44 PARALLELOGRAM DIVIDED TO SUSTAIN 



From the angle d, set off dn and dt respectively equal to one third 

 of da and df, and through the points n and t, draw nm and tm 

 parallel to df and da the sides of the triangle, and meeting each 

 other in the point m ; then, according to what has been demonstrated 

 in Problem (B), m is the centre of gravity of the triangle adf. 



Produce tm directly forward, meeting a b at right angles in the 

 point r, and through the point r and in the plane of the fluid surface, 

 draw rs also at right angles to a b, and demit ms meeting rs perpen- 

 dicularly in s; then is mrs the inclination of the plane to the horizon, 

 and sm the perpendicular depth of the centre of gravity of the triangle 

 ad/* below the upper surface of the fluid. 



Put b ab, the horizontal breadth of the parallelogram abed, 

 I ad, the immersed length tending downwards, 

 p zn the pressure perpendicular to the surface of the triangle a df. 

 P= the pressure on the entire parallelogram abed, 

 = mrs, the angle which the immersed plane makes with the 



horizon , 

 d^ism, the perpendicular depth of the centre of gravity of the 



triangle adf, 



s =z the specific gravity of the fluid, 

 and x zz df, the distance between d and the point through which the 



line of division passes. 



Then, according to the principles of Plane Trigonometry, the per- 

 pendicular depth of the centre of gravity of the triangle adf, becomes 



6?rr-|Zsin.0; 

 consequently, the pressure on its surface, is 



p %l*xs sin.0 ; J see equation (10) J . 



But according to equation (7) under the 3rd problem, (art. 33), the 

 pressure on the entire rectangle, is 



and by the conditions of the present problem, the pressure on the 

 triangle, is equal to one half the pressure on the entire parallelogram ; 

 therefore, we have 



/?== !P; that is 



from which, by expunging the common terms, we get 



4x = 3b; 



consequently, by division, we obtain 

 _3b 



~ 4 ' 



