Answer:
Since the weight cannot take a negative value, it is concluded that the minimum weight required for the gate is zero, because the same pressure force will rotate the gate.
Explanation:
Given:
L = 12 m
F = 60 kN
w = 4 m
H = 9 m
The pressure force is:
[tex]F_{p} =\rho gAx=1000*9.8*\frac{9}{sin60} *4*\frac{9}{2} =1833.203kN[/tex]
The center of pressure is:
[tex]h=\frac{\frac{(\frac{9}{sin60})^{3} *4 }{12}*sin^{2}60 }{\frac{9}{sin60}*4*\frac{9}{2} } +\frac{9}{2} =6[/tex]
If the moment is equal to zero,
[tex]F_{p} *(9-6)+W*\frac{L}{2} cos60=F*9\\1833.203*3+W*6*cos60=60*9\\W=-1653.2kN[/tex]
