in equilibrium by three forces acting in one plane, either their directions are parallel, and one force is equal to the sum or difference of the other two; or their directions meet in a point, and each force is as the sine of the angle between the other two. (B). AB is a rod capable of turning freely about its extremity A, which is fixed; CD is another rod equal to 2AB, and attached at its middle point to the extremity B of the former, so as to turn freely about this point; a given force acts at C in the direction CA: find the force which must be applied at D in order to produce equilibrium. |