Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap8 Tension, shear and bending moment 10 Dynamics in 1D

Chapter 9 Hydrostatics

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Hydrostatics concerns the equivalent force and moment due to distributed pressure on a surface from a still fluid. Pressure increases with depth. With constant pressure, the equivalent force has magnitude = pressure times area, acting at the centroid. For linearly-varying pressure on a rectangular plate the equivalent force is the average pressure times the area acting somewhere between 1/2 and 2/3 of the way down. The net force acting on a totally submerged object in a constant density fluid is the displaced fluid’s weight, acting at the centroid.

Hydrostatics is primarily concerned with finding the net force and moment of still fluid on a surface. The surfaces are typically the sides of a pool, dam, container, or pipe, or the outer surfaces of a floating object such as a boat or of a submerged object like a toilet bowl float. Finally, one is sometimes concerned with the force on an imagined surface that separates some water of interest from the other water. Although the hydrostatics of air helps explain the floating of hot air balloons, dirigibles, and chimney smoke; and the hydrostatics of oil is important for hydraulics (hydraulic brakes for example), often the fluid of concern for engineers is water. So, as in the title of the chapter (‘hydro’), we often use the word ‘water’ as an informal synonym for ‘fluid.’

Besides the utility of the subject in applications, hydrostatics is also a good introduction to distributed forces and continuum mechanics.

9.1 Fluid pressure

Besides the basic laws of mechanics that you already know, elementary hydrostatics is based on the following two constitutive assumptions (see page 0.1):

1)

The force of water on a surface is perpendicular to the surface; and

2)

The density of water, $\rho$ (pronounced ‘row’) is a constant (doesn’t vary with depth or pressure),

Sometimes we use the weight density $\gamma=g\rho$ (pronounced ‘gammuh equals gee row’), the weight per unit volume . The first assumption, that all static water forces are perpendicular to surfaces on which they act, can be restated:

Still water cannot carry any shear stress.

For near-still water, this constitutive assumption is abnormally accurate (compared to most constitutive assumptions for materials), approximately as good as the laws of mechanics.

The assumption of constant density is called incompressibility because it corresponds to the idea that water does not change its volume (compress) much under pressure. This assumption is reasonable for most purposes. At the bottom of the deepest oceans, for example, the extreme pressure (about 800 atmospheres) causes water to increase its density only about 4% from that of water at the surface

^†^†margin: That fluid density does depend on salinity, temperature and pressure is sometimes important in hydrostatics, in particular for determining which water floats on which other water. This density variation is important in the ecology of lakes, the effects of the oceans on climate, and in air for the stability of the atmosphere, and the mechanics of fireplace chimneys.

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Figure 9.1: A bit of area

\Delta A

on a surface on which pressure

p

acts. The outward (into the water) normal of the surface is

\hat{\boldsymbol{n}}

so the force is

\Delta{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-p\hat{\boldsymbol{n}}\Delta A

We also assume that the direction and magnitude of the local gravitational ‘constant’ is, well, constant. This assumption becomes inaccurate when considering, say, the hydrostatics of whole oceans (the direction of the gravity force changes as you go around the world, this helps keep the Australians in place). When considering the upper atmosphere one might consider that the magnitude of the gravity decays with distance from the center of the earth. And it’s more subtle than that if you care about fine measurement (see page 4.1). However, for most purposes taking gravity as constant in magnitude and direction is good enough.

Surface area $A$ , outward normal $\hat{\boldsymbol{n}}$ , pressure $p$ , and force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$

We will generalize the high-school physics fact

\mbox{force}=\mbox{pressure}\times\mbox{area}

to take account that force is a vector, that pressure varies with position, and that not all surfaces are flat. So we need a clear notation and sign convention. The area of a surface is $A$ which we can think of as being the sum of the bits of area $\Delta A$ that compose it:

A=\sum\Delta A=\int dA.

Every bit of surface area has an outer normal $\hat{\boldsymbol{n}}$ that points from the surface out into the fluid. The (scalar) force per unit area on the surface is called the pressure $p$ , so that the force on a small bit of surface is

\Delta{\accentset{\rightharpoonup}{\boldsymbol{F}}}=p\,(-\hat{\boldsymbol{n}})% \,(\Delta A)

pointing into the surface, assuming positive pressure, and with magnitude proportional to both pressure and area. Thus the total force and moment due to pressure forces on a surface :

\begin{array}[]{cccclc}{\accentset{\rightharpoonup}{\boldsymbol{F}}}&=&\int d{% \accentset{\rightharpoonup}{\boldsymbol{F}}}&=&-\int_{A}p\,\hat{\boldsymbol{n}% }\ \ dA\\ \\ {\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}&=&\int_{A}d{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/C}}&=&-\int_{A}{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/C}}\times(p\,\hat{% \boldsymbol{n}})\ \ dA\end{array}

(9.1)

Hydrostatics is the evaluation of the (intimidating-at-first-glance) integrals 9.1 and their role in equilibrium equations. In the rest of this section we consider various important special cases.

Water in equilibrium with itself

Before we worry about how water pushes on other things, let’s first understand what it means for water to be in static equilibrium. These first important facts about hydrostatics follow from drawing free-body diagrams of various chunks of water and assuming static equilibrium (see box 9.1 on page 9.1).

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Figure 9.2: A small prism of water in equilibrium is isolated from other water. The free-body diagram does not show the forces in the

z

direction. Force balance applied to this free-body diagram shows that

p_{x}=p_{y}=p

, pressure is the same in all directions.

1.

Pressure is the same in every direction, $p_{x}=p_{y}=p$ .
2.

Pressure doesn’t vary with side to side position, $p(x,y,z)=p(y)$ .
3.

Pressure varies linearly with depth, $p=\rho gh=\gamma h$ .

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Figure 9.3: Free-body diagrams of aligned boxes of water cut out of a bigger body of water. a) a horizontally aligned box, b) a vertically aligned box. (To avoid visual clutter the self-cancelling horizontal forces are not shown on the second free-body diagram.) Force balance applied to these free-body diagrams shows that

p=p(y)

The buoyant force of water on water.

Imagine a chunk of water the shape of a sea monster. Where? In a place under water in a calm swimming pool where there is nothing but water. Now draw a free-body diagram of that water. Because your sea monster is in equilibrium, force balance and moment balance must apply. The only forces are the complicated distribution of pressure forces and the weight of water. The pressure forces must exactly cancel the weight of the water and, to satisfy moment balance, must pass through the center-of-mass of the water monster. So, in static equilibrium:

The pressure forces acting on a surface enclosing a volume of water are equivalent to the negative weight passing through the center-of-mass of the water.

