Chapter 10, Problem 39AP

An elevator system in a tall building consists of a 800-kg car and a 950-kg counterweight joined by a light cable of constant length that passers over a pulley of mass 280 kg. The pulley, called a sheave, is a solid cylinder of radius 0.700 m turning on a horizontal axle. The cable does not slip on the sheave. A number n of people, each of mass 80.0 kg, are riding in the elevator car, moving upward at 3.00 m/s and approaching the floor where the car should stop. As an energy-conservation measure, a computer disconnects the elevator motor at just the right moment so that t he sheave–car–counterweight system then coasts freely without friction and comes to rest at the floor desired. There it is caught by a simple latch rather than by a massive brake. (a) Determine the distance d the car coasts upward as a function of n. Evaluate the distance for (b) n = 2, (c) n = 12, and (d) n = 0. (e) For what integer values of n does the expression in part (a) apply? (f) Explain your answer to part (e). (g) If an infinite number of people could fit on the elevator, what is the value of d?

To determine

The distance $d$ of the car coasts upward as a function of $n$.

Answer to Problem 39AP

The distance $d$ of the car coasts upward as a function of $n$ is $\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$.

Explanation of Solution

The mass of car is $800\text{kg}$, mass of counterweight is $950\text{kg}$, mass of pulley is $280\text{kg}$, radius of pulley is $0.700\text{m}$, mass of each people is $80.0\text{kg}$, speed of elevator car is $3.00\text{m}/\text{s}$.

From the law of energy conservation,

    $E_{\text{i}}=E_{\text{f}}$        (1)

Here, $E_{\text{i}}$ is the total initial energy of the system and $E_{\text{f}}$ is the total final energy of the system.

Formula to calculate the total initial energy of the system is,

    $E_{\text{i}}=K_{\text{ei}}+K_{\text{ci}}+K_{\text{si}}$        (2)

Here, $K_{\text{ei}}$ is the initial translational kinetic energy of the elevator, $K_{\text{ti}}$ is the initial translational kinetic energy of the counterweight and $K_{\text{si}}$ is the initial rotational kinetic energy of the sheave pulley.

Write the expression for the initial translational kinetic energy of the elevator is,

    $K_{\text{ei}}=\frac{1}{2}{m}_{\text{e}}{v}^2$

Here, $m_{\text{e}}$ is the mass of the elevator and $v$ is the initial speed of the system.

Write the expression for the initial translational kinetic energy of the counterweight is,

    $K_{\text{ci}}=\frac{1}{2}{m}_{\text{c}}{v}^2$

Here, $m_{\text{c}}$ is the mass of the counterweight.

Write the expression for the initial rotational kinetic energy of the sheave is,

    $K_{\text{si}}=\frac{1}{2}{I}_{\text{s}}{\omega }^2$        (3)

Here, $I_{\text{s}}$ is the moment of inertia of the sheave pulley and $\omega$ is the angular speed of the sheave pulley.

Write the expression for the moment of inertia of the pulley is,

    $I=\frac{1}{2}{m}_{\text{s}}{r}^2$

Here, $m_{\text{s}}$ is the mass of the sheave pulley.

Write the expression for the initial angular speed of the pulley is,

    $\omega =\frac{v}{r}$

Here, $r$ is the radius of the sheave pulley.

Substitute $\frac{1}{2}{m}_{\text{s}}{r}^2$ for $I$ and $\frac{v}{r}$ for $\omega$ in equation (3).

    $\begin{array}{c}{K}_{\text{si}}=\frac{1}{2}\left(\frac{1}{2}{m}_{\text{s}}{r}^2\right){\left(\frac{v}{r}\right)}^2\\ =\frac{1}{4}{m}_{\text{s}}{v}^2\end{array}$

Substitute $\frac{1}{2}{m}_{\text{e}}{v}^2$ for $K_{\text{ei}}$, $\frac{1}{2}{m}_{\text{c}}{v}^2$ for $K_{\text{ci}}$ and $\frac{1}{4}{m}_{\text{s}}{v}^2$ for $K_{\text{si}}$ in equation (2).

