IPhO 2025 · Theory problem 2

Cox's Timepiece

In 1765, British clockmaker James Cox invented a clock whose only source of energy is the fluctuations in atmospheric pressure. Cox's clock used two vessels containing mercury. Changes in atmospheric pressure caused mercury to move between the vessels, and the two vessels to move relative to each other. This movement acted as an energy source for the actual clock.

We propose an analysis of this device. Throughout, we assume that

  • the Earth's gravitational field \(\vec g=-g\vec u_z\) is uniform with \(g=9.8\,\mathrm{m\cdot s^{-2}}\) and \(\vec u_z\) a unit vector;
  • all liquids are incompressible and their density is denoted \(\rho\);
  • no surface tension effects will be considered;
  • the variations of atmospheric pressure with altitude are neglected;
  • the surrounding temperature \(T_a\) is uniform and all transformations are isothermal.
Figure 1

Fig. 1. Artistic view of Cox's clock \({}^{1}\)

Part A - Pulling on a submerged tube

We first consider a bath of water that occupies the semi-infinite space \(z\leq0\). The air above it is at a pressure \(P_a=P_0\). A cylindrical vertical tube of length \(H=1\,\mathrm m\), cross-sectional area \(S=10\,\mathrm{cm^2}\) and mass \(m=0.5\,\mathrm{kg}\) is dipped into the bath. The bottom end of the tube is open, and the top end of the tube is closed. We denote \(h\) the altitude of the top of the tube and \(z_\ell\) that of the water inside the tube. The thickness of the tube walls is neglected.

Figure 2

Fig. 2. Sketch of the tube in different configurations

We start from the situation where the tube in Fig. 2 contains no gas and its top is at the bath level: in other words, \(h=0\) and \(z_\ell=0\) (case a). The tube is then slowly lifted until its bottom end reaches the bath level. The pulling force exerted on the tube is denoted \(\vec F=F\vec u_z\).

A.1 — 0.2pt

For the configuration shown in Fig. 2 (case b), express the pressure \(P_w\) in the water at the top of the tube. Also express the force \(\vec F\) necessary to maintain the tube at this position. Expressions must be written in terms of \(P_0,\rho,m,S,h,g\) and \(\vec u_z\).

Three experiments are performed. In each, the tube is lifted from the initial state shown in Fig. 2(a) under the conditions specified in Table 1.

Experiment Liquid \(T_a\) (°C) \(\rho\) (\(\mathrm{kg\cdot m^{-3}}\)) \(P_{\mathrm{sat}}\) (Pa)
1 Water 20 \(1.00\times10^3\) \(2.34\times10^3\)
2 Water 80 \(0.97\times10^3\) \(47.4\times10^3\)
3 Water 99 \(0.96\times10^3\) \(99.8\times10^3\)

Table 1. Experimental conditions and numerical values of physical quantities for each experiment

(\(P_{\mathrm{sat}}\) designates the saturated vapour pressure of the pure fluid)

In each case, we study the evolution of the force \(F\) that must be applied in order to maintain the tube in equilibrium at an altitude \(h\), the external pressure being fixed at \(P_a=P_0=1.000\times10^5\,\mathrm{Pa}\). Two different behaviours are possible

Figure 3

A.2 — 0.8pt

For each experiment, complete the table in the answer sheet to indicate the expected behaviour and the numerical values for \(F_{\max}\) and for \(h^\star\) (when pertinent), where \(F_{\max}\) and \(h^\star\) are defined in the figures illustrating the two behaviours.

Figure 4

When we replace the water with liquid mercury (whose properties are given below), behaviour B is observed.

Liquid \(T_a\) (°C) \(\rho\) (\(\mathrm{kg\cdot m^{-3}}\)) \(P_{\mathrm{sat}}\) (Pa)
Mercury 20 \(13.5\times10^3\) \(0.163\)

A.3 — 0.3pt

Express the relative error, denoted \(\varepsilon\), committed when we evaluate the maximal force \(F_{\max}\) neglecting \(P_{\mathrm{sat}}\) compared to \(P_0\). Give the numerical value of \(\varepsilon\).

Part B - Two-part barometric tube

From now on, we work with mercury (density \(\rho=13.5\times10^3\,\mathrm{kg\cdot m^{-3}}\)) at the ambient temperature \(T_a=20^\circ\mathrm C\) and we take \(P_{\mathrm{sat}}=0\).

