FOR 



whence the magnitude and position of the resultant are known. 



Cor. 1. By putting- the values of X and Y in the expression for r, we 

 shall get 



r- ij $p* + 2pq cos. (* /3) -f q* 



which agrees with the result obtained in Cor. Art. 1. 



Cor. 2. If we call $ and ^ the /s. P A R and Q A R, we shall have 



- - -- ^ 



y [^ a 4- 2^ ? COS. (-/3) + e? 2 j 



& sin . 4 = _^ -- Ptinj.*^ __ 

 y [^ 2 4- 2^? cos. (-/3) -f j*j 



5. To find the resultant of any number of forces, p> p, p, ......... p, ia 



123 n 



the same plane ; their directions making with the line A x angles 



, , at, ......... ce, respectively. 



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J5y proceeding precisely as before, we shall have, by putting 



p cos. -f p cos. os. -f p cos. a. ...... + p cos. a, = X 



'2233 n n 



p sin. a + j sin. + j sin. ...... +p sin. * = Y 



/ / a 2 3 s n n 



r=v(X8 + Y)j tan.fl=~. 



6. To find the resultant of forces, whose directions are not all in the 

 same plane. 



In the preceding case, the forces were resolved in the directions of two 

 lines at right /s. to each other. In this case we must resolve them in 

 the directions of three lines each at right /s. to the other two, and meet- 

 ing together in a point. Let us suppose these three lines to be A A*, Ay 

 A z, and let p be a force, and , /3, y the ^s. which it makes with A x t 

 Ay, A x ; the force will then be equivalent to three forces 

 p cos. a. in A JT, p cos. ft in Ay t p cos. -y in Az. 



Hence if we have forces p t p, p ...... p 



making with A x angles , , ...... 



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123 



