Purchase of this book includes free trial access to www.million-books.com where you can read more than a million books for free. This is an OCR edition with typos. Excerpt from book: essentially the same, in-so-far as equilibrium is concerned, for the lines of direction of the three forces must pass through one and the same point if the forces are in equilibrium. It is quite evident that the three forces must lie in the same plane, for each of the three must be opposite to and in the same straight line with the resultant of the other two. Moreover, the resultant or vector sum of the three forces must be zero, if the forces are in equilibrium. Let OA, OB, OC, (see Fig. 3) represent three forces f1, ft, fz, which are in equilibrium, the first two being mutually perpendicular to each other. The third force f3, to produce equilibrium, must be equal and opposite to and in the same straight line with the resultant of the other two. Since the diagonal OC' represents the resultant of f1 and /2, it follows that the line OC which represents f3 must be equal and opposite to and in the same straight line with OC. This is the point of view considered and verified by Exp. 6. Since the forces are in equilibrium, their resultant effect in any direction must be zero, that is, the algebraic sum of the projections of the forces in that direction must be zero. In the line DA the effect of ft is f1 cos o, the effect of / is/2 cos 90, and the effect of / is / cos o. But the resultant effect along DA is zero, hence /! cOS 0 + /2 c0S OX)0 -f - ft cos 0 = 0. Since cos o = 1 and cos 90 = o, the equation becomes 11) /x + / cos 0 = 0. Similarly, the resultant effect along EB must be zero, and hence the sum of the projections of the forces into that line must be zero, or /! sin o + f2 sin 9o + f3 sin a = o. Since sin o = o and sin 9o = 1, this equation becomes (2) /2 + /s sin a = o. (It should be noted that sin a and cos a are negative for an angle in the thi...