Hydrostatics

Metacentre

When a floating body is tilted by a small angle, the centre of buoyancy $B$ shifts to a new position $B'$ because the shape of the displaced volume changes. The vertical line through $B'$ meets the original vertical axis of the body at a point $M$, the metacentre. The distance $GM$ is the metacentric height.

        upright               heeled
         |  M                  M
         |                    /|
         G                   / G
         B                  B'  W
   weight W down, buoyancy W up through B'

• If $M$ is above $G$ ($GM>0$): a righting couple $W\cdot GM\sin\theta$ acts, so the body is stable.
• If $M$ is below $G$: overturning couple, unstable.
• If $M$ coincides with $G$: neutral equilibrium.

$$GM=BM-BG=\frac{I}{V}-BG$$

where $I$ is the second moment of the waterline area about the axis of tilt and $V$ the displaced volume. Larger $GM$ gives greater stability but a shorter rolling period.

Laboratory determination of $GM$ of a rectangular vessel (pontoon)

Apparatus: a rectangular pontoon, a water tank, a movable weight $w$ on a cross-bar, a plumb line and a graduated scale to read the tilt angle.

Steps:

1. Weigh the pontoon with its fittings; its total weight is $W$.
2. Float it in water in the tank and check it floats level, with the plumb line at zero on the scale.
3. Place the movable weight $w$ at the centre of the cross-bar, then shift it a measured distance $x$ to one side.
4. The pontoon heels; read the angle $\theta$ from the plumb line on the scale (or $\tan\theta$ from the deflection).
5. Repeat for several positions $x$ on both sides, and for different values of $w$.
6. Compute $GM$ for each reading, then average:

$$w\,x=W\,GM\tan\theta\ \Rightarrow\ GM=\frac{w\,x}{W\tan\theta}$$

7. As a check, calculate theoretically $GM=\dfrac{I}{V}-BG$ with $I=\dfrac{LB^{3}}{12}$ for the rectangular waterline, and compare with the experimental value.

(Alternatively, the same heel observed at different heights of a vertical bar mass lets $G$ be located by the rolling-period method.)

Reading of the figure: the conical surface ABC hangs below the plane AC, which is 3 m across (radius 1.5 m); the apex B is 6 m below AC and the water stands 7 m above B, so the free surface is 1 m above AC. The air above the water has a gauge pressure of 100 kPa. The water fills the cone, so it lies on the inside of the surface.

   gauge 100 kPa
  ------------------- free surface
        1 m
  A -------+------- C   (3 m)
    \      |      /
     \     | 6 m /
      \    |    /
       \   |   /
          B

By symmetry the horizontal components cancel; the net force is vertical (downward). It equals the pressure force on the plane AC plus the weight of water in the cone.

Pressure at AC:

$$p_{AC}=100+9.81(1)=109.81\ \text{kPa}$$

Area of AC:

$$A=\pi(1.5)^{2}=7.069\ \text{m}^2$$

Pressure force on AC:

$$F_1=109.81\times7.069=776.2\ \text{kN}$$

Weight of water in the cone ($V=\tfrac13\pi r^{2}h$):

$$V=\tfrac13(7.069)(6)=14.137\ \text{m}^3,\qquad F_2=9.81\times14.137=138.7\ \text{kN}$$ $$F_{net}=F_1+F_2=776.2+138.7=914.9\ \text{kN}$$

Answer: net hydrostatic force on the conical surface $\approx915$ kN, acting vertically downward along the axis. (If the free surface is 7 m above AC instead, use $p_{AC}=100+9.81\times7$ in the same method.)

Significance of metacentric height

The metacentric height $GM$ is the distance between the centre of gravity $G$ and the metacentre $M$ of a floating or submerged body. It is the measure of stability against overturning:

• $GM>0$ (M above G): stable; the righting moment is $W\,GM\sin\theta$ for a small heel $\theta$.
• $GM=0$: neutral equilibrium.
• $GM<0$: unstable; the body capsizes.

A large $GM$ means a stiff, very stable vessel with a short rolling period (uncomfortable); a small positive $GM$ gives a gentle roll but less safety margin. It is also used in the experiment $GM=\dfrac{wx}{W\tan\theta}$ and in ship design.

When do the centre of gravity and the centre of pressure coincide?

The centre of pressure is at depth

$$h_{cp}=\bar h+\frac{I_G\sin^{2}\theta}{A\bar h}$$

This equals $\bar h$ (the depth of the centroid) when the second term is zero, i.e.