Box 9.1 Adding forces to derive Archimedes’ principle

(One can do most hydrostatics calculations, say typical homework problems, without being able to reproduce the derivations here.)

Archimedes’ principle follows from adding up all the pressure forces on the outer surfaces of an arbitrarily shaped, submerged solid, say something potato-shaped.

First we find the answer by cutting the potato into french fries. This approach is effectively a derivation of a theorem in vector calculus. After that, for those who have the appropriate math background, we quote the vector calculus directly.

First cut the potato into horizontal french-fries (horizontal prisms) and look at the forces on the end caps (there are no water forces on the sides since those are inside the potato).

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The pressure on two ends is the same (because they have the same water depth). The areas on the two ends are probably different because your potato is probably not box shaped. But the area is bigger at one end if the normal to the surface is more oblique compared to the axis of the prism. If the cross sectional area of the prism is $\Delta A_{0}$ then the area of one of the prism caps is

\par\Delta A=\Delta A_{0}/(\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\lambda}})\par

where $\hat{\boldsymbol{\lambda}}$ is along the axis of the prism and $\hat{\boldsymbol{n}}$ is the outer unit normal to the end cap (Note $\Delta A\geq\Delta A_{0}$ because $\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\lambda}}\leq 1$ ).

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So the net force on the cap is $-p\Delta A_{0}\hat{\boldsymbol{n}}/(\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{% \lambda}})$ . The component of the force along the prism is $\left[-p\Delta A_{0}\hat{\boldsymbol{n}}/(\hat{\boldsymbol{n}}\cdot\hat{% \boldsymbol{\lambda}})\right]\cdot\hat{\boldsymbol{\lambda}}$ which is $-p\Delta A_{0}$ . An identical calculation at the other end of the french fry gives minus the same answer. So the net force of the water pressure along the prism is zero for this and every prism and thus the whole potato. Likewise for prisms with any horizontal orientation. Thus the net sideways force of water on any submerged object is zero.

To find the net vertical force on the potato we cut it into vertical french fries. The net forces on the end caps are calculated just as in the above paragraph but taking account that the pressure on the bottom of the french fry is bigger than at the top. The sum of the forces of the top and bottom caps is an upwards force that is

\begin{array}[]{ccccc}\mbox{net upwards force on vertical french fry}&=&\Delta p% \Delta A_{0}\\ &=&(\gamma h)\Delta A_{0}\\ &=&\gamma(h\Delta A_{0})\\ &=&\gamma\,\Delta V_{0}\\ \end{array}

where $\Delta V_{0}$ is the volume of the french fry. Adding up over all the french fries that make up the potato one gets that the net upwards force is $\gamma V$ .

The net result, summarized by the figure below, is that the resultant of the pressure forces on a submerged solid is an upwards force whose magnitude is the weight of the displaced water. The location of the force is the centroid of the displaced volume. (Note that the centroid of the displaced volume is not necessarily at the center of mass of the submerged object.)

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A vector calculus derivation

Here is a derivation of Archimedes’ principle, at least the net force part, using multi-variable integral calculus. Only read on if you have taken a math class that covers the divergence theorem. The net pressure force on a submerged object is

\begin{array}[]{cclcccl}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{buoyancy}}&=&-\int_{A}&p&\hat{\boldsymbol{n}}&dA&\\ &=&-\int_{S}&p&\hat{\boldsymbol{n}}&dS&\\ &=&-\int_{S}&(H-z)\gamma&\hat{\boldsymbol{n}}&dS&\\ &=&-\int_{V}&\quad${{\accentset{\rightharpoonup}{\nabla}}\left((H-z) \gamma% \right)}$&dV&\\ &=&-\int_{V}&\qquad$(-\hat{\boldsymbol{k}}) \, \gamma$&dV&\\ &=&\phantom{-}\int_{V}&\qquad\qquad$ \gamma$&dV&\hat{\boldsymbol{k}}\\ &=&{\quad(\mbox{weight of displaced water})}&\hat{\boldsymbol{k}}.\\ \\ \end{array}

In this derivation we first changed from calling bits of surface area $d A$ to $d S$ because that is a common notation in calculus books. The depth from the surface, of a point with vertical component $z$ from the bottom, is $H-z$ . The ${\accentset{\rightharpoonup}{\nabla}}$ symbol indicates the gradient and its place in this equation is from the divergence theorem:

\int_{S}(\mbox{any scalar})\hat{\boldsymbol{n}}\ dS=\int_{V}{\accentset{% \rightharpoonup}{\nabla}}(\mbox{the same scalar})\ dV.

The gradient of $(H-z)\gamma$ is $-\hat{\boldsymbol{k}}\gamma$ because $H$ and $\gamma$ are constants. Note, where we write $\int_{S}$ some books would write $\int\!\!\!\int_{S}$ , and where we write $\int_{V}$ some books would write $\int\!\!\!\int\!\!\!\int_{V}$ .

Box 9.2 Pressure doesn’t depend on direction or horizontal position and increases linearly with depth

We assume that the pressure $p$ does not vary too wildly from point to point, thus if we look at a small enough region we can think of the pressure as constant in that region. If we draw a free-body diagram of a little triangular prism of water the net forces on the prism must add to zero (see fig. 9.2 on page 9.2). For each surface the magnitude of the force is the pressure times the area of the surface and the direction is minus the outward normal of the surface. We assume, for the time being, that the pressure is different on the differently oriented surfaces. So, for example, because the area of the left surface is $a\cos\theta w$ and the pressure on the surface is $p_{x}$ , the net force is $a\cos\theta wp_{x}\hat{\boldsymbol{\imath}}$ . Calculating similarly for the other surfaces:

\begin{array}[]{lclccc}{\accentset{\rightharpoonup}{\boldsymbol{0}}}&=&\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\\ &=&\underbrace{(a\cos\theta)w\,p_{x}\,\hat{\boldsymbol{\imath}}+(a\sin\theta)w% \,p_{y}\,\hat{\boldsymbol{\jmath}}-a\,w\,p\,\hat{\boldsymbol{n}}}_{\mbox{% pressure terms}}\\[25.0pt] &&\qquad\qquad\qquad-\ \underbrace{{a^{2}\cos\theta\sin\theta w\over 2}\rho g}% _{\mbox{weight}}\hat{\boldsymbol{\jmath}}\\ &=&aw\left(\cos\theta\,p_{x}\,\hat{\boldsymbol{\imath}}+\sin\theta\,p_{y}\,% \hat{\boldsymbol{\jmath}}-p\,(\underbrace{\cos\theta\hat{\boldsymbol{\imath}}+% \sin\theta\hat{\boldsymbol{\jmath}}}_{\hat{\boldsymbol{n}}})\right.\\[5.0pt] &&\qquad\left.\phantom{(\underbrace{\cos\theta\hat{\boldsymbol{\imath}}+\sin% \theta\hat{\boldsymbol{\jmath}}}_{\hat{\boldsymbol{n}}})}-{a\cos\theta\sin% \theta\over 2}\rho g\hat{\boldsymbol{\jmath}}\right).\end{array}

If $a$ is arbitrarily small, the weight term drops out compared to the pressure terms. Dividing through by $a w$ we get

{\accentset{\rightharpoonup}{\boldsymbol{0}}}=\cos\theta p_{x}\hat{\boldsymbol% {\imath}}+\sin\theta\,p_{y}\,\hat{\boldsymbol{\jmath}}-p\,(\cos\theta\hat{% \boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}}).

