276

SEAMANSHIP AND NAUTICAL KNOWLEDGE

(2) and (3) Draw DE and AF parallel to the span and EF parellel to
AD to meet AF. The two forces acting at A, viz., the resultant AD
and the tension on the span AF, are now represented graphically by
the parallelogram ADEF, its diagonal AE being the magnitude of the
thrust on the derrick (\1\ tons) and the length of AF or DE gives the
tension on the span (9J tons).

Fig 19—Hauling part led to Masthead.
AB = 10 tons
AC ss BD sa tension on fall = 5 tons.
AD the resultant 10 J tons
AE the thrust on derrick 12J tons.
AF tension on span 4 tons
(4:) To find the stress on the leading block, make XG and XH each
equal to the pull on the hauling part of the fall 5 tons, and complete
the parallelogram XQKH. Join XK.
The length of the diagonal XK represents the stress on the leading
block (7£ tons) which, is the resultant of the two forces acting at X,
viz.* the fall leading up the derrick and down to fcbe winch.