• when the plane surface is horizontal ($\theta=0$), since the pressure is uniform over the surface; or
• (approximately) when the surface is at a very large depth, since $\dfrac{I_G}{A\bar h}\to0$ relative to $\bar h$.

The figure values are only partly legible, so the reading used is stated. Pressures are gauge, $g=9.81$.

Assumed reading: above the plate AB the fluids are, from the top, air at $p_1=500$ kPa, oil ($S=0.8$) 3 m thick, then water 2 m thick down to AB. Below AB is water with $p_2=-360$ kPa at the plate level. The cylinder ($D=0.3$ m) extends 0.2 m above and 0.2 m below the plate and ends in hemispheres of radius $r=0.15$ m.

      air  p1 = 500 kPa
      oil  S=0.8   3 m
      water        2 m
   ---+===[ cap ]===+--- plate AB
         | cyl 0.2m|
         [ lower  ]
      water, p2 = -360 kPa

Pressures at the bases of the two hemispheres (flat plane at the end of each cylinder portion):

$$p_{plate,top}=500+0.8(9.81)(3)+9.81(2)=543.16\ \text{kPa}$$ $$p_u=543.16-9.81(0.2)=541.20\ \text{kPa},\qquad p_l=-360+9.81(0.2)=-358.04\ \text{kPa}$$

Vertical forces (the cylinder side has no vertical component). Projected area $A=\pi(0.15)^{2}=0.07069\ \text{m}^2$, hemisphere volume $V_h=\tfrac23\pi(0.15)^{3}=0.007069\ \text{m}^3$.

• Downward force on the upper hemisphere: $F_u=p_uA-\gamma_wV_h$
• Upward force on the lower hemisphere: the pressure there is lower, giving a net downward effect

$$F_{net}=(p_u-p_l)A-(\gamma_w+\gamma_w)V_h$$ $$F_{net}=(541.20+358.04)(0.07069)-2(9.81)(0.007069)$$ $$F_{net}=63.57-0.14=63.4\ \text{kN}$$

Answer: the fluid exerts a net vertical force of about 63.4 kN downward on the cylinder, and this is the load carried by the weld to the plate AB. The result is dominated by $(p_u-p_l)A$; use the same method with the exact figure heights.

Data: sea-water $\rho=1025\ \text{kg/m}^3$; cone vertex down, top diameter 1.2 m, vertex angle $60^\circ$ (half-angle $30^\circ$); buoy mass 300 kg, $G_1$ at 0.75 m from the vertex; light mass 55 kg at height $y$ above the top.

      light at y
      ____
     \    /   top dia 1.2 m
      \  /
 water \/  vertex (down)

Floating depth $d$ (vertex to waterline). Total mass $M=355$ kg:

$$V=\frac{355}{1025}=0.3463\ \text{m}^3=\frac13\pi\,(d\tan30^\circ)^{2}d$$ $$d^{3}=\frac{3V}{\pi\tan^{2}30^\circ}=0.9922\ \Rightarrow\ d=0.9974\ \text{m}$$

(The buoy height is $0.6/\tan30^\circ=1.0392$ m, so it floats with the top just above water.)

Waterline radius: $r_w=d\tan30^\circ=0.5758$ m.

Centre of buoyancy: $KB=\tfrac34d=0.7480$ m from the vertex.

$$I=\frac{\pi r_w^{4}}{4}=0.0863\ \text{m}^4,\qquad BM=\frac{I}{V}=0.2493\ \text{m}$$ $$KM=KB+BM=0.7480+0.2493=0.9974\ \text{m}$$

Limit of stability: $KG\le KM$. Combined centre of gravity from the vertex:

$$KG=\frac{300(0.75)+55(1.0392+y)}{355}\le0.9974$$ $$225+55(1.0392+y)\le354.07\ \Rightarrow\ 1.0392+y\le2.3468$$ $$y\le1.308\ \text{m}$$

Answer: the centre of gravity of the light unit may be at most about 1.31 m above the top of the buoy (2.35 m from the vertex) for the assembly not to be unstable.