Taking the dot product of both sides of this equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ gives that

p=p_{x}=p_{y}.

Since $\theta$ could be anything, force balance for the free-body diagram of a small prism tells us that for a fluid in static equilibrium

pressure is the same in every direction.

[Other free-body diagrams can be used. That pressure has to be the same in any pair of directions could also be found by drawing a prism with a cross section which is an isosceles triangle. The prism is oriented so that two surfaces of the prism have equal area and have the desired orientations. Force balance along the base of the triangle gives that the pressures on the equal area surfaces are equal. The argument that pressure must not depend on direction in 3D is generally based on equilibrium of a small tetrahedron.]

Pressure doesn’t vary with side-to-side position

Consider the equilibrium of a horizontally aligned box of water cut out of a bigger body of water (fig. 9.3a on page 9.3). The forces on the end caps at A and B are the only forces along the box. Therefore they must cancel. Since the areas at the two ends are the same, the pressure must be also. This box could be anywhere and at any length and any horizontal orientation. Thus for a fluid in static equilibrium

pressure doesn’t depend on horizontal position.

If we take the $\hat{\boldsymbol{\jmath}}$ or $y$ direction to be up (fig. 9.3 on page 9.3), then we have

p(x,y,z)=p(y).

Pressure increases linearly with depth Consider the vertically aligned box of fig. 9.3b.

\begin{array}[]{lcrclc}\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}% }_{\!\!\mathrm{i}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot% \hat{\boldsymbol{\jmath}}&\Rightarrow&\underbrace{p(y)a^{2}-p(y+h)a^{2}}_{% \mbox{pressure terms}}-\underbrace{\rho ga^{2}h}_{\mbox{weight}}&=&0\\ &\Rightarrow&p_{\rm bottom}-p_{\rm top}=\rho gh.\\ \end{array}

So the pressure increases linearly with depth. If the top of a lake, say, is at atmospheric pressure $p_{a}$ then we have that

p=p_{a}+\rho gh=p_{a}+\gamma h=p_{a}+(H-y)\gamma

where $h$ is the distance down from the surface, $H$ is the depth to some reference point underwater and $y$ is the distance up from that reference point (so that $h=H-y$ ). Neglecting atmospheric pressure at the top surface, we have the useful and easy to remember formula:

p=\gamma h.

(9.2)

Because the pressure at equal depths must be equal and because the pressure at the top surface must be equal to atmospheric pressure, the top surface must be flat and level. Thus waves and the like are a definite sign of static disequilibrium as are any bumps on the water surface even if they don’t seem to move (as for a bump in the water where a stream goes steadily over a rock).

The force of water on submerged and floating objects

The net pressure force and moment on a still object surrounded by still water can be found by a clever argument credited to Archimedes. The pressure at any one point on the outside of the object does not depend on what’s inside. The pressure is determined by how far the point of interest is below the surface by eqn. 9.2^†^†margin: If there is no column of water from the point up to the surface it is still true that the pressure is $\gamma h$ , as you can figure out by tracking the pressure changes along on a staircase-like path from the surface to that point. . So if you can find the resultant force on any object that is the shape of the submerged object, but replacing the submerged object, it tells you what you want to know.

The clever idea is to replace your object with water. In this new system the water is in equilibrium, so the pressure forces exactly balance the weight. We thus obtain Archimedes’ Principle:

The resultant of all pressure forces on a totally submerged object is an upwards force with the same magnitude as the weight of the displaced water. The resultant acts at the centroid of the displaced volume:

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\mbox{buoyancy}}=\gamma V\hat{% \boldsymbol{\jmath}}\qquad\mbox{acting at}\qquad{\accentset{\rightharpoonup}{% \boldsymbol{r}}}=\frac{\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{/0}}\ dV}{V}.

The result can also be found by adding the effects of all the pressure forces on the outside surface (see box 9.1 on page 9.1).

For floating objects, the same argument can be carried out, but since the replaced fluid has to be in equilibrium we cannot replace the whole object with fluid, but only the part which is below the level of the water surface.

Displaced fluid

Sometimes people discuss Archimedes’ principle in terms of the displaced fluid. A floating object in equilibrium displaces an amount of fluid with the same weight as the object; this is also the amount of volume of the floating object that is below the water level. On the other hand an object that is totally under water, for whatever reason (it is resting on the bottom, or it is being held underwater by a string, etc), displaces as much fluid as the space it occupies. Putting these two ideas together one can remember that

A floating object displaces its weight, a submerged object displaces its volume.

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Figure 9.4: The resultant force from a constant pressure

p

on a flat plate is

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-pA\hat{\boldsymbol{n}}

acting at the centroid of the plate.

The force of constant pressure on a totally immersed object

When there is no gravity, or gravity is neglected, the pressure in a static fluid is the same everywhere. Exactly the same argument we have just used shows that the resultant of the pressure forces is zero. We could derive this result just by setting $\gamma=0$ in the formulas above.

That is, it is not pressure that causes buoyancy, but the gravity-induced gradient of pressure. If there is no gravity there is no buoyancy.

The force of constant pressure on a flat surface

The net force of constant pressure on a flat surface (not all the way around a submerged volume) is the pressure times the area acting normal to the surface at the centroid of the surface:

\begin{array}[]{ccc}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm% {net}}&=&\int_{A}-p\,\hat{\boldsymbol{n}}\ dA\\ &=&-pA\hat{\boldsymbol{n}}.\\ \end{array}

That this force acts at the centroid can be checked by calculating the moment of the pressure forces relative to the centroid C,

\begin{array}[]{ccl}{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/C,% net}}&=&\int_{A}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/C}}% \times\left(-p\,\hat{\boldsymbol{n}}\ dA\right)\\ &=&\underbrace{\left(\int_{A}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{/C}}dA\right)}_{0}\times\left(-p\,\hat{\boldsymbol{n}}\right)\\ &=&0\ \ ,\\ \end{array}

where the zero is because the position of the center-of-mass, relative to the center-of-mass, is zero.

