69 



' 4 c2 — y2 

 2 c2-y2 



2 c2 



(4) 

 (5) 



Let ns now suppose that the frame has vertical legs (as in Fig. 2) extending 

 downward into the liquid. Let / be the length between the legs. 







JTiz 



Equation (1) becomes 



w = 2r(/ — t) sin n -\- ptly, 



a 

 =^ 2 /)c2(/ — t) sin ft -)- 2 Itpc sin-^' 



(6) 



dw 



When «j is a maximum, -r- =o. Let t be very small compared with /, then 



2 c cos a -\- t cos -^ ^ o. 



Eliminating a by (4) and (5), and inserting the value of c, 



2r t- ;6^2^ 



When < is small, a near approximation is 



\ p 



Supplying this value of y in (6), and solving for T, 

 ^ w id'H- It 



2,[i — tyA[i—ty' i{i—ty- 



^/p2/2^2 _^4.((, i—t)ll 



(7) 



(8) 



Table II gives the value of Tcalculated by the above formula for mica frames 

 varying in thickness from 0.0013 cm. to 0.02067 cm. 