The figure is only partly legible. Assumed reading: a circular door of diameter 3 m (radius $R=1.5$ m, length 3 m along the slope), inclined at $\theta=60^\circ$ to the horizontal, with its upper edge 1.5 m below the water surface. The air above the water has gauge pressure $p_0=800$ kPa. $g=9.81$.

   air 800 kPa
  ---------------- water surface
   1.5 m
    \
     \  door 3 m along slope
      \   60 deg

Depth of centroid:

$$\bar h=1.5+R\sin60^\circ=1.5+1.5(0.866)=2.799\ \text{m}$$

Pressure at the centroid:

$$p_c=p_0+\gamma\bar h=800+9.81(2.799)=827.46\ \text{kPa}$$

Area: $A=\pi R^{2}=\pi(1.5)^{2}=7.069\ \text{m}^2$.

Force:

$$F=p_cA=827.46\times7.069=5849\ \text{kN}$$

Position. The air pressure is equivalent to an extra head $p_0/\gamma=81.55$ m of water above the surface. Distance of the centroid from this imaginary free surface along the slope:

$$\bar y=\frac{81.55+2.799}{\sin60^\circ}=97.40\ \text{m}$$ $$\bar y_{cp}-\bar y=\frac{I_G}{A\bar y}=\frac{\pi R^{4}/4}{\pi R^{2}\,\bar y}=\frac{R^{2}}{4\bar y}=\frac{2.25}{4(97.40)}=0.0058\ \text{m}$$

Answer: $F\approx5.85\times10^{3}$ kN (5849 kN), acting normal to the door at about 5.8 mm below the centre of the door, measured along the slope. The large air pressure makes the centre of pressure almost coincide with the centroid.

Reading of the figure: AB is the inclined side wall, vertical height 5 m and horizontal run 2 m, so its length is $L=\sqrt{5^{2}+2^{2}}=5.385$ m; width of tank $b=6$ m; A at the bottom, B at the top. The oil surface (level of B) is under a gauge pressure $p_0=19.62$ kPa. Oil $S=0.8$, $\gamma=0.8(9.81)=7.848$ kN/m³.

   B  p0 = 19.62 kPa
    \
     \   AB, L = 5.385 m
      \
   A   \ 5 m vertical

Equivalent oil head of the gauge pressure: $\dfrac{19.62}{7.848}=2.5$ m of oil above B.

Total pressure (centroid is 2.5 m below B):

$$A_{AB}=5.385\times6=32.31\ \text{m}^2$$ $$p_c=19.62+7.848(2.5)=39.24\ \text{kPa}$$ $$F=p_cA_{AB}=39.24\times32.31=1267.9\ \text{kN}$$

Centre of pressure. Distance of the centroid from the equivalent free surface along the slope, with $\sin\theta=5/5.385$:

$$\bar y=\frac{2.5+2.5}{\sin\theta}=5.385\ \text{m}$$ $$\bar y_{cp}-\bar y=\frac{L^{2}}{12\,\bar y}=\frac{29}{12(5.385)}=0.449\ \text{m}$$

The centre of pressure is $L/2+0.449=3.141$ m from B along AB (vertical depth below B $=3.141\sin\theta=2.917$ m).

Check with the pressure diagram. Pressures: at B, 19.62 kPa; at A, $19.62+7.848(5)=58.86$ kPa (trapezium):

$$F=\frac{19.62+58.86}{2}\times32.31=1267.9\ \text{kN}$$ $$x_{B}=\frac{L}{3}\cdot\frac{2(58.86)+19.62}{58.86+19.62}=\frac{5.385}{3}\times1.7500=3.141\ \text{m from B}$$

Answer: $F\approx1268$ kN, acting 3.14 m from B along the slope (2.92 m vertically below B).

Data: circular end, $D=2$ m ($R=1$ m). Upper half oil ($\rho_o=880$, $\gamma_o=8.633$ kN/m³), lower half water ($\gamma_w=9.81$ kN/m³); the top of the tank is at atmospheric pressure. Depths are measured from the top.

   _______ top, p = 0
  /  oil  \
 |---------|  interface at depth 1 m
  \ water /
   -------

For a semicircle of radius $R$: $A=\dfrac{\pi R^{2}}{2}=1.5708\ \text{m}^2$, centroid $\dfrac{4R}{3\pi}=0.4244$ m from the diameter, $I_G=\left(\dfrac{\pi}{8}-\dfrac{8}{9\pi}\right)R^{4}=0.1098\ \text{m}^4$.