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Figure 9.5: The resultant force from a constant depth-increasing pressure on a rectangular plate.

The force of water on a rectangular plate

Consider a rectangular plate with width (into the page) $w$ and length $\ell$ . There is water on one side and nothing on the other. Assume the water-side normal to the plate is $\hat{\boldsymbol{n}}$ and that the top edge of the plate is horizontal. Take $\hat{\boldsymbol{\jmath}}$ to be the up direction with $y$ being distance up from the bottom and the total depth of the water is $H$ . Thus the area of the plate is $A=\ell w$ . If the bottom and top of the plate are at $y_{1}$ and $y_{2}$ the net force on the plate can be found as:

\begin{array}[]{cclc}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{net}}&=&-\int_{A}p\hat{\boldsymbol{n}}\ dA\\ &=&-\int_{A}\gamma(H-y)\hat{\boldsymbol{n}}\ dA\\ &=&-w\int_{0}^{\ell}\gamma(H-y(s))\hat{\boldsymbol{n}}\ ds\\ &=&-w\int_{0}^{\ell}\gamma\left(H-(y_{1}+\hat{\boldsymbol{n}}\cdot\hat{% \boldsymbol{\jmath}}\ s)\right)\hat{\boldsymbol{n}}\ ds\\ &=&-w\gamma\left(H\ell-y_{1}\ell-\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{% \jmath}}\ \ell^{2}/2\right)\hat{\boldsymbol{n}}\\ &=&-w\ell\gamma\left(H-(y_{1}+\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\jmath% }}\ \ell/2)\right)\hat{\boldsymbol{n}}\\ &=&-w\ell\gamma\left(H-(y_{1}+(y_{2}-y_{1})/2)\right)\hat{\boldsymbol{n}}\\ &=&-w\ell\left(\gamma(H-y_{1})/2+\gamma(H-y_{2})/2)\right)\hat{\boldsymbol{n}}% \end{array}

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle-w\ell\frac{p_{1}+p_{2}}{2}\ \hat{\boldsymbol{n}}.$
		$\displaystyle=$	$\displaystyle-(\mbox{area})(\mbox{average pressure})(\mbox{outwards normal % direction}).$

The net water force is the same as that of the average pressure acting on the whole surface. To find where it acts, it is easiest to think of the pressure distribution as the sum of two different pressure distributions. One is a constant over the plate at the pressure of the top of the plate. The other varies linearly from zero at the top to $\gamma(y_{2}-y_{1})$ at the bottom.

p=\gamma(H-y)=\underbrace{\gamma(H-y_{2})}_{\text{Constant pressure, the % pressure at the top edge.}}+\underbrace{\gamma(y_{2}-y)}_{\text{Varies linearly from 0 at % the top to $\gamma(y_{2}-y_{1})$ at the bottom.}}

The first corresponds to a force of $w\ell\gamma(H-y_{2})$ acting at the middle of the plate. The second corresponds to a force of $w\ell\gamma\frac{y_{2}-y_{1}}{2}$ acting a third of the way up from the bottom of the plate.

SAMPLE 9.1

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Figure 9.6:

A uniform solid cylinder of mass $m=12\,\mathrm{kg}$ , diameter $d=0.1\,\mathrm{m}$ and height $h=2\,\mathrm{m}$ floats in water (density $\rho=1000\,\mathrm{kg}/{\,\mathrm{m}}^{3}$ ).

1.

Assuming the cylinder floats vertically, find the submerged height of the cylinder.
2.

If the cylinder floats longitudinally (its longitudinal axis parallel to the water surface), what will be the submerged section of the cylinder?

Solution

Cylinder floating vertically:

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Figure 9.7:

Let $h_{s}$ be the submerged height of the cylinder and $r=d/2$ be its radius. Then the force of buoyancy $F_{B}$ is equal to the weight of water replaced by the submerged volume of the cylinder. Thus,

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}=\overbrace{\pi r% ^{2}h_{s}}^{\rm volume}\gamma\,\hat{\boldsymbol{\jmath}}.

From the force balance on the cylinder (see the free-body diagram in fig. 9.7),

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}-mg% \hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}(\pi r^{2}h_{s}\gamma-mg% )\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}h_{s}$	$\displaystyle=$	$\displaystyle\frac{mg}{\pi r^{2}\gamma}=\frac{m}{\pi r^{2}\rho}$
	$\displaystyle=$	$\displaystyle\frac{12\,\mathrm{kg}}{\pi\cdot(0.05\,\mathrm{m})^{2}\cdot 1000\,% \mathrm{kg}/{\,\mathrm{m}}^{3}}=1.53\,\mathrm{m}.$

Answer: $h_{s}=1.53\,\mathrm{m}$

2.

Cylinder floating horizontally: No matter how the cylinder floats, the force of buoyancy has to equal the weight of the cylinder. This force is equal to the weight of the displaced water. Thus, the volume of displaced water has to be the same no matter what the orientation of the cylinder is with respect to the water surface. Therefore, the submerged volume of the cylinder while floating longitudinally must equal the volume submerged while floating vertically. That is (see fig. 9.8),

$\mbox{area of BCD}\cdot h=\pi r^{2}h_{s}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}\mbox{area of BCD}=\pi r^{2}(h_{s}/h)=0.006\,\mathrm{m}^{2}.$

From the mass and volume we can calculate that the density of the cylinder is .764 that of water. Now we can figure out what $d_{s}$ should be so that the submerged cross-sectional area is 76% of the total cross-sectional area. This is an exercise in geometry. Since, $\mbox{area of BCD}=\pi r^{2}-\mbox{area of ABD}$ , $\mbox{area of ABD}=\pi r^{2}-\mbox{area of BCD}=0.018\,\mathrm{m}^{2}$ .

But the area of ABD is the area of the circular sector OBAD ( $r^{2}\theta$ ) minus the area of triangle OBD ( $\frac{1}{2}\cdot r\cos\theta\cdot 2r\sin\theta$ ). Thus,

$\displaystyle\mbox{area of ABD}\equiv r^{2}\theta-\frac{1}{2}r^{2}\sin 2\theta$ $\displaystyle=$ $\displaystyle 0.018\,\mathrm{m}^{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta-\frac{1}{2}\sin 2% \theta-0.738$ $\displaystyle=$ $\displaystyle 0.$

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Figure 9.8:

We need to solve this nonlinear equation. Using trial and error or root finding on a computer or a graphical method, we find $\theta=1.126\ {\rm rad}=64.5^{\circ}$ . Using this value, we get, $d_{s}=r+r\cos\theta=0.07\,\mathrm{m}$ .

