# How much work to sink top of a conical buoy to surface level?

A conical buoy weighing $$B$$ lbs with its vertex $$A$$ ft below the surface has its top sticking up $$\frac{A}{3}$$ ft above the surface. How much work to lower the top of the buoy to surface level??

If I add $$\frac{37B}{27}$$ lbs of weight to the top it causes a displacement of $$\frac{A}{3}$$. Work = (force)*(displacement) = $$\frac{37AB}{81}$$ ft-lbs. But I have the wrong answer. Any ideas?

Non-OP edit: This is problem $$65$$ of Additional Problems for Chapter $$7$$ of Simmons Calculus, with answer given of $$\frac{67}{324}AB$$.

• Submerged volume is 27/64 the total volume. Total volume minus submerged volume is equal to exposed volume which is 37/64 the total volume. Exposed volume is 37/27 the submerged volume. B = (water density)*(submerged volume)*(gravitational acceleration). Apr 16, 2023 at 11:17
• (37/27)B = (water density)*(37/27)*(submerged volume)*(gravitational acceleration) Apr 16, 2023 at 11:19
• Force of (37B/27)lbs causes a displacement of A/3 where the exposed volume is brought to surface level. multiply them together for 37AB/81 ft-lbs Apr 16, 2023 at 11:25
• What am I doing wrong? The book has the answer 67AB/324 Apr 16, 2023 at 11:27
• I thought the change in height of the water would be negligible in a big enough pool of water? Apr 16, 2023 at 11:46

Using integration tag provided by OP, the work W is expressed as $$W=\int_0^{\frac A3}F(x)dx$$

The issue is that the force is not constant: $$F(x)=\rho_wV(x)g-Bg, \rho_w:\text{ water density }, g:\text {gravitational acceleration}$$

The volume V(x) is expressed as $$V(x)=\pi R^2(x)h(x)=\pi R^2(x)(x+A), h(x)=x+A$$

At equilibrium, the submerged inverted cone has the radius r and the height A. Via scale ratio on similar triangles, the radius R(x) of the forcibly submerged cone becomes $$\frac{R(x)}{x+A}=\frac rA$$

At equilibrium, the volume of displaced water weighs as much as the buoy: $$\rho_w\pi r^2A=B\Rightarrow \frac rA=\sqrt{\frac{B}{\pi \rho_w A^3}}$$

Wrapping it back up: $$R(x)=(x+A) \sqrt{\frac{B}{\pi \rho_w A^3}}$$

$$V(x)=\pi R^2(x)(x+A)=(x+A)^3\frac{B}{\rho_w A^3}$$

$$F(x)=\rho_wV(x)g-Bg=Bg[\frac {(x+A)^3}{A^3}-1]$$

$$W=\int_0^{\frac A3}{F(x)dx}=Bg\int_0^{\frac A3}{\frac {(x+A)^3}{A^3}-1}dx$$

$$W=\frac{67}{324}ABg$$

• Just one minor correction: the final answer is $\frac{67}{324}AB,$ not $\frac{67}{324}ABg.$ In the problem statement, $B$ is weight (force exerted by gravity), not mass. Apr 17, 2023 at 3:50

If you assume that the change in height of the surface of the water is negligible (which is a reasonable assumption for the problem) then the work to fully submerge the buoy is the work required to lift the additional displaced water to the surface.

The additional displaced water is the water between a cone with its vertex $$A$$ feet below the surface and a similar cone with its vertex $$\dfrac43 A$$ below the surface.

We also know that the water displaced by the smaller cone has weight $$B.$$

The work required to lift a weight $$W$$ a vertical distance $$h$$ is $$Wh.$$ Let $$h$$ be the distance from a point in the cone to the surface. Then the total work required to lift a conical region of water with its vertex $$H$$ feet below the surface is

$$\int_0^H h \frac{\mathrm dw}{\mathrm dh} \, \mathrm dh$$

where $$\dfrac{\mathrm dw}{\mathrm dh}$$ is the rate at which we accumulate the weight of the cone of water as we account for its deeper portions. That is, $$\dfrac{\mathrm dw}{\mathrm dh}$$ is proportional to the square of the distance upward from the vertex of the cone, $$\dfrac{\mathrm dw}{\mathrm dh} = k(H - h)^2,$$ and integrating this over the height of the cone we should get the mass of water in the cone. For the smaller cone of water, $$H = A$$, so

$$B = \int_0^A \frac{\mathrm dw}{\mathrm dh} \, \mathrm dh = \int_0^A k(A - h)^2 \, \mathrm dh = k\frac{A^3}3$$

and therefore $$k = \dfrac{3B}{A^3}$$ and $$\dfrac{\mathrm dw}{\mathrm dh} = \dfrac{3B(A - h)^2}{A^3}.$$ Then the work required to lift the smaller cone of water to the surface is

$$\int_0^A h \frac{3B(A - h)^2}{A^3} \, \mathrm dh = \frac{3B}{A^3} \int_0^A h (A - h)^2 \, \mathrm dh = \frac{3B}{A^3} \cdot \frac{A^4}{12} = \frac14 AB,$$

which we could also have intuited by observing that the mass of water is $$B$$ and the average distance from the base of a cone is $$\dfrac14$$ the height of the cone, in this case $$\dfrac14 A.$$

We can perform similar integrations to obtain the work performed to lift the larger cone of water, or observe that the total weight of water is $$\dfrac{64}{27}B$$ and the average depth is $$\dfrac 14 \cdot \dfrac43 A = \dfrac13 A,$$ so the work to raise that volume of water is

$$\frac13 A \cdot \frac{64}{27}B = \frac{64}{81}AB.$$

The work to lift the water in the region between these two cones (which is what we were trying to compute in the first place) is simply the difference between these two quantities of work,

$$\frac{64}{81}AB - \frac14 AB = \frac{175}{324}AB.$$

So this is the total amount of work done on the water while pushing the buoy down. But the buoy itself does work equal to its own weight times the distance moved downward. The work done by the buoy is therefore $$\dfrac13AB.$$ The answer to the question is the additional work that must be done to displace the water,

$$\frac{175}{324}AB - \frac13AB = \frac{67}{324}AB.$$