Let us consider a tube with a reservoir on top, modeled as two superposed cylinders of different dimensions, as shown in Fig. 3.

  • the bottom part (still called the tube) has cross-sectional area \(S_t\) and height \(H_t=80\,\mathrm{cm}\);
  • the top part (called the bulb) has cross-sectional area \(S_b>S_t\) and height \(H_b=20\,\mathrm{cm}\).

This two-part tube is dipped into a semi-infinite liquid bath.

Figure 5

Fig. 3. Sketch of the two-part barometric tube

As in Part A, the system is prepared such that the tube contains no air. We identify the vertical position of the tube by the altitude \(h_t\) of the junction between the tube and the bulb. The height of the column of mercury is again denoted \(z_\ell\). The force \(\vec F\) that must be exerted to maintain the tube in equilibrium in the configuration shown in Fig. 3 can now be written as

\[ \vec F=(m_{\mathrm{tb}}+m_{\mathrm{add}})g\vec u_z \tag{1} \]

where \(m_{\mathrm{tb}}\) is the total mass of the two-part tube (when empty of mercury).

B.1 — 0.3pt

On the answer sheet, color the area corresponding to the volume of liquid mercury that is responsible for the term \(m_{\mathrm{add}}\) appearing in equation (1).

Figure 6

The mass \(m_{\mathrm{add}}\) depends both on the height \(h_t\) and the atmospheric pressure \(P_a\). For the next question, assume that the atmospheric pressure is fixed at \(P_a=P_0=1.000\times10^5\,\mathrm{Pa}\). Starting from the situation where the system is completely submerged, the tube is slowly lifted until its base is flush with the liquid bath.

B.2 — 1.4pt

Sketch the evolution of the mass \(m_{\mathrm{add}}\) as a function of \(h_t\) for \(h_t\in[-H_b,H_t]\). On the graph, provide the expression for the slopes of the different segments, as well as the \(h_t\) analytical value of any angular points, in terms of \(P_0,\rho,g,S_b,S_t,H_b\) and \(H_t\).

Figure 7

As the system is lifted while \(P_a=P_0=10^5\,\mathrm{Pa}\), we stop when the free surface of the liquid is in the middle of the bulb. The value of \(h_t\) is fixed and then we observe variations in the mass \(m_{\mathrm{add}}\) due to variations in the atmospheric pressure described by

\[ P_a(t)=P_0+P_1(t) \tag{2} \]

where \(P_0\) designates the average value and \(P_1\) is a perturbative term. We model \(P_1\) by a periodic triangular function of amplitude \(A=5\times10^2\,\mathrm{Pa}\) and period \(\tau_1\) of 1 week.

Figure 8

Fig. 4. Simplified model of the perturbative term \(P_1(t)\)

B.3 — 0.3pt

Given that \(S_t=5\,\mathrm{cm^2}\) and \(S_b=200\,\mathrm{cm^2}\), express the amplitude \(\Delta m_{\mathrm{add}}\) of the variations of the mass \(m_{\mathrm{add}}\) over time, then give its numerical value. Assume that the liquid surface always stays in the bulb.

Part C - Cox's timepiece

The real mechanism developed by Cox is complex (Fig. 5). We study a simplified version, depicted in Fig. 6, and described below

  • a cylindrical bottom cistern containing a mercury bath;
  • a two-part barometric tube identical to that studied in part B, which is still completely emptied of any air, is dipped into the bath;
  • the cistern and the two-part tube are each suspended by a cable. Both cables (assumed to be inextensible and of negligible mass) pass through a system of ideal pullies and finish attached to either side of the same mass \(M\), which can slide on a horizontal surface;
  • the total volume of liquid mercury contained in the system is \(V_\ell=5\,\mathrm L\).

The height, cross-section and masses of each part are given in Table 2. The position of mass \(M\) is referenced by the coordinate \(x\) of its center of mass. We consider solid friction between the horizontal support and the mass \(M\), without distinction between static and dynamic coefficients; the magnitude of this force when sliding occurs is denoted \(F_s\).

