Solved Problem on Center of Mass

A man of mass m is sitting on the stern of a boat at rest on a lake. The mass of the boat is M = 3m, and its length is L. The man get up and walks toward the bow. Neglecting the water resistance, determine the distance D that the boat displaces during the displacement of the man from stern to bow.

Problem data:

Mass of man: m;
Mass of boat: M = 3m;
Length of boat: L.

Solution

The position of the center of mass for a system of two bodies is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}} \tag{I} \end{gather} \]

In the first situation (Figure 1), we have a reference frame pointing to the right, with the origin at the back of the boat where the man is sitting. "Forgetting" the boat and the man, and considering only its center of mass, m to the center of mass of the man, and B to the center of mass of the boat. The center of mass of the man is at the origin of the reference frame, x_h = 0, and the center of mass of the boat, length L, is at half its length \( x_{B}=\frac{L}{2} \). Thus substituting these values and the masses given in the problem in expression (I), we have the center of mass of the man-boat system in the initial situation

\[ \begin{gather} x_{i}=\frac{mx_{h}+Mx_{B}}{m+M}\\[5pt] x_{i}=\frac{m\times 0+3m\dfrac{L}{2}}{m+3m}\\[5pt] x_{i}=\frac{3\cancel{m}\dfrac{L}{2}}{4\cancel{m}}\\[5pt] x_{i}=\frac{3L}{8} \tag{II} \end{gather} \]

Figure 1

In the second situation (Figure 1), the man walked the length L of the boat, while the boat displaces a certain distance D backward. Thus the center of mass of the man is at the position \( x_{h}=L-D \), and the center of mass of the boat is in \( x_{B}=\dfrac{L}{2}-D \), again substituting these values in the expression (I) for the final situation

\[ \begin{gather} x_{f}=\frac{mx_{h}+Mx_{B}}{m+M}\\[5pt] x_{f}=\frac{m(L-D)+3m\left(\dfrac{L}{2}-D\right)}{m+3m}\\[5pt] x_{f}=\frac{mL-mD+3m\dfrac{L}{2}-3mD}{4m}\\[5pt] x_{f}=\frac{\dfrac{2mL-2mD+3mL-6mD}{2}}{4m}\\[5pt] x_{f}=\frac{5\cancel{m}L-8\cancel{m}D}{8\cancel{m}}\\[5pt] x_{f}=\frac{5L-8D}{8} \tag{III} \end{gather} \]

The problem tells us that the boat is at rest in the initial situation, as the center of mass of the man-boat system does not move, equating expressions (II) and (III)

\[ \begin{gather} x_{i}=x_{f}\\[5pt] \frac{3L}{\cancel{8}}=\frac{5L-8D}{\cancel{8}}\\[5pt] 3L=5L-8D\\[5pt] 8D=5L-3L\\[5pt] 8D=2L\\D=\frac{2L}{8} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {D=\frac{L}{4}} \end{gather} \]

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