320 
MINUTES OF PROCEEDINGS OF 
Draw a vertical from 0, meeting the line IID in K ; then it is evident 
that DK, KO, OD, the sides of the triangle DKO, taken in order, are pro¬ 
portional to the forces which keep D at rest. Also that the sides DA, AC, 
CD, of the triangle DAC are respectively proportional to the three forces 
R, F, and N, which keep the foot A at rest. 
I. By Construction . (See Pigs. 11 and 12). 
Given the length of each spar, the weights, the distance between the end 
of rear guy and feet of sheers, and the distance from feet to the vertical 
through centre of gravity of weight. 
Having chosen some convenient scale, we must first find the length of 
OD, the axis of the sheers, which is best done separately. In Pig. 11, take 
oa, half the distance between the feet of the spars, draw od perpendicular to it, 
and from a, set off along the perpendicular a length ad equal to l; this 
gives us od. 
Then, in Pig. 12, starting with a horizontal line, mark off HO, the distance 
between end of back guy and the feet of the sheers, 0 being the bisection of 
the space between the feet; set off OC, the distance measured from 0 to the 
foot of the vertical through the centre of gravity of the weight; at C erect a 
perpendicular to the horizontal line, and set off from 0 along this perpen¬ 
dicular the length OD already found, thus getting the inclination of the 
sheers. If this angle be already known or given, Pig. 11 is not required, 
for we can find the length of OD by drawing OA at right angles to OD, 
measuring along it the semi-distance between the feet, and setting off l from 
A along OD; then let fall the perpendicular DC upon the horizontal line. 
In either case, the triangle OAD must be described. 
Join DH , draw OK vertical, and along OK, or OK produced, set off OK', 
representing on any convenient scale the number of cwt. in [W + w), and 
through K' draw a line parallel to KD, meeting OD produced in D'; then 
K'D' and D'O give, on the same scale, the tension T, and the counter¬ 
balancing force along OD. Bisect D O in M, and draw MG parallel to 
DA) then MG represents the force R, D'O being 2 R resolved along the 
axis of the sheers. 
To find N and F, we have the triangle DAC, which, however, lies in a 
place oblique to that of the paper, so that to get its proper dimensions we 
must turn it doim, about DC as an axis, upon the plane of the paper. We can 
do this by describing an arc of a circle, with centre D and radius l, or DA, 
which cuts the horizontal line in A'. Join A'D, then the triangle DA'C gives 
the acUial magnitude, according to scale, of the triangle DAC. Set off from 
D along DA', produced if necessary, DS, representing to scale the number of 
cwt. in R, and through S draw SR parallel to A'C, meeting DC, or DC pro¬ 
duced, in L) then SR, RD, are respectively equal to the forces F and N 
upon the scale chosen. 
II. Analytically. (Prom Pig. 10, or Pig. 12). 
Prom the triangle DKO, 
T KD sin DOK 
W~~ KO~ sin KDO 5 
W cos 0 
sin (6 — 0) ‘ 
T = 
( 1 ) 