Two stops limit the displacement of the mass \(M\) such that \(-X\leq x\leq X\) (with \(X>0\)). Assume that the value of \(X\) guarantees that

  • the bottom of the two-part tube never touches the bottom of the cistern nor comes out of the liquid bath;
  • the altitude \(z_\ell\) of the mercury column is always in the upper bulb.
Figure 9

Fig. 5. Real Cox's timepiece \({}^{2}\) (without mercury)

Figure 10

Fig. 6. Sketch of the system modeling the timepiece

Reference Name Height Cross section area Empty mass
\(\boxed{1}\) cistern \(H_c=30\,\mathrm{cm}\) \(S_c=210\,\mathrm{cm^2}\) \(m_c\)
\(\boxed{2}\) tubular part of the barometric tube \(H_t=80\,\mathrm{cm}\) \(S_t=5\,\mathrm{cm^2}\) total mass of the barometric tube: \(m_{\mathrm{tb}}\)
\(\boxed{2'}\) bulb of the barometric tube \(H_b=20\,\mathrm{cm}\) \(S_b=200\,\mathrm{cm^2}\)

Table 2. Dimensions and notations for the model system

The system evolves in contact with the atmosphere, whose pressure fluctuates as in Fig. 4 (still with amplitude \(A=5\times10^2\,\mathrm{Pa}\) and period \(\tau_1=1\) week). At the start \(t=0\), the mass \(M\) is at rest at \(x=0\) and the tensions exerted by the two cables on either side of the mass \(M\) are in balance while \(P_1(0)=0\). We define

\[ \xi= \frac{S_b+S_c-S_t}{S_bS_c}\frac{F_s}{A} \simeq \frac{S_b+S_c}{S_bS_c}\frac{F_s}{A} \tag{3} \]

where the last expression uses that \(S_t\ll S_b,S_c\) (which we will assume is valid until the end of the problem).

C.1 — 1pt

Determine the threshold \(\xi^\star\) such that \(M\) remains indefinitely at rest when \(\xi>\xi^\star\).

For the next question only, suppose that the mass \(M\) is temporarily blocked at \(x=X\).

C.2 — 1pt

Give an expression for the total tension force \(\vec T=T\vec u_x\) acting on the mass \(M\) due to the tension in two cables at this position, when \(P_1=0\), in terms of \(\rho,g,X\) and pertinent cross-sections.

When \(\xi<\xi^\star\), starting again from \(x=0\) and \(P_1=0\), two different behaviours can be observed for \(t\geq0\). To distinguish them, we need to introduce another parameter

\[ \lambda= \frac{2(S_b-S_t)}{S_b}\frac{\rho gX}{A} \simeq \frac{2\rho gX}{A} \tag{4} \]

C.3 — 2pt

Complete the table in the answer sheet to indicate the condition under which each regime is obtained. Conditions must be expressed as inequalities on \(\xi\) and/or \(\lambda\). In addition, sketch the variations of \(x(t)/X\) for \(t\in[0,3\tau_1]\) that are consistent with the variations of \(P_1(t)/A\) already present. Specification of remarkable points coordinates is not required.

Figure 11

In the real Cox's timepiece, energy provided by the mechanism is stored using a system of ratchets and used to raise a counterweight, like in a traditional clock. In the simplified model studied here, the energy recovered by the clock corresponds to the energy dissipated by the friction force exerted by the horizontal surface on the mass \(M\). From now on, we assume that the system is dimensioned such that to work in the regime that allows the clock to recuperate energy. We also assume that the permanent regime is established. We denote \(W\) the energy dissipated by the solid friction force during a period \(\tau_1\), which can be expressed only in terms of \(F_s\) and \(X\).

All else equal, \(F_s\) and \(X\) can be adjusted to maximize the energy \(W\); we denote \(F_s^\star\) and \(X^\star\) their respective values in the optimal situation.

C.4 — 1pt

Considering \(S_b\simeq S_c\) and \(S_t\ll S_b\), determine the expressions for \(F_s^\star\) and \(X^\star\) as functions of \(\rho,g,S_c\) and \(A\). Express the corresponding maximum energy \(W^\star\), then calculate its numerical value with \(A=5\times10^2\,\mathrm{Pa}\).

We denote \(W_{\mathrm{pr}}^\star\) the work of atmospheric pressure forces received by the system in the optimal situation during a period \(\tau_1\).

C.5 — 1.7pt

Express \(W_{\mathrm{pr}}^\star\), then calculate the ratio \(W^\star/W_{\mathrm{pr}}^\star\). It could be useful to represent the evolution of the system in a \((P,V)\) diagram, where \(V\) is the system's volume.

