Chapter 2, Problem 2.121P

To determine

The criteria for the beam to float exactly on its diagonal.

Answer to Problem 2.121P

The beam will float on its surface for $D={(\frac{Lhb}{(\pi (SG-1))})}^{\frac{1}{3}}$, ${\gamma }_b=\frac{\gamma }{3}$.

Explanation of Solution

Given Information:

Let diameter of the sphere be D

Dimensions of the box − length-L, breadth − b, height -h

Specific weight of the box = γ_b

Specific weight of the water = γ

Specific gravity of the sphere = SG > 1

$D={(\frac{Lhb}{(\pi (SG-1))})}^{\frac{1}{3}}$

. $\begin{array}{l}Volumeofthebox=Lbh\\ Volumeoftheboxinthefluid=\frac{1}{2}Lbh\\ Weightofthebox=Volumeofbox\times {\gamma }_b\\ \text{ =}Lbh{\gamma }_b\\ Buoyancyforce=Volumeofthefluiddisplaced\times \gamma \\ =\frac{1}{2}Lbh\gamma \\ Tensionintherope=(Weightofthesphere-BuoyanceforceontheSphere)\\ =V_s\times \gamma \times (SG-1)\\ Forboxtobeinequilbrium=Totalupwardforces=Totaldownwardforces\end{array}$

Calculation:

$VolumeoftheSphere=\frac{4\pi }{24}D$

$=\frac{4\pi }{24}\frac{Lhb}{(\pi (SG-1))}$

$\begin{array}{l}Totalupwardforces=\frac{1}{2}Lbh\gamma \\ Totaldownwardforces=Lbh{\gamma }_b+V_s\times \gamma \times (SG-1)\\ \text{ =}Lbh{\gamma }_b+(\frac{4\pi }{24}\frac{Lhb}{(\pi (SG-1))}\times \gamma \times (SG-1))\\ =Lbh({\gamma }_b+(\frac{1}{6}\times \gamma )\\ \frac{1}{2}Lbh\gamma =Lbh({\gamma }_b+(\frac{1}{6}\times \gamma )\\ {\gamma }_b=\gamma (\frac{1}{2}-\frac{1}{6})\\ {\gamma }_b=\frac{\gamma }{3}\end{array}$

Conclusion:

Hence, proved for the given diameter of the sphere ${\gamma }_b=\frac{\gamma }{3}$.