THE ROYAL ARTILLERY INSTITUTION. 
319 
It will, however, be more convenient to express F and R in terms of the 
length of each spar, and the distance between their feet. 
AH „ AFj o a sin 60 ° 
* ma ~ AD ~ AJ)~ * ' 
whence 
tan a 
f 3P — a 3 
; and sec a 
l \/3 . I 
V3 . I 
therefore we have 
f 3 1 * — o? 
W 
F^ -C . 
and R = 
3 
W l 
7F ‘ ~ a? ' 
(4) 
(5) 
Sheers. 
Prob. —In sheers, to find (1) the tension of the bach guy ; (2), the hori¬ 
zontal force required to keep the feet from sliding ; (3), the normal reaction 
at the foot of each spar ; and (4), the whole pressure on the ground. 
AD, BD (see Pig. 10), are the spars, lashed together at D; DII represents 
the line of action of the back guy, taken in the vertical plane DCH, perpen¬ 
dicular to DAB, the plane of the sheers; then the intersection DO of these 
planes bisects the fixed angle ADB, and the centre of gravity of the weight 
W will also lie in the plane DHC, as it will be in the vertical through D. 
The feet A and B are securely lashed to pickets. 
Let l = effective length of each spar, 
W — weight of gun and main tackle, 
w = weight of each spar, 
Z 2a = inclination of spars to one another, 
z. 9 = inclination of OD, axis of sheers, to horizon, 
Z 0 = inclination of rear guy to horizon, 
F = horizontal force (ho'wever applied) required to keep each foot 
from sliding, 
N — normal reaction at foot of each spar, 
R = resultant, or whole pressure on the ground at foot of each spar, 
T = tension of back guy. 
Whether we neglect the weight of the spars, or consider \w to act at either 
extremity, it is evident that the apex D may be considered as kept at rest by 
three forces acting in the plane DHC —viz., W, T, and a force equal and 
opposite to their resultant, and whose line of action must lie in OD produced, 
and which force is plainly the sum of the resolved parts of the two equal 
forces, R, acting along AD and BD. 
Again, the foot A is at rest through the action of three forces— N, F, 
and R. Of these, N acts vertically at A, R acts along the line DA (see 
Triangle Gyn), and both lie in the plane DAC; hence the third force F 
must also act in that plane—that is, along the line AC, its intersection with 
the horizontal plane. Similarly, the corresponding force F at B will act in 
the line BC, 