Answer: $d_{s}=0.07\,\mathrm{m}$

SAMPLE 9.2

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Figure 9.9:

The force due to varying hydrostatic pressure: The hydrostatic pressure distribution on the face of a wall submerged in water up to a height $h=10\,\mathrm{m}$ is shown in the figure. Find the net force on the wall from water. Take the length of the wall (into the page) to be 1 $\,\mathrm{m}$ .

Solution

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Figure 9.10:

Since the pressure varies across the height of the submerged part of the wall, let us take an infinitesimal strip of height $d y$ along the full length $\ell$ of the wall as shown in fig. 9.10. Since the height of the strip is infinitesimal, we can treat the water pressure on this strip to be essentially constant and equal to $p_{0}\frac{y}{h}$ . Then the force on the strip (of area $\ell dy$ ) due to the constant water pressure $p(y)=p_{0}y/h$ is

d{\accentset{\rightharpoonup}{\boldsymbol{F}}}=(p(y)\cdot\overbrace{\ell dy)}^% {\rm area}\hat{\boldsymbol{\imath}}\\ =p_{0}\frac{y}{h}\,\ell\,dy\,\hat{\boldsymbol{\imath}}.

The net force due to the pressure distribution on the whole wall can now be found by integrating $d{\accentset{\rightharpoonup}{\boldsymbol{F}}}$ along the height of the wall.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle\int d{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\int_{0}^{h}p% _{0}\frac{y}{h}\,\ell\,dy\,\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\left(p_{0}\frac{\ell}{h}\int_{0}^{h}y\,dy\right)\hat{\boldsymbol% {\imath}}=p_{0}\frac{\ell}{h}\,\frac{h^{2}}{2}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}p_{0}h\ell\,\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\cdot(100\frac{\,{\rm kN}}{\,\mathrm{m}^{2}})\cdot(10% \,\mathrm{m})\cdot(1\,\mathrm{m})\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle(500\,{\rm kN})\,\hat{\boldsymbol{\imath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=500\,{\rm kN}\,\hat{\boldsymbol{% \imath}}$

Alternatively, the net force can be computed by calculating the area of the pressure triangle and multiplying by the unit length ( $\ell=1\,\mathrm{m}$ ), i.e.,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle\overbrace{\frac{1}{2}\cdot h\cdot p_{0}}^{\mbox{triangle area}}% \ell\,\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\left(\frac{1}{2}\cdot 10\,\mathrm{m}\cdot 100\frac{\,{\rm kN}}{% \,\mathrm{m}^{2}}\cdot 1\,\mathrm{m}\right)\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle 500\,{\rm kN}\,\hat{\boldsymbol{\imath}}.$

SAMPLE 9.3

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Figure 9.11:

Forces on a submerged sluice gate: A rectangular plate is used as a gate in a tank to prevent water from draining out. The plate is hinged at A and rests on a frictionless surface at B. Assume the width of the plate to be 1 $\,\mathrm{m}$ . The height of the water surface above point A is $h$ . Ignoring the weight of the plate, find the forces on the hinge at A as a function of $h$ . In particular, find the vertical pull on the hinge for $h=0$ and $h=2\,\mathrm{m}$ .

Solution Let $\gamma=\rho g$ be the weight density (weight per unit volume) of water. Then the pressure due to water at point A is $p_{A}=\gamma h$ and at point B is $p_{B}=\gamma(h+\ell\sin\theta)$ . The pressure acts perpendicular to the plate and varies linearly from $p_{A}$ at A to $p_{B}$ at B. The free-body diagram of the plate is shown in fig. 9.12.

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Figure 9.12: The free-body diagram of the gate (plate AB) with distributed water pressure.

Let ˆ λ be a unit vector along BA and ˆ n be a unit vector normal to BA. For computing the reaction forces on the plate at points A and B, we first replace the distributed pressure on the plate by two equivalent concentrated forces $F_{1}$ and $F_{2}$ by dividing the pressure distribution into a rectangular and a triangular region and finding their resultants.

\displaystyle F_{1}=p_{A}\ell=\gamma h\ell,\quad F_{2}=(p_{B}-p_{A})\,\frac{% \ell}{2}=\frac{1}{2}\gamma\ell^{2}\sin\theta.

Now, we carry out moment balance about point A, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , which gives

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}% \times{\accentset{\rightharpoonup}{\boldsymbol{B}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/A}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\!\mathrm{C/A}}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{1}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle-\ell\hat{\boldsymbol{\lambda}}\times B_{n}\hat{\boldsymbol{n}}-% \frac{2\ell}{3}\hat{\boldsymbol{\lambda}}\times(-F_{1}\hat{\boldsymbol{n}})-% \frac{\ell}{2}\hat{\boldsymbol{\lambda}}\times(-F_{2}\hat{\boldsymbol{n}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle-B_{n}\ell\hat{\boldsymbol{k}}+F_{1}\frac{2\ell}{3}\hat{% \boldsymbol{k}}+F_{2}\frac{\ell}{2}\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}B_{n}=\frac{2F_{1}}{3}+% \frac{F_{2}}{2}=\gamma\ell(\frac{2}{3}h+\frac{1}{4}\ell\sin\theta)$

and, from force balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , we get

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{A}}}$	$\displaystyle=$	$\displaystyle-B_{n}\hat{\boldsymbol{n}}+F_{1}\hat{\boldsymbol{n}}+F_{2}\hat{% \boldsymbol{n}}$
	$\displaystyle=$	$\displaystyle\left(-\gamma\ell(\frac{2}{3}h+\frac{1}{4}\ell\sin\theta)+\gamma h% \ell+\frac{1}{2}\gamma\ell^{2}\sin\theta\right)\hat{\boldsymbol{n}}$
	$\displaystyle=$	$\displaystyle\left(\frac{1}{3}\gamma h\ell+\frac{1}{2}\gamma\ell^{2}\sin\theta% \right)\hat{\boldsymbol{n}}=\gamma\ell(\frac{1}{3}h+\frac{1}{2}\ell\sin\theta)% \hat{\boldsymbol{n}}.$

Filename:sfig8-5-wiper-a

Figure 9.13: The free-body diagram of the pin at A.

