274 
MINUTES OE PROCEEDINGS 
Substituting in (1), 
Tension of rope x OB sin a = JF. OB cos 0 + w\ OB cos 0. 
.\ Tension of rope = (W + \w) ... (2) 
This divided by 8 or by 10, as the case may be, for the mechanical advantage 
gained by the arrangement of the guy and runner tackle will give the force 
to be applied at the running end of the tackle of the guy. 
Lever Sheers . 
If the spar be inclined at an angle a to the horizon, to find breaking 
strain. 
AB is the part of the spar bearing the strain acting vertically in the 
direction AC, it may be resolved into two forces, 
AB tending to break the spar, and BC tending to force the end at JS into 
the ground. 
The triangles ABC and BOB are similar, 
and AB = AC cos BAC 
= W cos a; 
breaking strain = W cos a. 
Note.—A ll these calculations have been made without taking into account the stiffness of ropes, 
or friction, which of course is very considerable and absorbs a great quantity of the work done, but 
is useful in assisting to sustain a machine in a state of rest; as for instance, passing a rope round 
the windlass of a gyn, capstan, &c. 
Prob. What will be the greatest strain upon the points of support to 
which the feet of sheers are attached, under the following conditions? 
Weight of each cylindrical spar, 4 cwt.—weight of gun and blocks hanging 
from the top of the sheers, 60 cwt.-—inclination of the plane containing the 
axis of the spars, 60° to the horizon—the back guy makes an angle of 30° 
to the horizon; and the spars make an angle of 60° with each other. The 
coefficient of friction between the feet of the spars and the platform T5, 
