In Figure a,4.5 kg dog stand on the 18 kg flatboat at distance D = 6.1 m from the shore. It walks 2.4 mThe distance between the dog and shore is . along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore. (Hint: See Figure b.)

The mass of the dog is$m_1=4.5kg$ .

The mass of the flatboat is$m_2=18kg$.

The distance between flatboat and the shore is $D=6.1m$.

The displacement of dog relative to the boat is d = 2.4 m.

The center of gravity is a position defined relative to an object or system of objects. It is the average position of all parts of the system, weighted by their weight. For simple rigid objects of uniform density, the center of mass is located at the center of gravity.

You can use the concept of the center of mass of the system.

Formula:

$\overrightarrow{R_{cm}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2}$

Here, $\overrightarrow{R_{cm}}$ is the distance to the center of mass, $\overrightarrow{r_1}$is the distance of mass $m_1$, and $\overrightarrow{r_2}$is the distance of mass $m_2$.

The center of mass of the system is not moving. Hence, the center of mass of the system is,

$$\begin{gathered} \overrightarrow{R_{cm}}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2} \\ 0=\frac{-m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}}{m_1+m_2} \\ -m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}=0 \\ {m}_1\overrightarrow{r_1}=m_2\overrightarrow{r_2} \\ \overrightarrow{r_2}=\left(\frac{m_1}{m_2}\right)\overrightarrow{r_1} \end{gathered}$$

Where, $\overrightarrow{r_2}$and $\overrightarrow{r_2}$are the distances of the dog and the boat from the center of mass of the system respectively.

The dog walks relative to the boat, hence

$$\begin{gathered} \overrightarrow{r_1}+\overrightarrow{r_2}=\overrightarrow{d} \\ \overrightarrow{r_1}+\left(\frac{m_1}{m_2}\right)\overrightarrow{r_1}=\overrightarrow{d} \\ \overrightarrow{r_1}+\left[1+\left(\frac{m_1}{m_2}\right)\right]=\overrightarrow{d} \\ \overrightarrow{r_1}=\frac{\overrightarrow{d}}{\left[1+\left(\frac{m_1}{m_2}\right)\right]} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} \overrightarrow{r_1}=\frac{2.4m}{\left[1+\left(\frac{4.5kg}{18kg}\right)\right]} \\ =1.92m \end{gathered}$$

The dog is $\overrightarrow{r_1}$closer to the shore than initially. Hence, the distance between the dog and shore is,

$$\begin{gathered} x=D-\overrightarrow{r_1} \\ =6.1m-1.92m \\ =4.2m \end{gathered}$$

Hence, the distance between the dog and shore is 4.2 m .