# Force on curved surface of submerged cylinder

A cylindrical wooden log of length L and radius r is floating on water(density d) completely submerge find the force on lower curved surface of the cylinder

I tried integrating the lower surface by breaking into layers of Ldh area (h the level from top layer,dh the elemental length) the force came dghLdh now it's symmetric on both sides of the lower surface so the horizontal components cancel and the vertical add . On finalizing i found $$F=\int{2dgLcosθhdh}$$ where cosθ=h/r and limit of h runs from r to 2r but the answer doesn't match. The given answer is $$dgLr²(\frac{π}{2}+2)$$. Can anyone please help in where i am wrong and help in finding the correct answer.

$$F_{buoyant}=d\pi r^2 Lg \\ F_{upper}=d\left(2r^2L-\frac{\pi}{2}r^2L \right)$$ ($$F_{upper}$$ is due to weight of water over the cylinder) $$F_{lower} = F_{buoyant}+ F_{upper}\\ =d\left(2r^2L+\frac{\pi}{2}r^2L \right) \\ =d r^2L \left(2+\frac{\pi}{2} \right)$$

• Can you clarify more on F(upper), in the given figure above the lower half of cylinder the upper half is present not any water.
– user262559
Commented Jan 2, 2021 at 8:39
• @AplusB - The cylinder in the diagram is placed as such the water surface is tangential to the upper curved surface of the cylinder. Or I should say, cylinder is just immersed in the water. So, in the cross-sectional view, cut a square of side = 2r enclosing the circular face of cylinder. The region enclosed within the circular boundaries would be cylinder's part, and rest would be water, stretched across length L. Some water would be above the horizontal diameter of cylinder, and a lot would be below it. As the buoyancy is the net force applied by the water, you can calculate the rest. Commented Jan 2, 2021 at 8:52
• Yes,i understood , this part now thanks for the answer and clarification
– user262559
Commented Jan 2, 2021 at 8:57

Probably try integrating over $$d\theta$$ instead of $$dh$$. So that you can write the differential area as $$lrd\theta$$, and $$h$$ as $$r +r\cos\theta$$, and then proceed the same way for force as you did.

• Hello Vivek, I've edited your answer to write the formulae in MathJax. For future reference, you may find this MathJax tutorial useful.
– Urb
Commented Jan 2, 2021 at 10:35

Your exposition was sufficiently disorganized that I couldn't really tell what you were trying to do. Here's my take on the problem: set up the coordinate system so that the $$z$$-axis runs along the central axis of the log, $$y$$ is up and $$x$$ to the right as shown in Figure $$1$$.
Here the pressure is $$p=\rho gd$$ marked in magenta in Figure $$1$$ where $$\rho$$ is the density of the fluid, $$g$$ is the acceleration of gravity, and the depth $$d=r-r\sin\theta$$ where $$r$$ is the log radius and also the depth of the origin of coordinates below the surface marked in blue. $$\theta$$ is the usual polar angle, $$-\pi$$ on the left edge, $$-\pi/2$$ at the bottom, and $$0$$ on the right edge.
Then the element of area is $$d^2A=rd\theta dz$$ so the force on this element is $$d^2F=pd^2A=\rho gr^2(1-\sin\theta)d\theta dz$$ and the upward force is $$d^2F_y=-\sin\theta d^2F=\rho gr^2(\sin\theta-1)\sin\theta d\theta dz$$ so now we have $$F_y=\int_0^L\int_{-\pi}^0\rho gr^2(\sin\theta-1)\sin\theta d\theta dz=\rho gr^2\left(\frac12\pi+2\right)L$$ Another way to is consider Archimedes' principle: the force on the top of the section is easy to compute because the pressure is a constant $$\rho gr$$ and the inward normal is always down, so the net pressure force on the log is $$F_y-\rho gr\cdot2rL=\rho g\frac{\pi}2r^2L$$ because if the log were just more water it wouldn't be accelerated so the net pressure force must be equal to the weight of the displaced water. The pressure force from the two ends cancels out of course. This checks with the answer obtained by direct integration.

• can you explain in the second part why have you taken $ρgr$ as pressure on top of lower curved surface because it's covered by upper half surface over which any layer of water would be present. Thanks for the solution
– user262559
Commented Jan 3, 2021 at 6:27
• My claim is that cutting off the top half of the log doesn't have any effect on the pressure force on the lower half, so I did just that to create a situation where the net pressure force on the bottom is easier to calculate via Archimedes' principle. Commented Jan 3, 2021 at 20:03