Credits:

[1]: Bruno Vacaro;
[2]: Victoria and Albert Museum, London.

Official English text of the problem.

Answers

Define \[ L=\frac{P_0}{\rho g},\qquad D=S_b+S_c-S_t,\qquad q=\lambda+\frac{\xi}{2}. \]

  • A.1 \[ \boxed{P_w=P_0-\rho gh},\qquad \boxed{\vec F=(mg+\rho gSh)\vec u_z}. \]

  • A.2

Experiment Behaviour \(h^\star\) (cm), when pertinent \(F_{\max}\) (N)
1 A — \(14.7\)
2 A — \(14.4\)
3 B \(2.13\) \(5.10\)
  • A.3 \[ \boxed{\varepsilon= \frac{SP_{\mathrm{sat}}}{mg+S(P_0-P_{\mathrm{sat}})} =1.55\times10^{-6}}. \] Neglecting mercury vapour pressure overestimates the force.

  • B.1 Shade all mercury inside the tube and bulb above the external bath level. In the depicted configuration, \[ \boxed{m_{\mathrm{add}}=\rho\bigl[S_th_t+S_b(z_\ell-h_t)\bigr]}. \]

  • B.2 \[ \boxed{ m_{\mathrm{add}}(h_t)= \begin{cases} \rho S_b(h_t+H_b),&-H_b\le h_t\le0,\\ \rho(S_bH_b+S_th_t),&0\le h_t\le L-H_b,\\ \rho\bigl[S_bL-(S_b-S_t)h_t\bigr],&L-H_b\le h_t\le L,\\ \rho S_tL,&L\le h_t\le H_t. \end{cases}} \] The successive slopes are \[ \boxed{\rho S_b,\quad \rho S_t,\quad-\rho(S_b-S_t),\quad0}. \] The angular points occur at \[ \boxed{h_t=0,\quad L-H_b\simeq0.556\,\mathrm m,\quad L\simeq0.756\,\mathrm m}. \]

  • B.3 \[ \boxed{\Delta m_{\mathrm{add}}=\frac{S_bA}{g}=1.02\,\mathrm{kg}}. \]

  • C.1 \[ \boxed{\xi^\star=2}. \]

  • C.2 \[ \boxed{ \vec T= \frac{4\rho gS_c(S_b-S_t)}{S_b+S_c-S_t}X\,\vec u_x \simeq \frac{4\rho gS_bS_c}{S_b+S_c}X\,\vec u_x}. \]

  • C.3, for \(\xi<2\):

Regime Condition Description of \(x(t)/X\)
1: remains at one stop \(q>1\), equivalently \(\xi+2\lambda>2\) Initially \(0\), then rises rapidly to \(+1\) and remains there.
2: repeated strokes \(q<1\), equivalently \(\xi+2\lambda<2\) Initially \(0\), then switches between plateaus at \(+1\) and \(-1\).

The first positive stroke begins when \(P_1/A=\xi/2\). Subsequent negative strokes begin at \(P_1/A=-q\) on falling pressure, and positive strokes at \(P_1/A=q\) on rising pressure. The boundary \(q=1\) is marginal; ideal static friction permits the mass to remain at the stop.

  • C.4 The ideal limiting optimum is \[ \boxed{F_s^\star=\frac{S_cA}{2}\simeq5.25\,\mathrm N}, \qquad \boxed{X^\star=\frac{A}{4\rho g}\simeq0.945\,\mathrm{mm}}, \] \[ \boxed{W^\star=\frac{S_cA^2}{2\rho g} \simeq1.98\times10^{-2}\,\mathrm J}. \]

  • C.5 \[ \boxed{W_{\mathrm{pr}}^\star=\frac{S_cA^2}{\rho g} \simeq3.97\times10^{-2}\,\mathrm J}, \qquad \boxed{\frac{W^\star}{W_{\mathrm{pr}}^\star}=\frac12}. \]

Solution

A.1

Hydrostatic equilibrium gives \[ \frac{dP}{dz}=-\rho g. \] The water is connected to the bath, where \(P=P_0\) at \(z=0\). Consequently, while the water remains in contact with the closed top, \[ P_w=P_0-\rho gh. \]

