Answer :
The force on one end of the trough is 5.4 X 10⁵ N
Explanation:
The triangle is equilateral which means all the interior angles are 60° and the sides are 6m long.
According to the figure,
AI / 8 = sin (60) = √3/2
AI = 4√3
The depth of the water is AI = 4√3
The interval becomes, | 0 , 4√3|
w = 2JK
[tex]w = \frac{2}{\sqrt{3} } (4\sqrt{3} - y')\\\\P_i = pgy\\\\F_i = pgy' X \frac{2}{\sqrt{3} } (4\sqrt{3} - y')dy[/tex] (the hydrostatic force acting on the strip is the product of the pressure and the area)
where.
ρ = 875 kg/m³
g = 9.8m/s²
d = depth ( d = y')
[tex]\lim_{n \to \infty} E^n_1 F_i\\\\ \lim_{n \to \infty} E^n_1 pgy' \frac{2}{\sqrt{3} } (4\sqrt{3} - y')dy\\\\= \int\limits^4_0 {pgy \frac{2}{\sqrt{3} }(4\sqrt{3} - y) } \, dy\\\\= \frac{2pg}{\sqrt{3} } \int\limits^4_0 {4\sqrt{3} - y) } \, dy[/tex]
limit is 0 → 4√3
On solving the equation, we get the value of limit as 32√3
[tex]F = \frac{2pg}{\sqrt{3} } X 32\sqrt{3} = 64pg\\\\F = 64 X 875 X 9.8\\\\F = 548800N[/tex]
Therefore, the force on one end of the trough is 5.4 X 10⁵ N
