THE ROYAL ARTILLERY INSTITUTION'. 
275 
Let AC, BC be the two spars, and CD the “guy,” or rope; then, if AC, 
BC be of the same length, ABC will be an equilateral triangle, since 
ACB = 60 °. 
The gu} 7- or cord will lie in a vertical plane bisecting AB perpendicularly. 
Let DCFE be this plane, intersecting the plane which contains the axes of 
the spars in EC, and the horizontal plane in DEE. Consequently the angle 
CEE, which measures the inclination of the plane ACB to the horizontal 
plane, is equal to 60°. Now the point B is kept at rest by the pressure 
along CB, the normal pressure E, and the friction, and because when three 
forces are in equilibrium, they are in the same plane, it follows that the 
normal pressure R, the friction /xR, and the pressure along CB, are propor¬ 
tional to CF, BE, and CB ; that is, the friction at B acts along BE. 
Similarly, the friction at A, acts along AF. Let T be the tension of the 
guy. Then resolving vertically and horizontally, 
or 
and 
or 
R + R + T sin 30 = 4 + 4 + 60, 
+ JT=68; ...(1) 
T cos 80 = /xR cos 6 + /xR cos 6 = 2fxR cos 0, 
or \TV'6 = 2{xR cos 0 .....(2) 
Eliminating T between (l) and (2), we get 
R __ 186 \/3 __ 34 \/J 
4 (jx cos 6 n/ 3) [x cos 0 + V 3 
It remains to find cos 0. 
Now because BEF is a right angle, EF being perpendicular to AB, 
