EQUAL FLUID PRESSURES. 45 



59. This equation is too simple in its arrangement to require any 

 formal directions for its resolution ; nevertheless, the following- rule 

 may be useful to many of our readers. 



RULE. {Take three fourths of that side of the given rec- 

 tangular parallelogram, in which the line of division terminates , 

 and the point thus discovered, is that through which the line 

 of division passes. 



60. EXAMPLE 10. A rectangular parallelogram, whose sides are 

 respectively equal to 24 and 42 feet, is immersed in a cistern full of 

 water, in such a manner, that its shorter side is coincident with the 

 surface of the fluid, and its plane inclined to the horizon in an angle 

 of 52 degrees ; it is required to determine a point in its lower side, 

 to which, if a straight line be drawn from one of the upper angles, the 

 parallelogram shall be divided into two parts sustaining equal pres- 

 sures ; and moreover, if a straight line be drawn from the same point 

 in the lower side, to the other upper angle, it is required to assign the 

 pressure on the triangle thus cut off? 



Here, by operating according to the rule, the point of division is 



x = % X 24= 18 feet. 



In the next place, to determine the pressure sustained by the 

 triangle bcf t cut off from the parallelogram abed, by means of the 

 line/6 drawn from the point f to the angle at b, we have according 

 to equation (12), (Problem 4), 



p z= i (b x) I 9 s sin.0, 



where (b #) in this equation, takes place of b in the one re- 

 ferred to. 



The natural sine of 52 degrees according to the Trigonometrical 

 Tables, is .78801 ; hence, by substituting the respective data in the 

 above equation, we shall have 



pl (2418) X 42 2 X .78801 2746.09928 cub. ft. of 

 water ; consequently, the pressure expressed in Ibs. avoirdupois, is 



p 1 = 2746.09928 X 62.5 = 181631.205 Ibs. 



This seems to be an immense pressure, on a triangle whose surface 

 is only 126 square feet ; it is however but one sixth part of the pres- 

 sure on the entire parallelogram ; this is manifest, for the pressure on 

 the triangle adf, is three times the pressure on the triangle bcf, since 

 the base df is equal to three times the base cf, and the altitudes of 

 the triangles, as well as the perpendicular depths of the centres of 

 gravity, are the same; but the pressure on the parallelogram abed, 



