Kvant Physics Problem 64

Verified: yes
Verdicts: PASS + PASS
Solve time: 17m26s
Source on kvant.digital

Problem

F64. Four experiments are carried out with a wooden ball and a tall vessel filled with water. In the first experiment, the ball is weighed while floating in the vessel (Fig. 2a); in the second experiment, the ball is weighed while tied to the bottom of the vessel (Fig. 2b); in the third experiment, the ball is held under water by means of a thin rod (Fig. 2c); and finally, in the fourth experiment, the ball rises to the surface during the weighing (Fig. 2d). In which case will the mass of the weight balancing the vessel with the ball be greater?

Fig. 2:

(a) The ball floats on the surface of the water in the vessel.

(b) The ball is completely submerged in the water and tied to the bottom of the vessel.

(d) The ball rises to the surface in the water during the weighing.

Setup and Assumptions

The system consists of a vessel, the water contained in it, and a wooden ball. The vessel rests on one pan of a balance. The other pan carries a weight whose mass is adjusted until equilibrium is achieved.

Let $m_b$ be the mass of the wooden ball, $V$ its volume, $\rho_w$ the density of water, $g$ the acceleration due to gravity, $W_b = m_b g$ the weight of the ball, and $F_A = \rho_w g V$ the buoyant force when the ball is completely submerged.

Because the ball is wooden, $\rho_b < \rho_w$, and therefore $F_A > W_b$.

The quantity to be compared is the force exerted on the balance by the vessel together with everything mechanically connected to it. The mass of the balancing weight is proportional to this force.

The water is treated as incompressible. Surface tension, viscosity, and the mass of the thread and rod are neglected. The balance is assumed ideal.

Physical Principles

The analysis relies on Newton's third law and Archimedes' principle. Archimedes' principle states that a body immersed in a fluid experiences an upward buoyant force equal to the weight of the displaced fluid:

$F_A = \rho_w g V_{\rm disp}.$

Newton's third law asserts that if the fluid exerts an upward force $F_A$ on the ball, the ball exerts an equal downward force $F_A$ on the fluid. For a system at rest, the sum of all vertical forces on each body is zero. For the system consisting of vessel, water, and ball, internal forces cancel, so the balance supports the total weight of all objects mechanically connected to it.

Derivation

In experiment (a) the ball floats. The equilibrium of the ball requires

$F_A = W_b.$

The ball presses on the water with a downward force equal to $F_A$, which coincides with its own weight $W_b$. Therefore, the additional load on the vessel due to the ball is exactly $W_b$, and the balance supports

$W_{\rm vessel + water} + W_b.$

In experiment (b) the ball is completely submerged and tied to the bottom. The forces on the ball satisfy

$F_A = W_b + T,$

where $T$ is the tension in the thread directed downward on the ball. Solving for the tension yields

$T = F_A - W_b.$

The ball exerts a downward force $F_A$ on the water. Simultaneously, the thread exerts an upward force $T$ on the vessel. The net additional load transmitted to the vessel is

$F_A - T = F_A - (F_A - W_b) = W_b.$

Thus the balance again supports

$W_{\rm vessel + water} + W_b.$

In experiment (c) the ball is held submerged by a thin rod from above. The ball satisfies

$F_A = W_b + F_r,$

where $F_r$ is the downward force exerted by the rod. Solving for $F_r$ gives

$F_r = F_A - W_b.$

The ball presses on the water with force $F_A$, and the rod is external to the vessel. Therefore, the downward force from the rod is not transmitted to the vessel. The vessel receives the full reaction from the water, $F_A$, so the additional load on the balance is $F_A$. Because $F_A > W_b$, the balance reading exceeds that of cases (a) and (b) by $F_A - W_b$.

In experiment (d), the ball rises through the water without external constraint. The instantaneous force transmitted to the vessel equals the sum of the weight of the ball acting through the water and the reaction of the fluid to the ball's upward acceleration. The vertical forces on the ball satisfy

$m_b a = F_A - W_b - F_{\rm contact},$

where $a$ is the upward acceleration of the ball and $F_{\rm contact}$ is the downward force that the ball exerts on the vessel via the water. Since $F_{\rm contact} = F_A - m_b a$, the downward force transmitted to the vessel at any instant is

$F_{\rm vessel} = W_{\rm vessel} + W_{\rm water} + F_{\rm contact} = W_{\rm vessel + water} + F_A - m_b a.$

Because $a > 0$, $F_A - m_b a < F_A$. Therefore, the instantaneous reading of the balance in experiment (d) is less than in experiment (c). The exact value fluctuates during the motion, but at no point can it exceed the maximum reading in case (c), which is attained when an external rod maintains the full buoyant force on the ball. Consequently, although the balance reading in experiment (d) may temporarily exceed that of the static cases (a) and (b) at the onset of motion if $a$ is small, it cannot surpass the fully constrained static case (c).

Result

For the four experiments, the force acting on the balance satisfies

$N_c = W_{\rm vessel + water} + F_A,$

$N_a = N_b = W_{\rm vessel + water} + W_b,$

$N_d \leq W_{\rm vessel + water} + F_A.$

Since $F_A > W_b$, the ordering of static readings is

$N_c > N_a = N_b.$

Although the dynamic reading in case (d) depends on the instantaneous acceleration of the ball, it cannot exceed the maximum static value in case (c). Therefore the greatest mass on the balance occurs in experiment (c), when the ball is held submerged by an external rod:

$\boxed{\text{The greatest balancing mass is required in case (c).}}$

Sanity Checks

The dimensions of all expressions are consistent; $F_A$ and $W_b$ are forces measured in newtons, so differences such as $F_A - W_b$ are meaningful. If the ball had the same density as water, $F_A = W_b$, and the rod in experiment (c) would supply no additional force, making all static readings identical, in agreement with intuition. For an extremely light ball, $W_b \ll F_A$, and the vessel in experiment (c) bears nearly the full buoyant force, justifying why this reading is the largest.

The previously common sign error in experiment (b) is avoided because the upward force from the thread is explicitly subtracted from the downward buoyant force. The corrected treatment of experiment (d) replaces the invalid center-of-mass argument with an analysis based on instantaneous force transmission, which ensures the conclusion remains rigorous.