    $\begin{array}{c}{E}_{\text{i}}=\frac{1}{2}{m}_{\text{e}}{v}^2+\frac{1}{2}{m}_{\text{c}}{v}^2+\frac{1}{4}{m}_{\text{s}}{v}^2\\ =\frac{1}{2}\left(m_{\text{e}}+m_{\text{c}}+\frac{1}{2}{m}_{\text{s}}\right)v^2\end{array}$        (4)

Since at the end the system comes to rest hence all the kinetic energies will be zero only potential energy remains in the system.

Formula to calculate the total final energy of the system is,

    $E_{\text{f}}=U_{\text{ef}}+U_{\text{ec}}+U_{\text{es}}$        (5)

Here, $U_{\text{ef}}$ is the final potential energy of the elevator, $K_{\text{ti}}$ is the final potential energy of the counterweight an d $K_{\text{si}}$ is the final potential energy of the sheave pulley.

Write the expression for the final potential energy of the elevator is,

    $U_{\text{ef}}=m_{\text{e}}gd$

Here, $g$ is the acceleration due to gravity and $d$ is the distance cover by the system.

Write the expression for the final potential energy of the counterweight is,

    $U_{\text{cf}}=-m_{\text{c}}gd$

Here, $g$ is the acceleration due to gravity and $d$ is the distance cover by the system.

Since the sheave pulley remains at its position so its final potential energy is zero.

    $U_{\text{sf}}=0$

Substitute $m_{\text{e}}gd$ for $U_{\text{ef}}$, $-m_{\text{c}}gd$ for $U_{\text{cf}}$ and $0$ for $U_{\text{sf}}$ in equation

    $\begin{array}{c}{E}_{\text{f}}=m_{\text{e}}gd-m_{\text{c}}gd+0\\ =\left(m_{\text{e}}-m_{\text{c}}\right)gd\end{array}$

Substitute $\frac{1}{2}\left(m_{\text{e}}+m_{\text{c}}+\frac{1}{2}{m}_{\text{s}}\right)v^2$ for $E_{\text{i}}$ and $\left(m_{\text{e}}-m_{\text{c}}\right)gd$ for $E_{\text{f}}$ in equation (1).

    $\frac{1}{2}\left(m_{\text{e}}+m_{\text{c}}+\frac{1}{2}{m}_{\text{s}}\right)v^2=\left(m_{\text{e}}-m_{\text{c}}\right)gd$        (6)

Formula to calculate the mass of the elevator is,

    $m_{\text{e}}=m_{\text{car}}+nm$

Here, $m_{\text{car}}$ is the mass of the car, $n$ is the number of people enter in the elevator and $m$ is the mass of each person.

Substitute $m_{\text{car}}+nm$ for $m_{\text{e}}$ in equation ()

    $\frac{1}{2}\left(\left(m_{\text{car}}+nm\right)+m_{\text{c}}+\frac{1}{2}{m}_{\text{s}}\right)v^2=\left(\left(m_{\text{car}}+nm\right)-m_{\text{c}}\right)gd$

Substitute $800\text{kg}$ for $m_{\text{car}}$, $80\text{kg}$ for $m$, $950\text{kg}$ for $m_{\text{c}}$, $280\text{kg}$ for $m_{\text{s}}$, $3.00\text{m}/\text{s}$ for $v$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above equation to find the expression for $d$ in terms of $n$.

    $\begin{array}{c}\frac{1}{2}\left[\begin{array}{l}\left(800\text{kg}+n\left(80\text{kg}\right)\right)+\\ 950\text{kg}+\frac{1}{2}\times 280\text{kg}\end{array}\right]{\left(3.00\text{m}/\text{s}\right)}^2=\left(\begin{array}{l}\left(800\text{kg}+n\left(80\text{kg}\right)\right)-\\ 950\text{kg}\end{array}\right)\left(9.8m/s^2\right)d\\ d=\frac{\left[1890+n\left(80\text{kg}\right)\right]\left(4.5{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}{\left[n\left(80\text{kg}\right)-150\text{kg}\right]\left(9.8m/s^2\right)}\\ =\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)\end{array}$

Conclusion:

Therefore, the distance $d$ of the car coasts upward as a function of $n$ is $\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$.

To determine

The distance $d$ for $n=2$.

