Question

An elevator system in a tall building consists of a $\mathbf{800}\mathbf{kg}$ car and a role="math" localid="1663753328337" $\mathbf{950}\mathbf{kg}$ counterweight joined by a light cable of constant length that passes over a pulley of mass $\mathbf{280}\mathbf{kg}$. The pulley, called a sheave, is a solid cylinder of radius$\mathbf{0}\mathbf{.}\mathbf{700}\mathbf{m}$ turning on a horizontal axle. The cable does not slip on the sheave. A number n of people, each of mass $\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{kg}$, are riding in the elevator car, moving upward at $\mathbf{3}\mathbf{m}\mathbf{/}\mathbf{s}$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 the 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$\mathbf{d}$ the car coasts upward as a function of $\mathbf{n}$. Evaluate the distance for (b) $\mathbf{n}\mathbf{=}\mathbf{2}$, (c) $\mathbf{n}\mathbf{=}\mathbf{12}$, and (d) $\mathbf{n}\mathbf{=}\mathbf{0}$. (e) For what integer values of $\mathbf{n}$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 $\mathit{d}$?

Short answer

(a) The expression for the distance car coasts upward as function of$\mathrm{n}$ is $\frac{360\mathrm{n}+8505}{784\mathrm{n}-1470}$.

(b) The value of the distance for$\mathrm{n}=2$ is $94.13\text{\hspace{0.17em}m}$.

(c) The value of the distance for$\mathrm{n}=12$ is $1.61\text{\hspace{0.17em}m}$.

(d) The value of the distance for$n=0$ is $-5.78\text{\hspace{0.17em}m}$.

(e) The integer value for$\mathrm{n}$ is $\mathrm{n}\ge 2$.

(f) Due to hight counterweight the distance can’t be negative and value of n is $\mathrm{n}\ge 2$.

(g) The distance for the infinite number of the people is$0.45\text{\hspace{0.17em}m}$ .

Step-by-step solution

Write the given data from the question.

Mass of the car,${\mathrm{m}}_{\mathrm{c}}=800\text{\hspace{0.17em}kg}$

Mass of counterweight,${\mathrm{m}}_{\mathrm{w}}=950\text{\hspace{0.17em}kg}$

Mass of pulley, ${\mathrm{m}}_{\mathrm{p}}=280\text{\hspace{0.17em}kg}$

Radius of pulley,$\mathrm{r}=0.700\text{\hspace{0.17em}m}$

There are$\mathrm{n}$ number of people each having mass,${\mathrm{m}}_{\mathrm{n}}=80\text{\hspace{0.17em}kg}$

The initial velocity of car,$\mathrm{v}=3\text{\hspace{0.17em}m}/\text{s}$

The final velocity of car,${\mathrm{v}}_{\mathrm{f}}=0$

Determine the distance the car coasts upward as a function of nand integer value of n.

The expression to calculate the angular velocity is given as follows.

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{v}}{\mathbf{r}}$

The expression to calculate the rotational kinetic energy is given as follows.

${\mathbf{K}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

Here,$\mathbf{I}$ is the moment of inertia.

The expression to calculate the potential energy is given as follows.

$\mathbf{U}\mathbf{=}\mathbf{mgh}$

Here,$\mathbf{g}$ is the acceleration due to gravity and$\mathbf{h}$ is the height.

Derive expression for the distance the car coasts upward as a function of n.

(a)

The total mass of the system,$\mathrm{m}={\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}$

The change in the kinetic energy of the system is given by,

$\mathrm{\Delta K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

Here, is the final kinetic energy and ${\mathrm{K}}_{\mathrm{i}}$is the initial kinetic energy.

$\mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}-\left(\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2\right)$

Substitute $\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}{\mathrm{r}}^2$for $I$and $\mathrm{v}/\mathrm{r}$for $\mathrm{\omega }$into above equation.

$\begin{array}{l}\mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}-\left[\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left(\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}{\mathrm{r}}^2\right){\left(\frac{\mathrm{v}}{\mathrm{r}}\right)}^2\right]\\ \mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}-\left[\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{4}{\mathrm{m}}_{\mathrm{p}}{\mathrm{v}}^2\right]\\ \mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}-\left[\frac{1}{2}\left(\mathrm{m}+\frac{{\mathrm{m}}_{\mathrm{p}}}{2}\right){\mathrm{v}}^2\right]\end{array}$

Substitute$0$for $v_f$,and${\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}$for$\mathrm{m}$into equation (i).

$\begin{array}{c}\mathrm{\Delta K}=\frac{1}{2}\mathrm{m}\left(0\right)-\left(\frac{1}{2}\left({\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}+\frac{{\mathrm{m}}_{\mathrm{p}}}{2}\right){\mathrm{v}}^2\right)\\ \mathrm{\Delta K}=-\frac{1}{2}\left({\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}+\frac{{\mathrm{m}}_{\mathrm{p}}}{2}\right){\mathrm{v}}^2\end{array}$ …… (ii)

The potential energy of the car and people increases and potential energy of the counterweight decreases.

Therefore, the total potential energy is given by,

$\mathrm{\Delta U}=({\mathrm{m}}_{\mathrm{c}}+{\mathrm{nm}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{w}})\mathrm{gd}$ …… (iii)

The law of conservation says the initial energy of system is remains the same as the final kinetic energy of the system.

