Question

The 6x5 m side of an open tank has a hinge at A and a horizontal cable at B The tank is to be filled with glycerin, with a density of 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 3.9 m. consider g = 9.8 m/s2

Answer 1

This is a classic hydrostatic problem, I will leave an attached image

of a possible sum of forces and the problem in question,

$dF = pdA = (\rho gy) (wdy)$

$F = \rho gw \int \limit ^ d_0 ydy = \rho gw \frac {d ^ 2} {2}$

For the point in question,

$\bar {Y} = \frac {\int ydF} {\int dF} = \frac {\rho gw \int \limit ^ d_0 y ^ 2dy} {\rho gw \int \limit ^ d_0 ydy}$

$\bar {Y} = \frac {d ^ 3/3} {d ^ 2/2} = \frac {2d} {3} = \frac {2 * 3.9} {3} = 2.6m$

From the bottom = 3.9-2.6 = 1.3

Well things

$\sum M_A = 1.3F-6T = 0$

$T = \frac {1.3} {6} * (1263 * 9.8 * 5 * \frac {3.9 ^ 2} {2})$

$T = 101974N$

$F_ {A} = F-T = 368676N$