448 MISCELLANEOUS HYDROSTATIC QUESTIONS, WITH THEIIl SOLUTIONS. 



579. QUESTION 10. Supposing the same piece of glass to weigh 9.996 

 Ibs. in air, but in water only 7.015 Ibs. ; what is its diameter? 



The specific gravity of water, when reduced to pounds per cubic 

 inch, is 0.03617, and that of air is 0.000045 ; therefore, by equation 

 (188), page 236, we have 



9.996 7.015 . 5.402 inches, the same as before. 



.5236 (0.03617 0.000045) 



580. QUESTION 1 1 . The lock of a canal is 40 feet wide, and the lock 

 gates being rectangular planes, stand 16 feet above the sill, with their 

 upper edges on a level with the surface of the water ; now, supposing 

 that the gates are found to meet each other in an angle of 141 35'; 

 what is the amount of pressure which they sustain, a cubic foot of 

 water weighing 62 \ Ibs. avoirdu poise ? 



Here the lock gates meet each other in an angle of 141 35' ; which, 

 according to Barlow, is the situation in which, with a given section 

 of timber, they obtain the greatest strength. But by the principles 

 of Plane Trigonometry, the length of each gate is 



20Xsec.l9 25'=21.206 feet; 



and by the question, the depth is 16 feet ; therefore, the whole surface 

 exposed to the pressure of the water, is 2 1. 206 X 32 = 678. 592 square 

 feet. Now the centre of gravity of each gate is 8 feet below the 

 surface of the water, the specific gravity of which is unity ; conse- 

 quently, by equation (8), page 19, the entire pressure upon the gates, is 



p=: 8X678.592 5428.736 cubic feet of water; 



or when reduced to Ibs. it is 

 5428.736X62J z= 339296 Ibs., or 151 tons 9 cwt. qrs. 48 Ibs. 



581. QUESTION 12. If the diameter of a cylindrical vessel be 20 

 inches; required its depth, so that when filled with a fluid, the pres- 

 sure on the bottom and sides may be equal to each other ? 



This question is resolved by the equations (59 and 60), page 100, 

 where it is manifest, from the construction of the equations, that 

 Bzz: \d, and consequently, by substitution, we obtain 



.7854D 2 dzz 3.1416Ddx Jd, and this expression is equivalent to 



.7854D 2 d 1.5708D<f, 



and by casting out the common factors, we have 2</~D, or by 

 division, c?m 10 inches ; hence when the depth of the vessel is equal 

 to half the diameter of the base, the concave surface and the bottom 

 of the vessel sustain equal pressures. 



NOTES. 



