Chapter 1, Problem 1.89PAE

1.89 Imagine that you place a cork measuring $1.30\text{ cm}\times \text{4}\text{.50 cm}\times \text{3}\text{.00 cm}$ in a pan of water. On top of this cork, you place a small cube of lead measuring 1.15 cm on a side. Describe how you would determine if the combination of the cork and lead cube will still float in the water. Note any information you would need to look up to answer the question.

Interpretation Introduction

Interpretation:

Write down the steps to determine the combination of the cork and lead cube that will still float in the water.

Concept introduction:

Answer to Problem 1.89PAE

Weight of system < maximum possible buoyant force

The system will float otherwise sink

Buoyant force $=$ weight of displaced water

Solution

$\begin{array}{l}\left(1.30\times {10}^{-2}\times 4.50\times {10}^{-2}\times 3.00\times {10}^{-2}\right)\times {\rho }_{cork}\times g\end{array}$

$\begin{array}{l}+{\left(1.15\times {10}^{-2}\right)}^3\times {\rho }_{lead}\times g<\end{array}$

$\begin{array}{l}\left(1.30\times {10}^{-2}\times 4.50\times {10}^{-2}\times 3.00\times {10}^{-2}{m}^3+{\left(1.15\times {10}^{-2}\right)}^3\right)\\ \times {\rho }_{water}\times g\end{array}$

If above relation is valid then cork and lead system will float otherwise sink

Information required:

1. Density of cork
2. Density of lead
3. Density of water

Given info:

Volume of cork $=1.30cm\times 4.50cm\times 3.00cm$

$=\left(1.30\times 4.50\times 3.00\right)\times {10}^{-6}{m}^3$

Volume of lead $={\left(1.15cm\right)}^3$

$\begin{array}{l}={\left(1.15\times {10}^{-2}\right)}^3{m}^3\\ ={\left(1.15\right)}^3\times {10}^{-6}{m}^3\end{array}$

Explanation of Solution

If weight of cork $+$ lead system < maximum buoyant force then float.

Weight of cork $=$ volume of cork $\times$ density of cork $\times$ g $=\left(1.30\times 4.50\times 3.00\right)\times {10}^{-6}{m}^3\times {\rho }_{cork}\times 9.80m/s^2$

Weight of lead $={\left(1.15\times {10}^{-2}m\right)}^3\times$ Density of lead $\times$ g

$={\left(1.15\right)}^3\times {10}^{-6}{m}^3\times {\rho }_{lead}\times 9.80m/s^2$

Maximum buoyant force $=\left(\text{Valume of cork + lead system}\right)$

$\times {\rho }_W\times g$

$=\left(1.30\times 4.50\times 3.00+{\left(1.15\right)}^3\right)\times {10}^{-6}{m}^3{\rho }_w\times 9.80m/s^2$

If maximum buoyant force is greater than weight of cork $+$ lead system, then lead $+$ cork system float otherwise sink.

System float only when.

$\begin{array}{l}\left(1.30\times 4.50\times 3.00\right)\end{array}$

$\begin{array}{l}\times {10}^{-6}{m}^3\times {\rho }_{cork}\times 9.80m/s^2+{\left(1.15\right)}^3\times {10}^{-6}{m}^3\end{array}$

$\begin{array}{}\times {\rho }_{lead}\times 9.80m/s^2<\left(1.30\times 4.50\times 3.00+{\left(1.15\right)}^3\right)\end{array}$

$\begin{array}{}\times {10}^{-6}{m}^3\times {\rho }_w\times 9.80m/s^2\end{array}$

To find this value you have to find out density of cork and lead.

Conclusion

If density of lead and cork are known then you can find out lead $+$ cork system will float or sink. 1