1. Po 
A hinged gate is placed at the bottom of a tank containing
water. Find the resultant force on the gate due to hydro-
static pressure, the torque on the gate about hinge B, and
the point of application of the resultant force.
D=10 ft, L= 6 ft and W = 1 ft D
Water x B L y
Force due to water: FR =  - p dA  FR i =  - p dA i Assume FR is in the plus x-direction.
A A
For the coordinate system specified, dA = dy dz. Our integral becomes:
W L
FR =   - p dy dz
o o
Since p does not vary with z, we can simplify this to: L
FR = W  - p dy o
However, p does depend on y, so we must find that relationship before integrating.
From the hydrostatic equation, we obtain p = Po + gh = Po + g(D + y) at some point
y on the gate.

2. Substituting for the pressure, L L
FR = - W  (Po + g(D + y) ) dy = - W[Poy + gDy + gy2/2] o o
FR = - [PoWL + gDLW + gL2W/2]
The first term represents the contribution due to atmospheric pressure; but this is also exerted on
the other side(PoWL). We drop this term.
FR = -gLW[ D + L/2 ] = - (62.4 lbf /ft3)(6 ft)(1 ft)[ 10 ft + 3 ft]
The final value for the resultant force is: FR = lbf What does the negative sign indicate?
What is the torque about hinge B caused by the hydrostatic force? x T FR
T =  r x dF =  y j x p dA (- i) = -(j x i)  y p dA
A A A
Start with the general relationship for torque. Making a smart choice of
coordinate system can greatly simplify the problem. What is r for this
problem? How is dF described? What is element dA, perpendicular to i?

3. Neglecting atmospheric contribution: p = g(D + y) We know dA = dy dz
Hence,
L W
T = k   y( D + y ) g dz dy = k g W ( yD + y2 ) dy
o o o
We carry out the integration. L
T = k g W[ - y2 D/2 + y3/3 ]o = k g W[ - L2 D/2 + L3/3 ]
T = k (62.4 lbf /ft3)(1 ft)[ (6 ft)2(10 ft)/2 + (6 ft)3/3]
T = 15,725 ft-lbf k
Now, where is the resultant force applied to give the same torque about hinge B? The location of the
unknown point will be designated r´ where r´ = y´ j.
r´ x FR = T  y´ j x FR (-i) = T k Recall j x (-i) = k
r´ = y´ = T / FR  = 15,725 ft-lbf /4870 lbf = 3.2 ft
Is this result reasonable?