520 PROFESSOR CHRYSTAL ON 
If X be the component of any force parallel to Oz, and &¢ those parallel 
to O€ On OZ, we have 
X=, E+Agn +3, and so on. 
In choosing the two sets of axes we have nine constants at our disposal. 
Let them be so chosen that 
Lc=p 2n=0 26=0 
S4r=0 Say=0_— sez —0 
2yz=0 B2ny=pg 2z=0 
Sa@=0°, 2y=0,,, =G=p 
With this understanding we get for the components of resultant force, and 
couple at O. 
P=ph, , Q=Ppy » R=pyr, , 
L=pgv,—php,;, M=phd, , N= —pgh, ; 
and for the equations to the single resultant when it exists, 
Spy —Y1 + Jv2—hpz=0 
LV, — Zy +hrd,=0 
Yr — py —G_=9. 
Multiplying by A, », v, and adding we get 
(Xyv2— Aor )9 + (Ag — aes) =0, 
In terms of Ap this becomes 
(gt+h)\=(g—h)pr. 
Multiplying the equations in (2) by zyz and adding we get 
(Gra —hps)t@+hrygy— gr z=9, 
which in terms of \wv becomes by the help of (4) 
(hy —gzd)h + (gz +hyd)v +2(g+h)xr=0. 
Similarly from the first of (2) by the help of (4) 
(y+) + (2—dAy)\v +2(g+h)r=0. 
The equations (4) (5) (6) determine \ywv when xyz are given. 
(1) 
(2) 
(4) 
Solving (5) and (6) for » and », and then eliminating by means of (4), we get 
{(g—Myz + 2(hy? + g2°)d—(g—h)y2r,* 
+4(9?—h*)r\4 (a—g)z—(a@t+h)yr} 4 (a—h)y t+ (a@+g)er} =0. 