The pressure on the outside of the top produces a downward force \(P_0S\), while the water underneath produces an upward force \(P_wS\). The vertical walls contribute no vertical pressure force, and their thickness is neglected. Equilibrium therefore requires \[ F+P_wS-P_0S-mg=0. \] Thus \[ \boxed{\vec F=(mg+\rho gSh)\vec u_z}. \]

A.2

A vapour cavity appears when the pressure at the top reaches the saturated vapour pressure: \[ P_0-\rho gh^\star=P_{\mathrm{sat}}. \] Hence \[ \boxed{h^\star=\frac{P_0-P_{\mathrm{sat}}}{\rho g}}. \]

Before this happens, \(F=mg+\rho gSh\). After a vapour cavity has formed, its pressure remains \(P_{\mathrm{sat}}\) at the prescribed temperature, so \[ F=mg+S(P_0-P_{\mathrm{sat}}). \] These expressions combine into \[ F(h)=mg+S\min\{\rho gh,\ P_0-P_{\mathrm{sat}}\}, \qquad 0\le h\le H. \]

Therefore:

  • if \(h^\star\ge H\), behaviour A occurs and \(F_{\max}=mg+\rho gSH\);
  • if \(h^\star<H\), behaviour B occurs and \(F_{\max}=mg+S(P_0-P_{\mathrm{sat}})\).

Using \(S=10^{-3}\,\mathrm{m^2}\), \(H=1\,\mathrm m\), and \(mg=4.90\,\mathrm N\):

Experiment Calculated saturation height Comparison with \(H\) Result
1 \(\dfrac{100000-2340}{1000(9.8)}=9.97\,\mathrm m\) Above \(H\) A; \(F_{\max}=14.7\,\mathrm N\)
2 \(\dfrac{100000-47400}{970(9.8)}=5.53\,\mathrm m\) Above \(H\) A; \(F_{\max}=14.406\,\mathrm N\simeq14.4\,\mathrm N\)
3 \(\dfrac{100000-99800}{960(9.8)}=0.02126\,\mathrm m\) Below \(H\) B; \(h^\star=2.13\,\mathrm{cm}\), \(F_{\max}=5.10\,\mathrm N\)

Both graphs begin at \(F_0=mg=4.90\,\mathrm N\).

A.3

For mercury, the actual maximum force and its approximation are \[ F_{\max}=mg+S(P_0-P_{\mathrm{sat}}), \qquad F_{\max}^{(0)}=mg+SP_0. \] The approximation overestimates the force by \(SP_{\mathrm{sat}}\). Relative to the actual force, \[ \varepsilon =\frac{F_{\max}^{(0)}-F_{\max}}{F_{\max}} =\frac{SP_{\mathrm{sat}}}{mg+S(P_0-P_{\mathrm{sat}})}. \] Numerically, \[ \boxed{ \varepsilon= \frac{10^{-3}(0.163)} {0.5(9.8)+10^{-3}(100000-0.163)} =1.55\times10^{-6}}. \] This is approximately \(0.000155\%\).

B.1

The additional load is the weight of the mercury raised above the external bath surface. Mercury below that level is supported hydrostatically by the bath.

Thus, on the answer-sheet diagram, shade:

  • the mercury in the narrow tube from the external bath level up to the tube–bulb junction;
  • all the mercury in the bulb, from its bottom up to its internal free surface.

Do not shade the mercury below the external bath level. For the configuration shown, \[ \boxed{m_{\mathrm{add}} =\rho\bigl[S_th_t+S_b(z_\ell-h_t)\bigr]}. \]

Since the pressure above the internal surface is zero, \[ z_\ell=\frac{P_a}{\rho g}. \] An equivalent useful expression is therefore \[ \boxed{ m_{\mathrm{add}} =\frac{S_bP_a}{g}-\rho(S_b-S_t)h_t } \] when the free surface lies in the bulb and the junction is above the external bath.

B.2

Introduce the barometric height \[ L=\frac{P_0}{\rho g} =\frac{10^5}{(13.5\times10^3)(9.8)} =0.75586\,\mathrm m. \] For the given dimensions, \[ 0<L-H_b<L<H_t. \]

Until the top reaches \(L\), the tube remains completely filled. Subsequently the internal free surface stays at \(z_\ell=L\), while the tube continues to rise.

There are four geometrically distinct intervals.