Oil part (upper semicircle)

Centroid depth $\bar h_o=1-0.4244=0.5756$ m.

$$F_o=8.633(0.5756)(1.5708)=7.805\ \text{kN}$$ $$h_{cp,o}=0.5756+\frac{0.1098}{1.5708(0.5756)}=0.5756+0.1214=0.697\ \text{m (from top)}$$

Water part (lower semicircle)

Pressure at the interface: $p_i=8.633(1)=8.633$ kPa, equivalent to $0.880$ m of water. Centroid is 0.4244 m below the interface.

$$F_w=(8.633+9.81\times0.4244)(1.5708)=12.797\times1.5708=20.10\ \text{kN}$$ $$h_{cp,w}=1+0.4244+\frac{0.1098}{1.5708(0.880+0.4244)}=1.4244+0.0536=1.478\ \text{m (from top)}$$

Net force and position

$$F=7.805+20.100=27.91\ \text{kN}$$ $$\bar h=\frac{7.805(0.697)+20.100(1.478)}{27.91}=1.260\ \text{m below the top}$$

Answer: net hydrostatic force $\approx27.9$ kN, acting 1.26 m below the top of the tank (0.74 m above the bottom) on the vertical centre line.

Answer: the depth remains 4 m; the water level does not change.

Reasoning (Archimedes' principle). Take water density $\rho$. The floating ice displaces a volume $V_d=0.3\ \text{m}^3$ (the part below the surface). By equilibrium, the weight of the ice equals the weight of water displaced:

$$W_{ice}=\rho g V_d=\rho g(0.3)$$

So the ice has a mass of $0.3\rho$. When it melts, it turns into water of the same mass, with volume

$$V_{melt}=\frac{W_{ice}}{\rho g}=0.3\ \text{m}^3$$

The melted water occupies exactly the 0.3 m³ that the submerged part of the ice used to displace, so the total volume of liquid below the original surface is unchanged. The 0.2 m³ of ice above the surface was already part of the ice's weight and is included in the 0.3 m³ of melt water.

$$h_{new}=4\ \text{m}$$

This holds because the ice floats in water of the same kind (pure water). If fresh-water ice melted in denser sea water, the level would rise slightly.

The number of steps is not legible, so this reading is used. Assumed: water depth equals the dam height $H=3$ m, with 10 steps, each 0.3 m high and 0.3 m deep, and the dam is $w=3$ m wide. The water stands against the upstream stepped face.

 water ____
        |_|__          each step 0.3 x 0.3 m
        |   |_|__
        |       |_|__

The horizontal force acts on the vertical projection and does not matter here. The vertical force equals the weight of water above the stepped surface up to the free surface. Only the horizontal treads carry vertical pressure.

Tread $k$ ($k=1,\ldots,10$) lies at depth $0.3k$ m below the surface, has horizontal area $0.3\times3=0.9\ \text{m}^2$, and carries pressure $\gamma(0.3k)$:

$$F_V=\gamma\,w\,(0.3)\sum_{k=1}^{10}0.3k=9810(3)(0.3)(0.3)(55)$$ $$F_V=9810\times3\times4.95=145.7\ \text{kN}$$

(Equivalently, the volume of water above the steps is $3\times4.95=14.85\ \text{m}^3$, so $F_V=9.81\times14.85=145.7$ kN.)

Answer: vertical force on the dam $\approx146$ kN downward (for 3 m depth and 10 steps). For $n$ steps, $F_V=\gamma w(0.3)^{2}\dfrac{n(n+1)}{2}$.

Data: cube side 0.5 m, oil $\rho=0.88\times1000=880\ \text{kg/m}^3$, slope $\tan\theta=\tfrac14$ ($\theta=14.04^\circ$, $\sin\theta=0.2425$, $\cos\theta=0.9701$), acceleration $a=5\ \text{m/s}^2$ up the slope.

Use axes fixed to the tank: $x$ up the slope, $y$ perpendicular to the base. The effective gravity (gravity minus acceleration) has components:

• $x$: $-(g\sin\theta+a)=-(2.379+5)=-7.379\ \text{m/s}^2$ (towards the rear)
• $y$: $-g\cos\theta=-9.517\ \text{m/s}^2$

         rear        front (up slope)
   rim -> |  \          |
          |    \        |
          |______\______|
        free surface slopes up to the rear

The free surface is perpendicular to this effective gravity, and its slope is

$$\tan\phi=\frac{7.379}{9.517}=0.7754$$

The surface rises towards the rear, so oil spills over the rear rim until the surface passes through that rim.

