Question

A 12m long gate that is 4 m wide is placed against a body of water at an angle of 60 degrees. The gate is being pulled by a rope placed 3 m from the top of the gate at 60 kN. If the water is 9 m deep, what is the minimum weight of the gate that will cause it to rotate clockwise

Answer 1

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:

$F_{p} =\rho gAx=1000*9.8*\frac{9}{sin60} *4*\frac{9}{2} =1833.203kN$

The center of pressure is:

$h=\frac{\frac{(\frac{9}{sin60})^{3} *4 }{12}*sin^{2}60 }{\frac{9}{sin60}*4*\frac{9}{2} } +\frac{9}{2} =6$

If the moment is equal to zero,

$F_{p} *(9-6)+W*\frac{L}{2} cos60=F*9\\1833.203*3+W*6*cos60=60*9\\W=-1653.2kN$