Question

Why is the following situation impossible? A barge is carrying a load of small pieces of iron along a river. The iron pile is in the shape of a cone for which the radius$r$of the base of the cone is equal to the central height h of the cone. The barge is square in shape, with vertical sides of length $2r$, so that the pile of iron comes just up to the edges of the barge. The barge approaches a low bridge, and the captain realizes that the top of the pile of iron is not going to make it under the bridge. The captain orders the crew to shovel iron pieces from the pile into the water to reduce the height of the pile. As iron is shoveled from the pile, the pile always has the shape of a cone whose diameter is equal to the side length of the barge. After a certain volume of iron is removed from the barge, it makes it under the bridge without the top of the pile striking the bridge.

Short answer

The situation is impossible because lowering the height of the iron pile on the barge while keeping the base radius the same results in the top of the pile rising higher above the water level.

Step-by-step solution

A concept:

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho }_{fluid}g{V}_{disp}$

Here, $B$is the buoyant force,${\rho }_{fluid}$ is the density of the fluid, $V_{disp}$is the volume displaced, and $g$is the gravitational acceleration constant.

Why is the given situation impossible:

Assume the top of the barge without the pile of iron has height $H_o$above the surface of the water. When a mass of iron $M_{Fe}$is added to the barge, the barge sinks a distance $\Delta H$until the buoyant force from the water equals the additional weight of the iron. The barge is a square with sides of length$L$,so the volume of displaced water is$L^2\Delta H$ , and the buoyant force supporting the extra weight is,

$B=\left({\rho }_w{L}^2\Delta H\right)g=M_{Fe}g$

Where, ${\rho }_w$is the density of water.

The scrap iron pile has the shape of a cone, and the volume of a cone of base radiusand central height$h$is$V_{cone}=\frac{\pi R^{2}h}{3}$ ; therefore, the mass of the iron is $M_{Fe}=\frac{{\rho }_{Fe}\pi R^{2}h}{3},$where ${\rho }_{Fe}$is the density of iron. We find the distance the barge sinks with a pile of iron:

role="math" localid="1663778348027" $$\begin{gathered} B=\left({\rho }_w{L}^2\Delta H\right)g=M_{Fe}g \\ \left({\rho }_w{L}^2\Delta H\right)g=\left(\frac{{\rho }_{Fe}\pi R^{2}h}{3}\right)g \\ \Delta H=\frac{{\rho }_{Fe}}{{\rho }_w}.\frac{\pi }{3}.\frac{R^2}{L^2}h \end{gathered}$$

If the iron is piled to a height$h$, the barge will sink by the distance$\Delta H$ , so the distance from the water level to the top of the iron pile is:

$$\begin{gathered} D_{top}=H_o-\Delta H+h \\ =H_o-\left(\frac{{\rho }_{Fe}}{{\rho }_w}\right)\frac{\pi }{12}r+r \end{gathered}$$

For the situation of the problem, side$L=2r$ , and the initial conical pile of scrap iron has radius$R=r$ and height is $h=r$.

The distance the barge sinks is,

$$\begin{gathered} D_{top}=H_o-\left(\frac{7.86\times {10}^3}{1.00\times {10}^3}\right)\frac{\pi }{12}r+r \\ =H_o-2.06r+r \\ =H_o-1.06r \end{gathered}$$

And the height of the top of the pile above the water is,

$D_{top}=H_o-1.06r⩾D_{bridge}$

When the pile is reduced to a height h′, but still with the same base radius $R=r$, the distance the barge sinks is,

$$\begin{gathered} \Delta H=\left(\frac{{\rho }_{Fe}}{{\rho }_w}\right)\frac{\pi }{12}h\text{'} \\ =\frac{7.86\times {10}^3}{1.00\times {10}^3}\times \frac{3.14}{12}h\text{'} \\ =2.06h\text{'} \end{gathered}$$

The height of the top of the pile above the water is now

$$\begin{gathered} D_{top}=H_o-\Delta H+h\text{'} \\ =H_o-2.06h\text{'}+h\text{'} \\ =H_o-1.06h\text{'} \end{gathered}$$

But this means the top of the pile is now higher! To check this, recall that the height of the pile is reduced, so $h\text{'}<r$:

$$\begin{gathered} D{\text{'}}_{top}>D_{top} \\ {H}_o-1.06h\text{'}>H_o-1.06r \\ 1.06h\text{'}>-1.06r \\ h\text{'}<r \end{gathered}$$

Which is true.

The situation is impossible because lowering the height of the iron pile on the barge while keeping the base radius the same results in the top of the pile rising higher above the water level.