A square gate, 1.5 m × 1.5 m, on one of the vertical sides of a f
A. 500 × 9.81 N
B. 600 × 9.81 N
C. 750 × 9.81 N
D. 1128 × 9.81 N
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
Concept:
Hydrostatic Forces on Submerged Surface
|
Case |
Force |
Center of pressure (h) |
|
Horizontal Position |
wAx̅ |
h = x̅ |
|
Vertical Position |
wAx̅ |
\(h = \bar x + \frac{{{I_G}}}{{A\bar x}}\) |
|
Inclined Position |
wAx̅ |
\(h = \bar x + \frac{{{I_G}}}{{A\bar x}}{\sin ^2}\theta \) |
\({I_G} = \frac{{b{d^3}}}{{12}}\) (For rectangular plate)
\({I_G} = \frac{{\pi {d^4}}}{{64}}\) (For circular plate)
Here,
A = Area of surface touching fluid = b × d
IG = Area moment of inertia about centroidal axis.
x̅ = Vertical distance of C.O.G. of body from free surface
ω = Specific weight
θ = Angle at which the surface is inclined with horizontal
Calculation:
The hydrostatic press. Distribution on gate AB,
Magnitude of hyd. Force (FH) = pg A̅ × h̅
h̅ = 0.75 m (CG of sq. plate)
FH = 1000 × 9.81 × (1.5)2 × 0.75 = 1687.5 × 9.81 N
Force will act at a distance of \({h_{c.p}} = \bar h = \frac{{{I_{CG}}}}{{A \times \bar h}}\)
\(\bar h = 0.75 + \frac{{1.5 \times {{\left( {1.5} \right)}^3}}}{{12 \times {{\left( {1.5} \right)}^2} \times 0.75}} = 1m\) (from top)
Taking moment about hinge A
FH × 0.5 = F × 0.75
⇒ 1687.5 × 9.81 × 0.5 = F × 0.75
\( \Rightarrow F = \frac{{1687.5\; \times \;0.5\; \times \;9.81}}{{0.75}} = 1125 \times 9.81\;N\)
