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Here the two push rays MN and KN of equal strength are striking the Anu N. The rays are equally inclined to the line joining the centers of A and B. Their resultant effect  on Anu N will be towards the centre of object B because their perpendicular resultants NG and NH will cancel each other. The total resultant effect of both rays MN and KN towards B will be times their total strength. Therefore, resultant of single ray MN towards B will be equal to times its total strength.

The element EFDC is just like the circumference of a circle which is forming base of a cone with its vertex at N. The force on the element EFDC by the push rays is concentrating at N because these are traveling towards N. Their resultant effect will be towards centre of B, and will be times the total strength of the rays.

Therefore, resultant force at N by the push rays passing through the element EFDC and striking at N:

Resultant force on N by all the push rays which are passing through the complete cap EF