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ -\frac{1}{2}\left({\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}+\frac{{\mathrm{m}}_{\mathrm{p}}}{2}\right){\mathrm{v}}^2+({\mathrm{m}}_{\mathrm{c}}+{\mathrm{nm}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{w}})\mathrm{gd}=0\\ \frac{1}{2}\left({\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{w}}+{\mathrm{nm}}_{\mathrm{n}}+\frac{{\mathrm{m}}_{\mathrm{p}}}{2}\right){\mathrm{v}}^2=({\mathrm{m}}_{\mathrm{c}}+{\mathrm{nm}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{w}})\mathrm{gd}\end{array}$

Substitute$800\text{\hspace{0.17em}kg}$for ${\mathrm{m}}_{\mathrm{c}}$,$950\text{\hspace{0.17em}kg}$for ${\mathrm{m}}_{\mathrm{w}}$, $280\text{\hspace{0.17em}kg}$for ${\mathrm{m}}_{\mathrm{p}}$, $80\text{\hspace{0.17em}kg}$for ${\mathrm{m}}_{\mathrm{n}}$,$9.8\text{\hspace{0.17em}m}/{\text{s}}^2$for $\mathrm{g}$and $3\text{\hspace{0.17em}m}/\text{s}$for$\mathrm{v}$into above equation.

$\begin{array}{c}\frac{1}{2}\left(800+950+\mathrm{n}\times 80+\frac{280}{2}\right){\left(3\right)}^2=(800+\mathrm{n}\times 80-950)\times 9.8\mathrm{d}\\ \frac{1}{2}(1890+80\mathrm{n})\times 9=(80\mathrm{n}-150)\times 9.8\mathrm{d}\\ (8505+360\mathrm{n})=(784\mathrm{n}-1470)\mathrm{d}\\ \mathrm{d}=\frac{360\mathrm{n}+8505}{784\mathrm{n}-1470}\end{array}$ …… (iv)

Hence the expression for the distance car coasts upward as function of $\mathrm{n}$is $\frac{360\mathrm{n}+8505}{784\mathrm{n}-1470}$.

Calculate the distance for the value of n=2.

(b)

Calculate the distance for $\mathrm{n}=2$.

Substitute for into equation (iv).

$\begin{array}{l}\mathrm{d}=\frac{360\left(2\right)+8505}{784\left(2\right)-1470}\\ \mathrm{d}=\frac{9225}{98}\\ \mathrm{d}=94.13\text{\hspace{0.17em}m}\end{array}$

Hence the value of the distance$n=2$ for is $94.13\text{\hspace{0.17em}m}$.

Calculate the distance for the value of n=12.

(c)

Calculate the distance for $n=12$.

Substitute$12$ for$n$ into equation (iv).

$\begin{array}{l}\mathrm{d}=\frac{360\left(12\right)+8505}{784\left(12\right)-1470}\\ \mathrm{d}=\frac{12825}{7938}\\ \mathrm{d}=1.61\text{\hspace{0.17em}m}\end{array}$

Hence the value of the distance for$n=12$ is $1.61\text{\hspace{0.17em}m}$.

Calculate the distance for the value of n=0.

(d)

Calculate the distance for $\mathrm{n}=0$.

Substitute $0$for $n$into equation (iv).

$\begin{array}{l}\mathrm{d}=\frac{360\left(0\right)+8505}{784\left(0\right)-1470}\\ \mathrm{d}=\frac{8505}{-1470}\\ \mathrm{d}=-5.78\text{\hspace{0.17em}m}\end{array}$

Hence the value of the distance for$n=0$ is $-5.78\text{\hspace{0.17em}m}$.

Determine the integers for which the part (a) is applicable.

(e)

We know that the distance is positive quantity, if we substitute$\mathrm{n}=1$ then the result would be negative which can’t be possible. Therefore, the expression is valid for$\mathrm{n}\ge 2$ and motor will not disconnect. For the integer $\mathrm{n}\ge 2$, the expression in part (a) apply.

Hence the integer value for$\mathrm{n}$ is $\mathrm{n}\ge 2$.

Explanation for the part (e).

(f)

The counterweight is hight therefore it always accelerates the car. But if the distance be negative then it will deaccelerate the car which is not possible. The elevator can be deaccelerated only when the number of people is$2$ or more than $2$.

Hence due to hight counterweight the distance can’t be negative and value of n is $\mathrm{n}\ge 2$.

Calculate the distance if the number of people tends to infinity.       

Recall the equation (iv).

$\mathrm{d}=\frac{360\mathrm{n}+8505}{784\mathrm{n}-1470}$

Take limit as,

$\begin{array}{c}\mathrm{d}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{360\mathrm{n}+8505}{784\mathrm{n}-1470}\\ \mathrm{d}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\mathrm{n}\left(360+\frac{8505}{\mathrm{n}}\right)}{\mathrm{n}\left(784-\frac{1470}{\mathrm{n}}\right)}\\ \mathrm{d}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{360+\frac{8505}{\mathrm{n}}}{784-\frac{1470}{\mathrm{n}}}\end{array}$

Apply the limits,

$\begin{array}{c}\mathrm{d}=\frac{360+\frac{8505}{\mathrm{\infty }}}{784-\frac{1470}{\mathrm{\infty }}}\\ \mathrm{d}=\frac{360+0}{784-0}\\ \mathrm{d}=\frac{360}{784}\\ \mathrm{d}=0.45\text{\hspace{0.17em}m}\end{array}$

Hence the distance for the infinite number of the people is $0.45\text{\hspace{0.17em}m}$.