The force ${\accentset{\rightharpoonup}{\boldsymbol{A}}}$ computed above is the force exerted by the hinge at A on the plate. Therefore, the force on the hinge, exerted by the plate, is $-{\accentset{\rightharpoonup}{\boldsymbol{A}}}$ as shown in fig. 9.13. From the expression for this force, we see that it varies linearly with $h$ .

Let the vertical pull on the hinge be ${A_{\rm hinge}}_{y}$ . Then

\displaystyle{A_{\rm hinge}}_{y}

\displaystyle=

\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{A}}}\cdot\hat{% \boldsymbol{\jmath}}=-\gamma\ell(\frac{1}{3}h+\frac{1}{2}\ell\sin\theta)% \overbrace{\hat{\boldsymbol{n}}\cdot\hat{\boldsymbol{\jmath}}}^{\cos\theta}=% \frac{1}{4}\gamma\ell\sin 2\theta+(\frac{1}{3}\gamma\ell\cos\theta)h.

Now, substituting $\gamma=9.81\,{\rm kN}/\,\mathrm{m}^{3},\,\ell=2\,\mathrm{m},\,\theta=30^{\circ}$ , the two specified values of $h$ , and multiplying the result (which is force per unit length) with the width of the plate (1 $\,\mathrm{m}$ ) we get,

Answer: ${A_{\rm hinge}}_{y}|_{h=0}=4.25\,{\rm kN},\quad{A_{\rm hinge}}_{y}|_{h=2\,% \mathrm{m}}=15.58\,{\rm kN}$

SAMPLE 9.4

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Figure 9.14:

Tipping of a dam: The cross-section of a concrete dam is shown in the figure. Take the weight-density $\gamma(=\rho g)$ of water to be $10\,{\rm kN}/\,\mathrm{m}^{3}$ and that of concrete to be $25\,{\rm kN}/\,\mathrm{m}^{3}$ . For the given design of the cross-section, find the ratio $h/H$ that is safe enough for the dam to not tip over (about the downstream edge E).

Solution Let us imagine the critical situation when the dam is just about to tip over about edge E. In such a situation, the dam bottom would almost lose contact with the ground except along edge E. In that case, there is no force along the bottom of the dam from the ground except at E.^†^†margin: This assumption is valid only if water does not leak through the lower right of the damn through to the bottom of the dam. If it did, there would be a force on the bottom due to the water pressure. See the following sample where we include the water pressure at the bottom in the analysis.

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Figure 9.15:

With this assumption, the free-body diagram of the dam is shown in fig. 9.15.

To compute all the forces acting on the dam, we assume the width $w$ (into the paper) to be unity (i.e., $w=1\,\mathrm{m}$ ). Let $\gamma_{w}$ and $\gamma_{c}$ denote the weight-densities of water and concrete, respectively. Then the resultant force from the water pressure is

F=\frac{1}{2}\,\gamma_{w}h\cdot h\cdot w=\frac{1}{2}\gamma_{w}h^{2}w.

This is the horizontal force (in the - ˆ ı direction) that acts through the centroid of triangle ABC.

To compute the weight of the dam, we divide the cross-section into two sections — the rectangular section CDGH and the triangular section DEF. We compute the weight of these sections separately by computing their respective volumes (we use the shape constant $\alpha=1/10$ ):

	$\displaystyle W_{1}$	$\displaystyle=$	$\displaystyle\underbrace{\alpha H^{2}\cdot w}_{\rm volume}\cdot\gamma_{c}=% \gamma_{c}\alpha H^{2}w$
	$\displaystyle W_{2}$	$\displaystyle=$	$\displaystyle\underbrace{\frac{1}{2}\cdot 3\alpha H\cdot 3\alpha H\tan\theta% \cdot w}_{\rm volume}\cdot\gamma_{c}=\frac{9}{2}\gamma_{c}\alpha^{2}H^{2}w\tan\theta.$

Now we apply moment balance about point E, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{E}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , which gives

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{1}}}% \times{\accentset{\rightharpoonup}{\boldsymbol{W}}}_{1}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G_{2}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{W}}}_{2}+{\accentset{\rightharpoonup}{\boldsymbol% {r}}}_{\!\mathrm{G_{3}}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle-(3\alpha H+\frac{1}{2}\alpha H)W_{1}\hat{\boldsymbol{k}}-\frac{2% }{3}(3\alpha H)W_{2}\hat{\boldsymbol{k}}+\frac{h}{3}F\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting this equation with ˆ k , we get

$\displaystyle\frac{h}{3}F$	$\displaystyle=$	$\displaystyle(3\alpha H+\frac{1}{2}\alpha H)\cdot\gamma_{c}\alpha H^{2}w+\frac% {2}{3}(3\alpha H)\cdot\frac{9}{2}\gamma_{c}\alpha^{2}H^{2}w\tan\theta$
$\displaystyle\frac{1}{2}\gamma_{w}\frac{h^{3}}{3}$	$\displaystyle=$	$\displaystyle 9\gamma_{c}\alpha^{3}H^{3}\tan\theta+\frac{7}{2}\gamma_{c}\alpha% ^{2}H^{3}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\left(\frac{h}{H}\right)% ^{3}$	$\displaystyle=$	$\displaystyle\frac{\gamma_{c}}{\gamma_{w}}(54\alpha^{3}\tan\theta+21\alpha^{2})$
	$\displaystyle=$	$\displaystyle 2.5(54\cdot 0.1^{3}\cdot\sqrt{3}+21\cdot 0.1^{2})=0.7588$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\frac{h}{H}$	$\displaystyle=$	$\displaystyle 0.91.$

Thus, for the dam to not tip over, $h\leq 0.91H\mbox{ or }91\%\mbox{ of }H$ .

Answer: $\frac{h}{H}\leq 0.91$

SAMPLE 9.5 Dam design: You are to design a dam of rectangular cross-section ( $b\times H$ ), ensuring that the dam does not tip over even when the water level $h$ reaches the top of the dam ( $h=H$ ). Take the specific weight of concrete to be 3. Consider the following two scenarios for your design.

1.

The downstream bottom edge of the dam is plugged so that there is no leakage underneath.
2.

The lower right edge is not plugged and water leaks under the dam bottom has full pressure across the bottom.

Solution Let $\gamma_{c}$ and $\gamma_{w}$ denote the weight densities of concrete and water, respectively. We are given that $\gamma_{c}/\gamma_{w}=3$ . Also, let $b/H=\alpha$ so that $b=\alpha H$ . We consider both scenarios to find a safe cross-sectional shape and size for the dam. In the calculations below, we consider unit length (into the paper) of the dam.

