Chapter 2, Problem 2.125P

To determine

The height of the cylindrical log visible above the water surface.

Answer to Problem 2.125P

Height of the log visible or above the water surface = 0.203 m = 7.99 inches.

Explanation of Solution

Given Information:

Diameter of the log = 24 inches = 0.61 m

Density of the log = 710 kg/m³

Density of the water = 1000 kg/m³

Log is floating in the fresh water.

In the adjacent figure assume the circle represents the cross section of the log, AB is the water surface.

Let H be the height of the log in the water

$\begin{array}{l}ForceofBuoyance=Weightofthelog\\ Weightofthe\log =(Volumeofthe\log \times density\times g)\\ ForceofBuoyance=(Volumeofthewaterdisplaced\times densityofwater\times g)\end{array}$

Calculation:

$\begin{array}{l}Volumeofthe\log =(\frac{\pi }{4}{D}^2\times L)\\ \angle ACB=2\theta \\ AB=2(\frac{D}{2})\sin \theta \\ Areaofcrosssetionoftheloginwater=(Areaofthe△ACB)+(AreaoftheArcAB)\\ Areaofthe△ACB=\frac{1}{2}\times AB\times (\frac{D}{2}\cos \theta )\\ AreaoftheArcAB=\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360})\\ Totalareaoftheloginwater=\frac{1}{2}\times AB\times (\frac{D}{2}\cos \theta )+\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360})\\ Totalareaoftheloginwater=\frac{D^2}{4}\cos \theta \sin \theta +\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360})\end{array}$

$\begin{array}{l}Volumeofthewaterdisplaced=\frac{D^2}{8}\sin 2\theta +\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360}))L\\ Buoyanceforce\text{=}(\frac{D^2}{8}\sin 2\theta +\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360}))L\times 1000\times 9.81\\ Weightofthe\log =(\frac{\pi }{4}{D}^2\times L)\times 710\times 9.81\\ (\frac{\pi }{4}{D}^2\times L)\times 710\times 9.81=(\frac{D^2}{8}\sin 2\theta +\frac{\pi }{4}{D}^2\times (\frac{360-2\theta }{360}))L\times 1000\times 9.81\\ 0.2074=0.0465\sin 2\theta +0.2921(1-0.00555\theta )\\ 0.0016211\theta =0.0465\sin 2\theta +0.0847\\ \theta =28.68\sin 2\theta +52.2454\\ \theta -28.68\sin 2\theta -52.2454=0\\ for\theta =60,60-28.68\sin 120-52.2454=-17.08\\ for\theta =70,70-28.68sin140-52.2454=-0.79\\ for\theta =70.4,70.4-28.68\sin 140.8-52.2454=0.027\\ \theta ={70.4}^o\\ H=\frac{D}{2}+\frac{D}{2}\cos \theta \\ H=0.305+0.305\cos 70.4\\ H=0.407m\\ Heightofthelogvisible=D-H\\ =0.61-0.407m\\ Heightofthelogvisible=0.203m\end{array}$

Conclusion:

Height of the log visible or above the water surface = 0.203 m = 7.99 inches.