CABOQ SU3STGS

279

(4) Through X and Y draw lines parallel to DO cutting DF and
DH produced at A and B respectively. AB then represents 16 feet,
the spread between the shackles.

D

.Y

Fig 21.

(5) The length of DA, or of DB, measured from the same scale,
gives the required lengths of the legs of the sling, viz., 10-7 feet.

Calculation.— First, find angle 6, which is the angle between the
legs of the sling and the perpendicular dropped from their intersection
at D. This might be done as a side issue, because in triangle DFE
DJP=3 tons, D^==2 tons, /.#=90°. Find £FDE or 8.

Nat. Bee. 6=— = -=1-5 .'. 6=48° 12'.

Having found 0, make £CAD equal to (90°— 48° 12')=41° 48'.
Then AD measured from the same scale as AB will give the rmnjrm^m
length of the leg. It is easy to calculate it because
AD=AG cosec 6=8 cosec 48° 12'=8xl-34=10-72 feet.
By making the legs longer their angle of intersection would be smaller
and the tension on each reduced. The same result could be got by
closing in the shackles A and B on the beam thus making the base of
the triangle ADB smaller. „
Example. — A beam weighing 3 tons is being lifted by means of a
sling, the legs of which are each 6 feet long. If the tension on each
leg is 2 tons, find the distance between the shackles, assuming that the
weight of the beam is evenly balanced on the sling.
Construction. — (1) Draw a vertical line DC and lay off DG=3 tons,
(2) Bisect DG at E* and draw EF perpendicular to DG.