  1. \(-H_b\le h_t\le0\).
    Only a height \(h_t+H_b\) of the bulb is above the bath: \[ m_{\mathrm{add}}=\rho S_b(h_t+H_b). \] The graph starts at \((-H_b,0)\), with slope \(\rho S_b\).

  2. \(0\le h_t\le L-H_b\).
    The entire bulb and a length \(h_t\) of narrow tube are above the bath, all filled with mercury: \[ m_{\mathrm{add}}=\rho(S_bH_b+S_th_t). \] The graph continues upward with the smaller slope \(\rho S_t\).

  3. \(L-H_b\le h_t\le L\).
    A vacuum forms above the mercury, whose surface remains at \(L\) inside the bulb: \[ m_{\mathrm{add}} =\rho\bigl[S_th_t+S_b(L-h_t)\bigr]. \] The slope is \(-\rho(S_b-S_t)\), so the graph decreases.

  4. \(L\le h_t\le H_t\).
    The bulb is empty and the internal free surface lies in the narrow tube: \[ m_{\mathrm{add}}=\rho S_tL=\frac{S_tP_0}{g}. \] The graph is horizontal.

The graph is continuous, with the following vertices and endpoint:

\(h_t\) \(m_{\mathrm{add}}\)
\(-H_b\) \(0\)
\(0\) \(\rho S_bH_b\)
\(L-H_b\) \(\rho[S_bH_b+S_t(L-H_b)]\), the maximum
\(L\) \(\rho S_tL\)
\(H_t\) \(\rho S_tL\)

Its three angular points have horizontal coordinates \[ \boxed{0,\quad \frac{P_0}{\rho g}-H_b,\quad\frac{P_0}{\rho g}}, \] numerically \(0\), \(55.6\,\mathrm{cm}\), and \(75.6\,\mathrm{cm}\).

B.3

At the prescribed fixed position, the free surface remains inside the bulb. Thus \[ m_{\mathrm{add}}(t) =\frac{S_b[P_0+P_1(t)]}{g} -\rho(S_b-S_t)h_t. \] Only the first term varies, so the amplitude is \[ \boxed{\Delta m_{\mathrm{add}}=\frac{S_bA}{g}}. \] With \(S_b=200\,\mathrm{cm^2}=0.0200\,\mathrm{m^2}\), \[ \boxed{\Delta m_{\mathrm{add}} =\frac{0.0200(500)}{9.8} =1.02\,\mathrm{kg}}. \]

This is the amplitude about the mean; the peak-to-peak variation is \(2.04\,\mathrm{kg}\). The corresponding surface-height amplitude is \[ \frac{A}{\rho g}=3.78\,\mathrm{mm}, \] well below the \(10\,\mathrm{cm}\) distance from the middle of the bulb to either end.

C.1

We first obtain the cable-force difference for arbitrary \(x\) and \(P_1\).

Let \(z_s\) denote the external mercury surface and \(z_c\) the cistern bottom. From the cable arrangement, displacement of the mass to the right by \(x\):

  • raises the cistern by \(x\);
  • lowers the barometric tube by \(x\).

Therefore \[ z_c=z_{c0}+x,\qquad h_t=h_{t0}-x. \]

Hydrostatic equilibrium gives \[ z_\ell-z_s=\frac{P_a}{\rho g}. \] The mercury above the external surface has mass \[ m_{\mathrm{add}} =\rho\bigl[S_t(h_t-z_s)+S_b(z_\ell-h_t)\bigr] =\frac{S_bP_a}{g}-\rho(S_b-S_t)(h_t-z_s). \]

Because the tube walls have negligible volume, mercury conservation can be written as \[ V_\ell=S_c(z_s-z_c)+\frac{m_{\mathrm{add}}}{\rho}. \] Writing \(\delta m=m_{\mathrm{add}}-m_{\mathrm{add},0}\), this implies \[ \delta z_s=x-\frac{\delta m}{\rho S_c}. \] Subtracting the initial expression for \(m_{\mathrm{add}}\) from its current value gives \[ \delta m=\frac{S_bP_1}{g} +\rho(S_b-S_t)(x+\delta z_s). \] Consequently, \[ \boxed{ \delta m= \frac{S_bS_c}{(S_b+S_c-S_t)g}P_1 +\frac{2\rho S_c(S_b-S_t)}{S_b+S_c-S_t}x }. \]