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Figure 9.16:
1.

No water pressure on the bottom: When there is no water pressure on the bottom of the dam, then the water pressure acts only on the downstream side of the dam. The free-body diagram of the dam, considering critical tipping (just about to tip), is shown in fig. 9.16 in which $F$ is the resultant force of the triangular water pressure distribution. The known forces acting on the dam are $W=\gamma_{c}\alpha H^{2},\qquad\mbox{and}\qquad F=(1/2)\gamma_{w}h^{2}$ . The moment balance about point A gives

$\displaystyle F\cdot\frac{h}{3}$ $\displaystyle=$ $\displaystyle W\cdot\frac{\alpha H}{2}$

$\displaystyle\frac{1}{2}\gamma_{w}\frac{h^{3}}{3}$ $\displaystyle=$ $\displaystyle\gamma_{c}\frac{\alpha^{2}H^{3}}{2}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha^{2}$ $\displaystyle=$ $\displaystyle(1/3)(\gamma_{w}/\gamma_{c})(h/H)^{3}.$

Considering the case of critical water level up to the height of the dam, i.e., $h/H=1$ , and substituting $\gamma_{c}/\gamma_{w}=3$ , we get

$\alpha^{2}=1/9\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha=1/3=0.333.$

Thus the width of the cross-section needs to be at least one-third of the height. For example, if the height of the dam is $9\,\mathrm{m}$ then it needs to be at least $3\,\mathrm{m}$ wide.

Answer: $b/H=0.33$

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Figure 9.17:
2.

Full water pressure on the bottom: In this case, the water pressure on the bottom is uniformly distributed and its intensity is the same as the lateral pressure at B, i.e., $p=\gamma_{w}h$ . The free-body diagram diagram is shown in fig. 9.17 where the known forces are $W=\gamma_{c}\alpha H^{2},\,F=(1/2)\gamma_{w}h^{2},\qquad\mbox{and}\qquad R=% \gamma_{w}\alpha hH$ . Again, we carry out moment balance about point A to get

$\displaystyle F\cdot\frac{h}{3}$ $\displaystyle=$ $\displaystyle(W-R)\cdot\frac{\alpha h}{2}$

$\displaystyle\gamma_{w}h^{3}$ $\displaystyle=$ $\displaystyle 3(\gamma_{c}\alpha H^{2}-\gamma_{w}\alpha hH)\alpha H$

$\displaystyle\alpha^{2}$ $\displaystyle=$ $\displaystyle\frac{(h/H)^{3}}{3(\gamma_{c}/\gamma_{w}-h/H)}.$

Once again, substituting the given values and $h/H=1$ , we get

$\alpha^{2}=1/6\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha=0.408.$

Thus the width in this case needs to be at least $0.41$ times the height $H$ , slightly wider than the previous case.

Answer: $b/H\geq 0.41$

Problems for 9.1 Net force and moments in hydrostatics

Preparatory Problems

More-Involved Problems

9.1.1 A balloon with volume $V$ , whose membrane has negligible mass, holds a gas with density $\rho_{2}$ . It is surrounded by a gas with density $\rho_{1}$ .

(a)

In terms of $\rho_{1}$ , $\rho_{2}$ , $g$ , and $V$ , find the tension in the string.
(b)

By some means look up the density of Helium and air at atmospheric temperature and pressure and calculate the volume, in cubic feet and in cubic meters, of a helium balloon that could lift $75\,\mathrm{kg}$ .

Filename:sfig8-6-2a

Figure 9.18:

Balloon

9.1.2 A spherical body of mass $m=10\,\mathrm{kg}$ and radius $R=100\,{\rm mm}$ hangs from a continuous string as shown in the figure. The body is partially submerged in water and angle $\alpha=45^{{}^{\circ}}$ (fixed). If the force of buoyancy is $\rho Vg$ where $\rho=1000\,\mathrm{kg}/\,\mathrm{m}^{3}=$ density of water, $V$ is the submerged volume of the body, and $g$ is the usual $g$ ; find the tension in the string as a function of the submerged volume $V$ . Find the maximum and the minimum tension corresponding to fully and zero submerged volume of the body respectively.

Filename:efig1-2-28

Figure 9.19:

9.1.3 A 4-meter-high ‘door’ holds back a stream ( $\gamma=10,000\,{\mbox{N}}/\,\mathrm{m}^{3}$ ) that is 3m deep and 12m wide. The door is hinged along its bottom and is propped up by a thin rod B that goes from a ball joint at H at (3,12,0) to a ball joint at the upper left corner B of the door at (0,0,4). Neglect the mass of the door. Find the axial-force in the rod BH. Answer: $(117\gamma/2)\,\mathrm{m}^{3}=5.85*10^{5}\,{\mbox{N}}$

Filename:pfigure-blue-118-2

Figure 9.20:

9.1.4 Water is held in a reservoir by a board with negligible weight that is 5 meters long. It is hinged 1 meter off the bottom at A and kept from leaking by a seal at B. Assume $\rho=1000\,\mathrm{kg}/\,\mathrm{m}^{3}$ , $g=10\,{\mbox{N}}/\,\mathrm{kg}$ .

(a)

What is $h$ when the board starts to pull away from the stop at B? Answer: Water starts to spill at $h=3r_{AB}=3\,\mathrm{m}$ .
(b)

At that $h$ what is the force of the hinge on the board? Answer: Assuming no friction at B, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}=2.25*10^{5}\hat{\boldsymbol{% \imath}}\,{\mbox{N}}$

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Figure 9.21:

9.1.5 The side of a pool is made of vertical boards which are stuck in the ground. Assuming that the boards, on average, get no support from their neighbors, and neglect the weight of the board itself,

(a)

calculate the force and moment from the ground on one board (answer in terms of some or all of $w$ , $h$ , $\rho$ , and $g$ .
(b)

For a one foot board and 8 foot deep pool, find the size of a force, and its location, so the force is equivalent to the water pressure on the board (answer in $\,{\rm lbf}$ ).

Filename:Danef94s1q2

Figure 9.22:

9.1.6 A sluice gate is a dam that can be opened. Sometimes it is just a board in a slot that is opened by pulling up the board. For water with density $\rho$ and depth $h$ pressing against a board with width $w$ pressing against one face of the slot (the face away from the water) with coefficient of friction $\mu$ .

(a)

find the force $F$ needed to pull up the board in terms of $g,\rho,\mu,h$ , and $w$ .
(b)

Find the force assuming $g=10\,\mathrm{m}/\mathrm{s}^{2},h=1\,\mathrm{m},w=1\,\mathrm{m}$ , $\mu=0.5$ and $\rho=1000\,\mathrm{kg}/\,\mathrm{m}^{3}$ .