Answer to Problem 39AP

The distance $d$ for $n=2$ is $94.1\text{m}$.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

Substitute $2$ for $n$ in above equation to find $d$.

    $\begin{array}{c}d=\left(1890+80\times 2\right)\left(\frac{0.459\text{m}}{80\times 2-150}\right)\\ =\frac{2050\times 0.459\text{m}}{10}\\ =94.1\text{m}\end{array}$

Conclusion:

Therefore, the distance $d$ for $n=2$ is $94.1\text{m}$.

To determine

The distance $d$ for $n=12$.

Answer to Problem 39AP

The distance $d$ for $n=12$ is $1.62\text{m}$.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

Substitute $12$ for $n$ in above equation to find $d$.

    $\begin{array}{c}d=\left(1890+80\times 12\right)\left(\frac{0.459\text{m}}{80\times 12-150}\right)\\ =\frac{2850\times 0.459\text{m}}{810}\\ =1.62\text{m}\end{array}$

Conclusion:

Therefore, the distance $d$ for $n=12$ is $1.62\text{m}$.

To determine

The distance $d$ for $n=0$.

Answer to Problem 39AP

The distance $d$ for $n=0$ is $-5.79\text{m}$.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

Substitute $0$ for $n$ in above equation to find $d$.

    $\begin{array}{c}d=\left(1890+80\times 0\right)\left(\frac{0.459\text{m}}{80\times 0-150}\right)\\ =\frac{1890\times 0.459\text{m}}{-150}\\ =-5.79\text{m}\end{array}$

Conclusion:

Therefore, the distance $d$ for $n=0$ is $-5.79\text{m}$.

To determine

The integral values of $n$ for which the expression in part (a) apply.

Answer to Problem 39AP

The expression in part (a) is valid only when $n\ge 2$.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

From the above expression, the distance $d$ is a function of $n$. The distance can never be negative so, the minimum integral values of $n$ should be greater than $2$.

Conclusion:

Therefore, the expression in part (a) is valid only when $n\ge 2$.

To determine

The explanation for the answer in part (e).

Answer to Problem 39AP

The mass of the elevator is less than the mass of the counterweight for the value of $n=0$ and for $n=1$, so if it released the car would accelerate in upward direction.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

Substitute $1$ for $n$ in above equation to find $d$.

    $\begin{array}{c}d=\left(1890+80\times 1\right)\left(\frac{0.459\text{m}}{80\times 1-150}\right)\\ =\frac{904.23\text{m}}{-70}\\ =-12.9\text{m}\end{array}$

Since the value of distance $d$ is negative for $n=0$ or $n=1$ that indicates car would accelerate in upward direction because the mass of the counterweight becomes greater than the mass of the elevator which is undesirable.

Conclusion:

Therefore, the mass of the elevator is less than the mass of the counterweight for the value of $n=0$ and for $n=1$, so if it released the car would accelerate in upward direction.

To determine

The value of $d$ for infinite number of people enter into the elevator.

Answer to Problem 39AP

The value of $d$ for which the infinite number of people enter into the elevator is $0.459\text{m}$.

Explanation of Solution

The expression for the distance $d$ for the $n$ from part (a) is given by,

    $d=\left(1890+80n\right)\left(\frac{0.459\text{m}}{80n-150}\right)$

Rearrange the above equation.

    $d=\left(\frac{1890}{n}+80\right)\left(\frac{0.459\text{m}}{80-\frac{150}{n}}\right)$

Substitute $\infty$ for $n$ in above equation to find $d$.

    $\begin{array}{c}d=\left(\frac{1890}{\infty }+80\right)\left(\frac{0.459\text{m}}{80-\frac{150}{\infty }}\right)\\ =\left(0+80\right)\left(\frac{0.459\text{m}}{80-0}\right)\\ =0.459\text{m}\end{array}$

Since the value of distance $d$ is negative for $n=0$ or $n=1$ but it becomes positive after the value of $n$ become greater than $2$ because distance cannot be negative. So, the expression for the part (a) is valid only for $n\ge 2$.

Conclusion:

Therefore, the value of $d$ for which the infinite number of people enter into the elevator is $0.459\text{m}$.