In a stationary configuration, the tube-side cable tension is \[ T_{\mathrm{tb}}=(m_{\mathrm{tb}}+m_{\mathrm{add}})g, \] whereas the cistern-side tension is \[ T_c=(m_c+\rho V_\ell-m_{\mathrm{add}})g. \] Their initial difference is zero. Hence the signed force on \(M\), positive to the right, is \[ T=T_{\mathrm{tb}}-T_c=2g\,\delta m. \]

Define \[ K=\frac{S_bS_c}{S_b+S_c-S_t}, \qquad \alpha=\frac{2\rho g(S_b-S_t)}{S_b}. \] The resulting force law is \[ \boxed{T=2K(P_1+\alpha x)}. \] The given parameters satisfy \[ F_s=KA\xi,\qquad \alpha X=A\lambda. \]

At \(x=0\), the maximum force magnitude over the atmospheric cycle is \(2KA\). Static friction can balance it throughout the cycle if \[ F_s\ge2KA \quad\Longleftrightarrow\quad \xi\ge2. \] Thus the requested threshold is \[ \boxed{\xi^\star=2}. \]

C.2

Setting \(P_1=0\) and \(x=X\) in the force law, \[ \boxed{ \vec T= \frac{4\rho gS_c(S_b-S_t)}{S_b+S_c-S_t}X\,\vec u_x }. \] Using \(S_t\ll S_b,S_c\), \[ \boxed{ \vec T\simeq \frac{4\rho gS_bS_c}{S_b+S_c}X\,\vec u_x }. \]

The positive sign is significant: a displacement to the right produces a further force to the right. The equilibrium is mechanically unstable once friction can no longer hold the mass.

C.3

For the sketches, use the slow atmospheric driving limit: transfer between stops is rapid compared with the one-week pressure period. The stops arrest the moving system.

Write \[ p=\frac{P_1}{A},\qquad y=\frac{x}{X}. \] Then \[ \frac{T}{KA}=2(p+\lambda y), \qquad \frac{F_s}{KA}=\xi. \]

Initially \(y=0\) and \(p\) increases from zero. Motion to the right starts as \(p\) exceeds \[ p=\frac{\xi}{2}. \] Since \(\partial T/\partial x=2K\alpha>0\), movement to the right increases the driving force. The mass rapidly reaches \(y=+1\).

At the right stop, release to the left requires \[ T<-F_s. \] Thus \[ 2(p+\lambda)<-\xi \quad\Longleftrightarrow\quad p<-\left(\lambda+\frac{\xi}{2}\right). \] Define \[ q=\lambda+\frac{\xi}{2}. \]

If \(q>1\), the pressure never becomes sufficiently low to release the mass. If \(q<1\), release occurs during every decreasing-pressure half-cycle. The analogous release condition at the left stop is \(p>q\).

Therefore the two regimes are:

Regime Condition Precise sketch description
1: permanent residence at the right stop \(\xi<2\), \(\xi+2\lambda>2\) Draw \(x/X=0\) initially, followed by a rapid rise to \(+1\) on the first rising pressure branch at \(p=\xi/2\). Continue horizontally at \(+1\) through \(3\tau_1\).
2: repeated motion between stops \(\xi<2\), \(\xi+2\lambda<2\) Start with the same initial rise. Then draw horizontal plateaus at \(+1\) and \(-1\), joined by rapid downward transitions whenever falling pressure reaches \(p=-q\), and upward transitions whenever rising pressure reaches \(p=q\).

For an explicit placement of the transitions, let \(s=t/\tau_1\). The first positive transition is at \[ s=\frac{\xi}{8}. \] In the cycling regime, the negative transitions within the requested interval are at \[ s=\frac12+\frac q4,\quad \frac32+\frac q4,\quad \frac52+\frac q4, \] and subsequent positive transitions are at \[ s=1+\frac q4,\quad 2+\frac q4. \] These specify the requested three-period sketch completely.

At \(q=1\), the release threshold is only touched at a pressure extremum. Exact ideal static friction can keep the mass at the stop; sustained cycling is guaranteed by the strict inequality \(q<1\).

