260 On the Composition and Resolution of Forces. 



To obtain further information as to the form of/0, let P, Q, 

 R be three forces in equilibrium, and therefore necessarily in 

 the same plane, and let a line be drawn through their point 

 of application also in the same plane, inclined at angles 0, 

 a+0, /3 + to P, Q, R respectively, then sum of resolvents 

 = gives 



P/0-FQ/(« + 0) + R/(/3 + 0)=o 



for all values of 0. 



On equating to zero the coefficients of the several powers 

 of 6, we have the conditions 



P/0+Q/a + R/j3 = (10 



P/'o 4-Q/'«+R/'|8=0 (2.) 



P/"o+Q/"a+R/"/3 = o (3.) 



P/'"o + Q/'" a + R/'"/3 = o (4.) 



&c. &c. 



O R 



But the ratios ^-, -p- must be determinate functions of a. 



and |3, since the forces proportionally altered will still be in 

 equilibrium. Hence the preceding conditions must be equi- 

 valent to two only. 



Now if (1.) and (2.) were identical, or equivalent equations, 

 all the rest would be equivalent to them ; since 3 is derived 

 from 2 in the same manner as 2 from 1, &c. The whole set 

 would then reduce to only one, which is insufficient for the 



determination of p-, p-. 



Hence (3.) must be a consequence of (I.) and (2.), so that 

 if A and [x. be certain determinate constants, we have 



/'' a +x/' a +p/ a =o, 



/"/3 + A/'/3+,*/-/3=o. 



And since these conclusions are independent of any specific 

 values of a and /3, we have generally 



f»$ + \f'l + pf6=0 (A.) 



And it is easy to see that from this (4.) will be a consequence 

 of (2.) and (3.), &c, and the whole system of conditions will be 

 equivalent to two only, videlicet the two first. 

 Now if a and b be the roots of 



the solution of (A.) is 



/0 = Ae a '-rB^; 



