318 
MINUTES OF PROCEEDINGS OF 
Now, if we neglect the weight of the spar (or if we assume half its weight 
to act at either end), the leg AD is kept in equilibrium by two sets of forces 
acting at its extremities, A and D; and hence the two resultants of these sets 
of forces act along the spar, and are equal in magnitude, each being R, but 
opposite in direction. 
Then we may consider the foot A to be kept at rest by the action of three 
forces—viz., N acting along the vertical AZ, R acting in the direction DA, 
and a horizontal force F. Since the lines of action of two of these forces, N 
and R, lie in the vertical plane EAZ, the third force must also act in that 
plane—that is, along the line AE, its intersection with the horizontal plane. 
It is plain that R will be equal and opposite to the resultant of N and F-, 
also that, resolving vertically, we get 
%N=W,o-cN=^ . ( 1 ) 
o 
If we take into account the weight of each spar (w) we have 
3 N- j = W + 
w 
or N = — + w. 
O 
If II be the point in which the vertical let fall from D cuts AE, we have 
the sides of the triangle DAII —viz., DA, AH, and HD —proportional to the 
three forces R, F, and N respectively. 
We can readily find F and R, either by construction or by analysis. 
I. Graphically . (Fig. 7). 
Construct to some convenient scale the equilateral triangle ABC of 
side (a). Bisect BC in E, join AE, and take AH = \AE; at II erect a 
perpendicular to AE, and set off from A along the perpendicular, AD = l. 
Then z ADII = a, and the three forces keeping the foot of the spar at rest 
are proportional to the sides of the triangle DAII. 
To find F and R, measure off from D along the perpendicular DII pro- 
TF 
duced, DK = N = — lbs. on any convenient scale, and draw KL parallel to 
o 
HA, to meet the side DA produced in L; then, on the same scale, KL will 
represent the force F, and DL the resultant pressure R. 
Or, we may measure off D'K' = \ W (Fig. 8) from D' along any line 
parallel to DII, and through D' and K draw lines respectively parallel to 
DA and HA. 
II. 
Analytically. 
(From Fig. 6, or Fig. 7). 
F AH 
N~ BH 3 
. •. F= N . tan a — 
W, 
— tan a ; 
3 
R _ AD 
N~ DH 3 
cos a 
W 
— — see a, 
3 
( 2 ) 
( 3 ) 