Filename:pfigure-blue-123-1

Figure 9.23:

9.1.7 A concrete (density = $\rho_{c}$ ) wall with height $\ell$ , width $w$ and length (into the paper) $d$ rests on a flat rigid floor and serves as a dam for water with depth $h$ and density $\rho_{w}$ . Assume the wall only makes contact at edges A and B.

(a)

Assume there is a seal at A, so no water gets under the dam. What is the coefficient of friction needed to keep the block from sliding?
(b)

What is the maximum depth of water before the block tips?
(c)

Assume that there is a seal at B and that water gets under the block. What is the coefficient of friction needed to keep the block from sliding?
(d)

What is the maximum depth of water before the block tips?

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Figure 9.24:

9.1.8 A door holds back the water at a lock on a canal. The water surface is at the top of the door. The rope AB keeps it from swinging open. The door has hinges at C and D. The height of the door is $h$ , the width $w$ . The point B is a distance $d$ above the top of the door and is set back a distance $L$ . The weight density of the water is $\gamma$ .

(a)

What is the total force of the water on the door?
(b)

What is the tension in the rope AB?

Filename:pfigure-s95q14

Figure 9.25:

9.1.9 This problem somewhat explains the workings of some toilet valves. Open the tank of a toilet and look at the rubber piece at the bottom that sits on the bottom but then floats after initially lifted by the turning of the flush lever. The puzzle this problem solves is this: Why does the valve stick to the bottom, but then float when lifted.

(a)

A hollow cylinder with an open bottom (like an upside down but open can) is filled with air but is under water. What force is required to hold it under water (in terms of $\rho,r,h,\ell$ and $g$ ? Answer: $\rho g\pi r^{2}\ell$
(b)

The same can is on the bottom of a tank of water and its edges are sealed. The bottom is open to atmospheric air. How much force is needed to hold the can down now (so there is no force from the bottom of the tank onto the edges of the cylinder)? Answer: $-\rho g\pi r^{2}(h-\ell)$ , note the minus sign, it now takes force to lift the can.

Filename:Danef94s3q2

Figure 9.26:

Some puzzles. The following puzzles are sometimes presented as brain teasers. You should be able to reason the answers carefully and irrefutably.

9.1.10 A person is in a boat in a pool with surface area $A$ . She is holding a ball with volume $V$ and mass $m$ in a still pool. The ball is then thrown into the pool, no water is splashed out and the pool comes to rest again.

(a)

Assuming the ball floats, by how much does the pool level go up or down?
(b)

Assuming the ball sinks to the bottom by how much does the pool level go up or down?

Filename:bikefork1-ang-accel

Figure 9.27:

9.1.11 A steel boat with mass $m$ and density $\rho_{s}$ is floating in a pool of water with density $\rho_{w}$ and cross sectional area $A$ . By how much does the pool level go up or down when the boat sinks to the bottom?

Filename:bikefork-ang-accel

Figure 9.28:

9.1.12 Two cups of water are balanced. You then gently stick your finger into one of them. Does this upset the balance? This experiment can be set up with two cups and a hexagonal-cross-section pencil. The cups need not be identical, they just need to be balanced at the start.

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Figure 9.29:

9.1.13 A tray of water is suspended and level.

(a)

A hand is gently placed in the tray but does not touch the edges or bottom. Is the level of the tray upset.
(b)

Challenge: Assuming the tray is massless with width $w$ and water depth $h$ , how high must be the hinge so the equilibrium is stable. That is, imagine the tray is rotated slightly about the hinge, the water pressure should cause a torque which tends to restore the vertical orientation shown.

Filename:bikefork-alt

Figure 9.30:

9.1.14 Challenge: This challenge problem is closely related to the challenge problem above, but is much more famous. It seems to have been first solved by Leonard Euler and Pierre Bouguer in about 1735. This solution seems to be the first mechanics problem in which the significance of the area moment of inertia was appreciated [this is a hint].

For simplicity assume that a boat is shaped like a box with width $h$ and and length into the paper of $b$ . Assume that the boat floats with its bottom a depth $d$ under water. Now rotate the boat about an axis at the surface of the water and along its length (into the paper). Imagine that giant hands hold the boat in this position. In this rotated position the effect of the water pressure on the boat is a buoyant force and moment. This is equivalent to a force that is displaced slightly sideways.

Your goal is to find the height of the point that the line of action of this force intersects a mast of the boat. For small angles of boat tip the location is independent of the amount of tip.

This point is called the metacenter of the hull, and its distance up from the centroid of the boat’s submerged volume is the hull’s metacentric height. The condition of boat stability is that the metacenter be above the center of mass of the boat (thus the moment of the buoyant forces about the center of mass will tend to restore the boat to level).

Euler and Bouguer did the calculation you are asked to do here after, e.g., , the launching of the ‘great’ Swedish ship Vasa, which capsized in the harbor on day 1. This was unfortunate for Sweden at the time, but fortunate now, because the brand-new 375 year old ship is a see-worthy tourist attraction in Stockholm.

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Figure 9.31:

Ship stability.

	$\displaystyle\mbox{area of ABD}\equiv r^{2}\theta-\frac{1}{2}r^{2}\sin 2\theta$	$\displaystyle=$	$\displaystyle 0.018\,\mathrm{m}^{2}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta-\frac{1}{2}\sin 2% \theta-0.738$	$\displaystyle=$	$\displaystyle 0.$

$\displaystyle F\cdot\frac{h}{3}$	$\displaystyle=$	$\displaystyle W\cdot\frac{\alpha H}{2}$
$\displaystyle\frac{1}{2}\gamma_{w}\frac{h^{3}}{3}$	$\displaystyle=$	$\displaystyle\gamma_{c}\frac{\alpha^{2}H^{3}}{2}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\alpha^{2}$	$\displaystyle=$	$\displaystyle(1/3)(\gamma_{w}/\gamma_{c})(h/H)^{3}.$

$\displaystyle F\cdot\frac{h}{3}$	$\displaystyle=$	$\displaystyle(W-R)\cdot\frac{\alpha h}{2}$
$\displaystyle\gamma_{w}h^{3}$	$\displaystyle=$	$\displaystyle 3(\gamma_{c}\alpha H^{2}-\gamma_{w}\alpha hH)\alpha H$
$\displaystyle\alpha^{2}$	$\displaystyle=$	$\displaystyle\frac{(h/H)^{3}}{3(\gamma_{c}/\gamma_{w}-h/H)}.$