Question

Problem 8.106

A 45.0-kg woman stands up in a 60.0-kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point 1.00 m from the other end (the figure (Figure 1) ).

Part A

If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

dcanoe =

m to the left

Answer 1

Concepts and reason

The concept of center of mass is required to solve the problem.

First, calculate the position of center of mass of the system when the woman stands at one point of the canoe. Then, calculate the position of center of mass of the system when the woman stands at the other end of the canoe. As there is no external force, the position of center of mass should remain constant. Finally, calculate the displacement of canoe by determining the shift in the position of center of mass.

Fundamentals

The position of center of mass of the system consists of two objects of mass ${m_1}$ and ${m_2}$ is determined by using the following equation:

${x_{{\rm{cm}}}} = \frac{{{x_1}{m_1} + {x_2}{m_2}}}{{{m_1} + {m_2}}}$

Here, ${x_1}$ is the position of mass ${m_1}$ and ${x_2}$ is the position of mass ${m_2}$ .

The center of mass shifts towards the more massive object.

Consider the origin at the left end of the canoe.

The position of center of mass is determined by using the following equation:

${x_{{\rm{cm1}}}} = \frac{{{m_{\rm{w}}}{x_{{\rm{w}}1}} + {m_{\rm{c}}}{x_{\rm{c}}}_1}}{{{m_{\rm{w}}} + {m_{\rm{c}}}}}$

Here, ${m_{\rm{w}}}$ and ${m_{\rm{c}}}$ is the mass of woman and mass of canoe respectively, ${x_{{\rm{w1}}}}$ is the position of woman from the left end, and ${x_{{\rm{c1}}}}$ is the position of center of mass of canoe from the left end.

Substitute 45 kg for ${m_{\rm{w}}}$ , 60 kg for ${m_{\rm{c}}}$ , 1.00 m for ${x_{{\rm{w1}}}}$ , and 2.50 m for ${x_{{\rm{c1}}}}$ in above equation.

$\begin{array}{c}\\{x_{{\rm{cm1}}}} = \frac{{\left( {45\,{\rm{kg}}} \right)\left( {1{\rm{ m}}} \right) + \left( {60{\rm{ kg}}} \right)\left( {2.5{\rm{ m}}} \right)}}{{45{\rm{ kg}} + 60{\rm{ kg}}}}\\\\ = 1.857{\rm{ m}}\\\end{array}$

Consider the origin at the left end of the canoe.

The position of woman from left end is,

${x_{{\rm{w2}}}} = L - x$

Here, L is the length of the boat and x is the position of woman from right end.

Substitute 5.00 m for L and 1.00 m for x in equation ${x_{{\rm{w2}}}} = L - x$ .

$\begin{array}{c}\\{x_{{\rm{w2}}}} = 5.00{\rm{ m}} - 1.00{\rm{ m}}\\\\ = 4.00{\rm{ m}}\\\end{array}$

Calculate position of center of mass when the woman stands at 1.00 m from right end of canoe by using the following equation:

${x_{{\rm{cm2}}}} = \frac{{{m_{\rm{w}}}{x_{{\rm{w2}}}} + {m_{\rm{c}}}{x_{\rm{c}}}_1}}{{{m_{\rm{w}}} + {m_{\rm{c}}}}}$

Here, ${x_{{\rm{w2}}}}$ is the position of woman from the left end, and ${x_{{\rm{c1}}}}$ is the position of center of mass of canoe from the left end.

Substitute 45 kg for ${m_{\rm{w}}}$ , 60 kg for ${m_{\rm{c}}}$ , 4.00 m for ${x_{{\rm{w2}}}}$ , and 2.50 m for ${x_{{\rm{c2}}}}$ in above equation.

$\begin{array}{c}\\{x_{{\rm{cm2}}}} = \frac{{\left( {45\,{\rm{kg}}} \right)\left( {{\rm{4 m}}} \right) + \left( {60{\rm{ kg}}} \right)\left( {2.5{\rm{ m}}} \right)}}{{45{\rm{ kg}} + 60{\rm{ kg}}}}\\\\ = 3.143{\rm{ m}}\\\end{array}$

Calculate the distance moved by the canoe.

The distance moved by the canoe is calculated as follows:

$d = {x_{{\rm{cm2}}}} - {x_{{\rm{cm1}}}}$

Substitute 3.143 m for ${x_{{\rm{cm2}}}}$ and 1.857 m for ${x_{{\rm{cm1}}}}$ in the above equation.

$\begin{array}{c}\\d = 3.143{\rm{ m}} - 1.857{\rm{ m}}\\\\ = 1.29{\rm{ m}}\\\end{array}$

Thus, the distance moved by the canoe is $1.29{\rm{ m}}$ to the left.

Ans:

The distance moved by the canoe is $1.29{\rm{ m}}$ to the left.