Question

A man whose mass is $\mathbf{80}\mathbf{}\mathit{k}\mathit{g}$ and a woman whose mass is $\mathbf{50}\mathbf{}\mathit{k}\mathit{g}$sit at opposite ends of a canoe $\mathbf{5}\mathbf{}\mathit{m}$ long, whose mass is $\mathbf{30}\mathbf{}\mathit{k}\mathit{g}$. (a) Relative to the man, where is the mass of the system consisting of man-woman, and canoe? (Hint: Choose a specific coordinate system with a specific origin.) (b) Suppose that the man moves quickly to the center of the canoe and sits down there. How far does the canoe move in the water? Explain your work and your assumptions.

Short answer

1. The center of mass of the system consisting of man-woman, and canoe, Relative to the man is $\mathbf{2}\mathbf{.}\mathbf{03}\mathbf{}\mathit{m}$
2. The canoe moves $1.72m$ in the downward direction.

Step-by-step solution

Identification of given data

• The mass of a man is $m_1=80\mathrm{kg}$
• The mass of a woman is $m_2=50\mathrm{kg}$
• The mass of a canoe is $m_3=30\mathrm{kg}$
• The length of a canoe is $5m$ long

Concept of the mass of the system

The mass of the system is defined by considering the average positions of all the objects acting in the system.

(a) Determination of the mass of the system consisting of man-woman, and canoe, Relative to the man

The center of mass of the system can be,

$\mathit{M}{\mathit{x}}_{\mathbf{C}\mathbf{M}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{m}}_{\mathbf{3}}{\mathit{x}}_{\mathbf{3}}$

Where,

$$\begin{gathered} x_1=0m,x_2=2.5m,x_3=5m \\ M=m_1+m_2+m_3 \\ =80+50+30 \\ =160kg \end{gathered}$$

Substitute these values in above expression,

$$\begin{gathered} M{x}_{CM}=m_1{x}_1+m_2{x}_2+m_3{x}_3 \\ =\left[(80\times 0)+(30\times 2.5)+(50\times 5)\right](kg·m) \\ =\frac{325}{M}·\left(\frac{1kg·m}{1kg}\right) \end{gathered}$$

Substituting M values in above expression

$$\begin{gathered} x_{CM}=\frac{325}{160}·\left(\frac{1kg·m}{1kg}\right) \\ =2.03·\left(1m\right) \\ =2.03m \end{gathered}$$

Hence, the center of mass of the system is$2.03m$

(b) Determination of the new center of mass

We know that,

$$\begin{gathered} M{x}_{CM}=m_1{x}_1+m_2{x}_2+m_3{x}_3 \\ =(80\times 2.5) \\ M{x}_{CM}=\frac{200}{M}·\left(\frac{1kg·m}{1kg}\right) \end{gathered}$$

Substituting M values in above expression

$$\begin{gathered} x_{CM}=\frac{200}{160}\left(\frac{1kg·m}{1kg}\right) \\ {x}_{CM}=1.25m \end{gathered}$$

The distanceto find how the canoe moved will be foundby;

$$\begin{gathered} =2.5m+1.25m-2.03m \\ =1.72m \end{gathered}$$

Here the negative sign indicates the downward movement of the canoe. The canoe moves $1.72m$in the downward direction in the water when man moves quickly to the center of the canoe and sits down there.

Hence, the canoe moves $1.72m$in the downward direction.