Chapter 10, Problem 78GP

To determine

To Find: The number of people to hold the raft.

Answer to Problem 78GP

The number of people required to hold the raft before they start getting their feet wet is $88$

Explanation of Solution

Given info:

Number of logs $\text{= }10$

Diameter of log $\text{, }d=56\text{ cm = }0.56\text{ m}$

Radius of log $r=\frac{d}{2}\text{ = }0.28\text{ m}$

Length of log $l=6.1\text{ m}$

Formula used:

Write the expression for the water displaced by the log

  $W_{w_{\log }}=W_{\text{people on log}}+W_{\log }$

Write the expression for the weight.

  $W=m\times g$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{W}_{w_{\log }}=W_{peopleon\log }+W_{\log }\\ W=m\times g\\ 10\times m_{w_{\log }}\times g=(N)\times m_p\times g+10\times m_{\log }\times g\\ N=\frac{\left(10\times m_{w_{\log }}\times g\right)-\left(10\times m_{\log }\times g\right)}{m_p\times g}\\ N=\frac{\left(10\times {\rho }_w\times V_{\log }\right)-\left(10\times V_{\log }\times {\rho }_{\log }\right)}{m_p}\\ N=\text{1}0\times V_{\log }\times \frac{\left({\rho }_w-{\rho }_{\log }\right)}{m_p}\\ N=10\times \pi \times r_{\log }^2\times l_{\log }\times {\rho }_w\times \frac{\left(1-S{G}_{wood}\right)}{m_p}\\ N=10\times \pi \times {0.28}^2\times 6.1\times 1000\times \frac{\left(1-0.6\right)}{68}\\ N=88\text{ peoples}\end{array}$

Conclusion:

Thus, total $88$ people required to hold the raft before they start getting their feet wet.