Chapter 7, Problem 95P

(a)

To determine

The distance between the raft and pier when the women walks toward the pier until she gets to the other end of the raft and stops there.

Answer to Problem 95P

The distance is $2.5\text{m}$.

Explanation of Solution

Mass of women is $60.0\text{kg}$, mass of raft is $120\text{kg}$, length of raft is $6.0\text{m}$, and the distance to the end of raft from the pier is $0.50\text{m}$.

Imagine that the initial position of women is $x=0$ and the position of pier is left to the women.

Write the equation for the center of mass for the woman, pier, and the raft system.

$x_{cm}=\frac{m_w{x}_{wi}+m_r{x}_{ri}}{m_w+m_r}$ (I)

Here, $x_{cm}$ is the distance from origin to the center of mass of the system, $m_w$ is the mass of women, $m_r$ is the mass of raft, $x_{wi}$ is the initial distance from the center of mass to the position of women, and $x_{ri}$ is the initial distance from the center of mass to the position of raft.

Initially, $x_{cm}$ will be right to $x_{ri}$.After the women reached the other end of raft, $x_{cm}$ will be the left of $x_{ri}$.

Write the relation between $x_{ri}$,$x_{cm}$, and the final distance between the raft from the center of mass arises due to the symmetry of system.

$\begin{array}{c}{x}_{cm}-x_{ri}=x_{cm}-x_{rf}\\ x_{rf}=2x_{cm}-x_{ri}\end{array}$

Write the equation for the final distance of raft from the pier.

$d_f=x_{rf}-\left(\frac{l}{2}\right)$

Here, $d_f$ is the final distance of raft from the pier and $l$ is the length of raft.

Rewrite the above relation by substituting $2x_{cm}-x_{ri}$ for $x_{rf}$.

$d_f=2x_{cm}-x_{ri}-\left(\frac{l}{2}\right)$ (II)

Write the equation for $x_{wi}$.

$x_{wi}=l+d_{rp}$ (III)

Here, $d_{rp}$ is the initial distance from the end of raft to the pier.

Write the equation to find $x_{ri}$.

$x_{ri}=\frac{l}{2}+d_{rp}$ (IV)

Conclusion:

Substitute $6.0\text{m}$ for $l$ and $0.5\text{m}$ for $d_{rp}$ in equation (IV) to find $x_{ri}$.

$\begin{array}{c}{x}_{ri}=\left(\frac{6.0\text{m}}{2}\right)+0.5\text{m}\\ \text{=3}\text{.5}\text{m}\end{array}$

Substitute $6.0\text{m}$ for $l$ and $0.5\text{m}$ for $d_{rp}$ in equation (III) to find $x_{wi}$.

$\begin{array}{c}{x}_{wi}=6.0\text{m}+0.5\text{m}\\ =6.5\text{m}\end{array}$

Substitute $6.5\text{m}$ for $x_{wi}$, $60.0\text{kg}$ for $m_w$, $\text{3}\text{.5}\text{m}$ for $x_{ri}$, and $120\text{kg}$ for $m_r$ in equation (I) to find $x_{cm}$.

$\begin{array}{c}{x}_{cm}=\frac{\left(60.0\text{kg}\right)\left(6.5\text{m}\right)+\left(120.0\text{kg}\right)\left(\text{3}\text{.5}\text{m}\right)}{\left(60.0\text{kg}\right)+\left(120.0\text{kg}\right)}\\ =\frac{810\text{kgm}}{180\text{kg}}\\ =4.5\text{m}\end{array}$

Substitute $4.5\text{m}$ for $x_{cm}$, $\text{3}\text{.5}\text{m}$ for $x_{ri}$, and $6.0\text{m}$ for $l$ in equation (II) to find $d_f$.

$\begin{array}{c}{d}_f=2\left(4.5\text{m}\right)-\text{3}\text{.5}\text{m}-\left(\frac{6.0\text{m}}{2}\right)\\ =2.5\text{m}\end{array}$

Therefore, the distance is $2.5\text{m}$.

(b)

To determine

The distance travelled by the woman relative to pier in part (a).

Answer to Problem 95P

The distance travelled is $4.0\text{m}$.

Explanation of Solution

Mass of women is $60.0\text{kg}$, mass of raft is $120\text{kg}$, length of raft is $6.0\text{m}$, and the distance to the end of raft from the pier is $0.50\text{m}$.

Imagine that the initial position of women is $x=0$ and the position of pier is left to the women.

Write the equation for the distance travelled by women relative to pier.

$\left|\Delta x_w\right|=\left|d_f-x_{wi}\right|$

Here. $\left|\Delta x_w\right|$ is the distance travelled by women relative to pier.

Conclusion:

Substitute $2.5\text{m}$ for $d_f$ and $6.5\text{m}$ for $x_{wi}$ in the above equation to find $\left|\Delta x_w\right|$.

$\begin{array}{c}\left|\Delta x_w\right|=\left|2.5\text{m}-6.5\text{m}\right|\\ =4.0\text{m}\end{array}$

Therefore, the distance is $4.0\text{m}$.