(a) Volume left

Depth at the rear wall $=0.5$ m. Depth at the front wall $=0.5-0.5(0.7754)=0.1123$ m (positive, so the surface does not reach the base).

$$V=\frac{0.5+0.1123}{2}\times0.5\times0.5=0.0765\ \text{m}^3$$

Oil left $=0.0765\ \text{m}^3$ (76.5 litres), i.e. 61 % of the 125 litres.

(b) Pressure at the lowest corners

The pressure gradient normal to the base is $\rho\,g\cos\theta$, so at the base:

$$p_{rear}=880(9.517)(0.5)=4187\ \text{Pa}$$ $$p_{front}=880(9.517)(0.1123)=941\ \text{Pa}$$

Answer: (a) 0.0765 m³ of oil remains; (b) gauge pressure at the rear bottom corners 4.19 kPa and at the front bottom corners 0.94 kPa.

The resultant force on a curved surface has a horizontal component (force on the vertical projection) and a vertical component (weight of the liquid vertically above the surface up to the equivalent free surface). The figure is not fully readable, so the geometry below is assumed.

Assumed geometry (water on the concave side, length $L = 10$ m, gauge at level of B reads $-19$ kN/m²):

   B |  gauge level, p = -19 kPa
     | BC = 2 m (vertical wall)
   C |
      \  CD, DE = two quarter circles,
       ) D  R = 2 m (semicircle C-D-E)
      /
   E |________ F   EF = 3 m (horizontal)

Vertical extent of B to E = 2 + 2 + 2 = 6 m, and EF lies at 6 m below B.

Pressure at B and equivalent free surface

$$\frac{p_B}{\gamma} = \frac{-19}{9.81} = -1.937\ \text{m}$$

The equivalent free surface is 1.937 m below B. Pressure at depth $z$ below B is $p = -19 + 9.81z$ kPa.

Horizontal component (vertical projection BE, height 6 m):

$$\begin{aligned} F_H &= L\int_0^6 (-19 + 9.81z)\,dz = L\left(-19\times 6 + 9.81\times\frac{6^2}{2}\right)\\ &= 10(-114 + 176.58) = 625.8\ \text{kN} \end{aligned}$$

Vertical component

• On the semicircle CDE: the upper quarter is pushed up and the lower quarter is pushed down by the liquid. The net force equals the weight of liquid in the half-disc enclosed between the two arcs:

$$F_{V1} = \gamma\left(\frac{\pi R^2}{2}\right)L = 9.81\times 6.283\times 10 = 616.4\ \text{kN (downward)}$$

• On the flat EF at 6 m depth: $p = -19 + 9.81\times 6 = 39.86$ kPa

$$F_{V2} = 39.86\times 3\times 10 = 1195.8\ \text{kN (downward)}$$ $$F_V = 616.4 + 1195.8 = 1812.2\ \text{kN}$$

Resultant

$$\begin{aligned} F &= \sqrt{625.8^2 + 1812.2^2} = 1917.2\ \text{kN}\\ \theta &= \tan^{-1}\frac{1812.2}{625.8} = 70.9^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 1917$ kN (about 1.92 MN), inclined at $70.9^\circ$ to the horizontal, with $F_H = 625.8$ kN and $F_V = 1812.2$ kN. If the actual figure differs, use the same method: $F_H = \int p\,dA_{proj}$ and $F_V$ = weight of liquid up to the equivalent free surface.

Hydrometer

A hydrometer is a floating instrument used to find the specific gravity (or density) of a liquid directly. It is a sealed glass tube with a weighted bulb at the bottom (so it floats vertically) and a graduated stem at the top.

Principle: it floats in equilibrium, so weight = buoyant force. Its weight $W$ is constant, so a denser liquid needs less displaced volume and the hydrometer floats higher; a lighter liquid makes it sink deeper.

$$W = \rho g V_{sub}\quad\Rightarrow\quad \rho = \frac{W}{g V_{sub}}$$

If the stem has area $a$ and sinks by an extra length $x$, then $V_{sub} = V_0 + a x$. The scale is calibrated in specific gravity, which is read at the liquid surface. The scale is not uniform: higher density is at the top, and the divisions are closer at the bottom.

