# Page:EB1911 - Volume 17.djvu/978

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STATICS]
959
MECHANICS

such coplanar system can in general be reduced to a single resulla-nt acting in a definite line. As exceptional cases the system may reduce to a couple, or it may be in equilibrium. The moment of a force about a point O is the product of the force into the perpendicular drawn to its line of action from ° O, this perpendicular being reckoned positive or negative according as O P lies on one side or other of the line of action. If we mark off a segment AB along the line of action so as to represent the force completely, the moment is represented as to magnitude by twice the area of the triangle OAB, and the usual convention as to be reckoned positive or negative according as the letters 0, A, B, occur in “ counter-clockwise ” or “ clockwise ” order. The sum of the moments of two forces about any point O is equal to the moment of their resultant (P. Varignon, 1687). Let AB, AC (fig. 16) represent the two forces, AD their resultant; we have to prove that the sum of the triangles OAB, OAC is equal to the triangle OAD, regard being had to signs. Since the side OA is common, we have ~." to prove that the sum of the perpendiculars from B and C on OA is equal to the perpendicular from D on OA, these perpendiculars being reckoned positive or negative according as they lie to the right or left of AO. Regarded as a statement concerning the orthogonal projections -9 -9 -5 9 /A FIG. 15. to sign is that the area is C D o A: FIG. 16. of the vectors AB and AC (or BD), and of their sum AD, on a line perpendicular to AO, this is obvious. It is now evident that in the process of reduction of a coplanar system no change is made at any stage either in the sum of the projections of the forces on any line or in the sum of their moments about any point. It follows that the single resultant to which the system in general reduces is uniquely determinate, i.e. it acts in a definite line and has a definite magnitude and sense. Again it is necessary and sufficient for equilibrium that the sum of the projections of the forces on each of two perpendicular directions should vanish, and (moreover) that the sum of the moments about some one point should be zero. T he fact that three independent conditions must hold for equilibrium is important. The conditions may of course be expressed in different (but equivalent) forms; e.g.-the sum of the moments of the forces about each of the three points which are not collinear must be zero. The particular case of three forces is of interest. If they are not all parallel they must be concurrent, and their vector sum must be zero. Thus three forces acting perpendicular Flo. 17. to the sides of a triangle at the middle points will be in equilibrium provided they are proportional to the respective sides, and act all inwards or all outwards. This result is easily extended to the case of a polygon of any number of sides; it has an important application in hydrostatics. Again, suppose we have a bar AB resting with its ends on two smooth inclined planes which face each other. Let G be the centre of gravity (§ II), and let AG=a, GB=b. Let o., 13 be the inclinations of the planes, and 0 the angle which the bar makes with the vertical. The position of equilibrium is determined by the consideration that the reactions at A and B, which are by hypothesis normal to the planes, must meet at a point I on the vertical through G. Hence IG/a =sin (0-a.)/sin a., IG/b =sin (H+/3)/sind, whence cot 6 7 - (6) If the bar is uniform we have a =Ig, and C0t0=% (cot a-cotH). (7) The problem of a rod suspended by strings attached to two points of it is virtually identical, the tensions of the strings taking the place of the reactions of the planes. 1 rn /Y~.' X / g, ,, .» IB .1 C | 1 / I/ I: o. I 1 I », gif *' . !: A a'.' ':H " C FIG. 18. just as a system of forces is in general equivalent to a single force, so a given force can conversely be replaced by combinations of other forces, in various ways. For instance, 'a given force (and consequently a system of forces) can be replaced in one and only one way by three forces acting in three assigned straight lines, provided these lin.es be not concurrent or parallel. Thus if the three lines form a triangle ABC, and if the given force F meet BC in H, then F can be resolved into two components acting in HA, BC, respectively. And the force in HA can be resolved into two components acting in BC, CA, respectively. A simple graphical construction is indicated in fig. 19, where A P R ~" ! B 'H R FIG. 19. the dotted lines are parallel. As an example, any system of forces acting on the lamina in fig. 9 is balanced by three determinate tensions (or thrusts) in the three links, provided the directions of the latter are not concurrent. If P, Q, R, be any three forces acting along BC, CA, AB, respectively, the line of action of the resultant is determined by the consideration that the sum of the moments about any point on it must vanish. Hence in “tr1hnear" co-ordinates, with ABC as fundamental triangle, its equation is Po.-l-Q/3+R-y=0. If P: Q: R= a: b: c, where a, b, c are the lengths of the sides, this becomes the “ line at infinity, ” and the forces reduce to a couple. The sum of the moments of the two forces of a couple is the same about any point in the plane. Thus in the figure the sum of the moments about O is P . OA-P OB or P AB, which is independent of the position of 0 O. This sum is called the ' moment of the couple; it must of course have the proper sign attributed to it. It easily ' follows that any two couples of the same moment are P A

ii, P equivalent, and that any number of couples can be replaced by a single couple whose moment is the sum of their moments. Since a couple is for our purposes sufficiently represented by its moment, it has been proposed to substitute the name torque (or twisting effort), as free from the suggestion of any special pair of forces. A system of forces represented completely by the sides of a plane polygon taken in order is equivalent to a couple whose FIG. 20. 