Chapter 5, Problem 36P

(a)

To determine

To Evaluate: The distance the boy from the dock when he reaches the bow.

Answer to Problem 36P

3.6 m

Explanation of Solution

Given:

Mass of boy is $60\text{ kg}$ .

Mass of canoe is $40\text{ kg}$ .

Distance between the boy and bow of canoe is $6.0\text{ m}$ .

Center of canoe is $3.0\text{ m}$ from the bow.

Formula used:

The x -coordinate position of center of mass can be calculated by the following formula:

  $X_{cm}=\frac{(x_b{m}_b+x_c{m}_c)}{(m_b+m_c)}$

Here $x_b\text{ and }{x}_c$ are the distances of boy from bow and distance of center of canoe from bow respectively, while $m_b\text{ and }{m}_c$ are the masses of boy and canoe respectively.

Calculation:

The x -coordinate for the center of mass is calculated as follows:

  $X_{cm}=\frac{(x_b{m}_b+x_c{m}_c)}{(m_b+m_c)}$

Substituting $m_b=60\text{ kg, }{x}_b=6.0\text{ m},m_c=40\text{ kg and }{x}_c=3.0\text{ m}$

  $\begin{array}{l}{X}_{cm}=\frac{(6.0\text{ m}\times 60.0\text{ kg)+(3}\text{.0 m}\times 40.\text{0 kg})}{(60.0\text{ kg + 40}\text{.0 kg})}\\ \text{ = 4}\text{.8 m}\end{array}$

Now assuming that the boy moves by a distance of $x$ , then the distance of bow from the center of canoe can be given by the following formula:

  $x_c$ ' $=3.0m+x$

as the boy moves towards the bow, the center of mass of the system would be as follows:

  $X_{cm}=\frac{(x_b{}^{\prime }{m}_b+x_c{}^{\prime }{m}_c)}{(m_b+m_c)}$

Substituting: $X_{cm}=4.8\text{ m},m_b=60.0\text{ kg},m_c=40.0\text{ kg},x=x_b{}^{\prime }\text{ and }{x}_c{}^{\prime }=(3.0\text{ m}+x)$

  $\begin{array}{l}4.8\text{ m}=\frac{(x\times 60.0\text{ kg})+(3.0\text{ m}+x)\times 40.0\text{ kg}}{(60.0\text{ kg}+40.0\text{ kg})}\\ 480\text{ kg}\cdot \text{m}=x(60.0\text{ kg})+(3.0\text{ m}+x)40.0\text{ kg}\\ x=\frac{480\text{ kg}\cdot \text{m}-120\text{ kg}\cdot \text{m}}{100\text{ kg}}\\ \text{ = 3}\text{.6 m}\end{array}$

Conclusion:

As the boy moves towards the bow, the distance between the boy and dock is 3.6 m.

(b)

To determine

Whether the canoe is moving or not.

Explanation of Solution

Introduction:

The x-coordinate position of center of mass can be calculated by the following formula:

  $X_{cm}=\frac{(x_b{m}_b+x_c{m}_c)}{(m_b+m_c)}$

Here $x_b\text{ and }{x}_c$ are the distances of boy from bow and distance of center of canoe from bow respectively, while $m_b\text{ and }{m}_c$ are the masses of boy and canoe respectively.

It has been shown that there is no external force acting on the system, and no friction between water and canoe. Hence, the canoe does not move, since the system’s position of center of mass does not change.

The coordinate for center of mass is calculated as follows:

  $X_{cm}=\frac{(x_b{m}_b+x_c{m}_c)}{(m_b+m_c)}$

Substituting $m_b=60\text{ Kg, }{x}_b=6.0m,m_c=40\text{ Kg and }{x}_c=3.0m$

  $\begin{array}{l}{X}_{cm}=\frac{(6.0\text{ m}\times 60.0\text{ kg)+(3}\text{.0 m}\times 40.\text{0 kg})}{(60.0\text{ kg + 40}\text{.0 kg})}\\ \text{ = 4}\text{.8 m}\end{array}$

As the center of mass of the system remains the same, the canoe does not move under the given conditions.