Uses: testing milk, acid, battery electrolyte, alcohol, sugar solutions and sea water.

      |  <- stem with scale (reads S)
 ~~~~~|~~~~~ liquid surface
      |
    (   )  <- bulb
    (___)  <- lead shots (ballast)

Stability of floating bodies

A body floats in equilibrium when weight = buoyant force and the centre of gravity $G$ and the centre of buoyancy $B$ lie on the same vertical line. For stability, a small tilt produces a restoring couple. The metacentre M is the point where the line of action of buoyancy through the new centre of buoyancy cuts the original vertical axis.

$$GM = \frac{I}{V} - BG$$

where $I$ is the second moment of the waterline area about the longitudinal axis, $V$ is the volume of liquid displaced and $BG$ is the distance between $B$ and $G$ ($G$ above $B$). The restoring moment is $W\cdot GM\sin\theta$. A body with a large metacentric height is more stable, but rolls quickly.

The horizontal component of the force equals the force on the vertical projection of the surface; the vertical component equals the weight of liquid vertically above the surface up to the free surface.

Assumed geometry (the figure is not fully clear): A is at the free surface, AB is a vertical wall 4 m high, BC is a quarter circle of radius 2 m (concave upward), and CDE is a horizontal bed 4 m long at 6 m depth. $S = 1.1$, $L = 20$ m, $\gamma = 9.81\times 1.1 = 10.79$ kN/m³.

 A |~~~~~~~~~~~~~~~~~~~~~~  free surface
   |
   | 4 m
 B |
    \   R = 2 m
     '-.
 C ------ D ------ E   (depth 6 m)
   |<---- 4 m ---->|

Horizontal component (vertical projection = 6 m, free surface to the bed):

$$\begin{aligned} F_H &= \gamma\,\frac{h^2}{2}\,L = 10.79\times\frac{6^2}{2}\times 20\\ &= 3884.8\ \text{kN} \end{aligned}$$

Vertical component (area of liquid above the surface, per metre length):

• Above arc BC: the depth of the arc below the free surface is $4 + \sqrt{R^2-(x-2)^2}$, so the area is a $2\times 4$ rectangle plus the quarter disc of area $\pi R^2/4 = \pi$: $8 + \pi = 11.14$ m²
• Above bed CE: $4\times 6 = 24$ m²

$$\begin{aligned} A_{total} &= 8 + \pi + 24 = 35.14\ \text{m}^2\\ F_V &= \gamma A_{total} L = 10.79\times 35.14\times 20 = 7584.3\ \text{kN} \end{aligned}$$

Resultant

$$\begin{aligned} F &= \sqrt{3884.8^2 + 7584.3^2} = 8521.3\ \text{kN}\\ \theta &= \tan^{-1}\frac{7584.3}{3884.8} = 62.9^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 8521$ kN (8.52 MN), at $62.9^\circ$ to the horizontal; $F_H = 3884.8$ kN, $F_V = 7584.3$ kN (downward).

Resultant force on a curved surface = $\sqrt{F_H^2 + F_V^2}$, where $F_H$ is the force on the vertical projection and $F_V$ is the weight of the liquid vertically above the surface up to the free surface.

Assumed geometry (the figure is not fully clear): A is at the free surface; AB is a vertical wall of 4 m; the surface then dips as a semicircular trough of radius 2 m from B to E through the lowest point D, so D is 6 m below the free surface. $S = 1.25$, $L = 10$ m, $\gamma = 1.25\times 9.81 = 12.26$ kN/m³.

 A |~~~~~~~~~~~~~~~~~~~~~~~~ free surface
   | 4 m
 B |                     E
    \         R = 2 m   /
      '--.   C   .--'   
           ' D '          D lowest (6 m)

Horizontal component

On the trough, the forces on the left and right quarters are equal and opposite horizontally, so they cancel. Only the wall AB (depth 4 m) gives a net horizontal force:

$$F_H = \gamma\frac{h^2}{2}L = 12.26\times\frac{4^2}{2}\times 10 = 981.0\ \text{kN}$$

Vertical component

Volume above the trough = rectangle of 4 m depth over the 4 m width + half-disc of radius 2 m:

$$\begin{aligned} A &= 4\times 4 + \frac{\pi\times 2^2}{2} = 16 + 6.283 = 22.28\ \text{m}^2\\ F_V &= \gamma A L = 12.26\times 22.28\times 10 = 2732.5\ \text{kN (downward)} \end{aligned}$$

Resultant

$$\begin{aligned} F &= \sqrt{981.0^2 + 2732.5^2} = 2903.2\ \text{kN}\\ \theta &= \tan^{-1}\frac{2732.5}{981.0} = 70.3^\circ \text{ below the horizontal} \end{aligned}$$

Answer: $F \approx 2903$ kN, inclined at $70.3^\circ$ to the horizontal; $F_H = 981$ kN, $F_V = 2732.5$ kN.