C.4

In the permanent cycling regime, the mass travels from one stop to the other and back during each period. Its total sliding distance is \(4X\), so \[ \boxed{W=4F_sX}. \]

With \(S_b\simeq S_c\) and \(S_t\ll S_b\), \[ \xi\simeq\frac{2F_s}{S_cA}, \qquad \lambda\simeq\frac{2\rho gX}{A}. \] The cycling condition becomes \[ \xi+2\lambda<2 \quad\Longleftrightarrow\quad F_s<S_c(A-2\rho gX). \]

For a fixed \(X\), the greatest attainable energy is approached as \(F_s\) approaches this upper bound. Its limiting envelope is \[ W(X)=4S_cX(A-2\rho gX). \] Differentiating, \[ \frac{dW}{dX}=4S_c(A-4\rho gX). \] The optimum is therefore \[ \boxed{X^\star=\frac{A}{4\rho g}}, \qquad \boxed{F_s^\star=\frac{S_cA}{2}}. \] Equivalently, \[ \xi_{\mathrm{opt}}=1,\qquad \lambda_{\mathrm{opt}}=\frac12. \] Thus \[ \boxed{W^\star=4F_s^\star X^\star =\frac{S_cA^2}{2\rho g}}. \]

With \(S_c=0.0210\,\mathrm{m^2}\), \[ F_s^\star=\frac{0.0210(500)}2=5.25\,\mathrm N, \] \[ X^\star=\frac{500}{4(13.5\times10^3)(9.8)} =9.45\times10^{-4}\,\mathrm m, \] and \[ \boxed{ W^\star= \frac{0.0210(500)^2}{2(13.5\times10^3)(9.8)} =1.98\times10^{-2}\,\mathrm J }. \]

Strictly, these values lie on the marginal switching boundary. They give the ideal limiting maximum, approached from the cycling side; reliable operation requires a small margin inside that boundary.

C.5

The volume relevant to atmospheric work includes the vacuum enclosed above the mercury. The liquid volume and the material volume of the vessels are constant, so \[ V=V_{\mathrm{constant}}+V_{\mathrm{vac}}, \qquad V_{\mathrm{vac}}=S_b(h_t+H_b-z_\ell). \]

Use the same approximation as in C.4: \[ S_b=S_c=S,\qquad S_t=0. \] The conservation and hydrostatic relations derived in C.1 then give \[ \delta z_s=-\frac{P_1}{2\rho g}, \qquad \delta z_\ell=\frac{P_1}{2\rho g}. \] Since \(\delta h_t=-x\), \[ \boxed{ V=V_{\mathrm{ref}}-Sx-\frac{S}{2\rho g}P_1 }. \]

The atmospheric work received by the system is \[ W_{\mathrm{pr}}=-\oint P_a\,dV. \] Substitution yields \[ W_{\mathrm{pr}} =S\oint P_a\,dx +\frac{S}{2\rho g}\oint P_a\,dP_1. \] The second integral vanishes over a closed cycle. Also \(P_0\oint dx=0\). Therefore \[ \boxed{W_{\mathrm{pr}}=S\oint P_1\,dx}. \]

At the limiting optimum, \(q=1\). The positive stroke occurs at \(P_1=+A\), with \(\Delta x=2X^\star\), while the negative stroke occurs at \(P_1=-A\), with \(\Delta x=-2X^\star\). Consequently, \[ W_{\mathrm{pr}}^\star =S\bigl[A(2X^\star)+(-A)(-2X^\star)\bigr] =4SAX^\star. \] Thus \[ \boxed{ W_{\mathrm{pr}}^\star=\frac{S_cA^2}{\rho g} =3.97\times10^{-2}\,\mathrm J }. \]

In a diagram with \(V\) horizontal and \(P_a\) vertical, the cycle is a parallelogram:

  • at each fixed stop, \(dV/dP_a=-S_c/(2\rho g)\), giving two parallel sloping branches;
  • their horizontal separation is \(2S_cX^\star\);
  • constant-pressure strokes connect them at \(P_0+A\) and \(P_0-A\);
  • the cycle is traversed counterclockwise, giving positive work received equal to its area, \[ (2A)(2S_cX^\star)=4S_cAX^\star. \]

Finally, \[ \boxed{ \frac{W^\star}{W_{\mathrm{pr}}^\star} = \frac{S_cA^2/(2\rho g)}{S_cA^2/(\rho g)} =\frac12 }. \] The other half becomes kinetic energy during the rapid strokes and is dissipated when the moving system is arrested at the stops.

The solution was checked against the official solution.