Chapter 3, Problem 73RQ

(a)

To determine

The power required when the cabin is fully loaded.

The power required when no counterweight is used.

Explanation of Solution

Given:

Counterweight of the elevator cabin that is partially balanced $\left(m\right)$ is $400\text{kg}$.

Constant speed of the elevator cabin $\left(V\right)$ is $1.2\text{m}/\text{s}$.

Friction force $\left(F_f\right)$ is $\text{800}\text{N}$.

Density of the water $\left(\rho \right)$ is $1000\text{kg}/{\text{m}}^3$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^2$.

Calculation:

Calculate the power required when the cabin is fully loaded $\left(\dot{W}\right)$.

  $\begin{array}{c}\dot{W}=\frac{mgz}{\Delta t}\\ =mgV\\ =\left[\begin{array}{l}\left(400\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(1.2\text{m}/\text{s}\right)\\ \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m}/\text{s}}\right)\end{array}\right]\\ =4.71\text{kW}\end{array}$

Thus, the power required when the cabin is fully loaded is $\underset{\_}{4.71\text{kW}}$.

When no counterweight is used, then the mass of the elevator cabin would be $800\text{kg}$.

  $m=800\text{kg}$

Calculate the power required when no counterweight is used $\left(\dot{W}\right)$.

  $\begin{array}{c}\dot{W}=mgV\\ =\left[\begin{array}{l}\left(800\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(1.2\text{m}/\text{s}\right)\\ \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m}/\text{s}}\right)\end{array}\right]\\ =9.42\text{kW}\end{array}$

Thus, the power required when the cabin is fully loaded is $\underset{\_}{9.42\text{kW}}$.

(b)

To determine

The power required when the empty cabin is descending.

The power required when the empty cabin is descending with the friction force.

Explanation of Solution

Calculation:

Calculate the weight of the elevator cabin when it is empty.

  $\begin{array}{c}m=400\text{kg}-150\text{kg}\\ =250\text{kg}\end{array}$

Calculate the power required when the empty cabin is descending $\left(\dot{W}\right)$.

  $\begin{array}{c}\dot{W}=mgV\\ =\left[\begin{array}{l}\left(250\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(1.2\text{m}/\text{s}\right)\\ \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m}/\text{s}}\right)\end{array}\right]\\ =2.94\text{kW}\end{array}$

Thus, the power required when the empty cabin is descending is $\underset{\_}{2.94\text{kW}}$.

Calculate the power developed during the friction between the cabin and the guide rails $\left({\dot{W}}_{\text{friction}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{friction}}=\frac{F_{f}z}{\Delta t}\\ =F_{f}V\\ =\left(800\text{N}\right)\left(1.2\text{m}/\text{s}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m}/\text{s}}\right)\\ =0.96\text{kW}\end{array}$

Calculate the total power required $\left({\dot{W}}_{\text{total}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{total}}=\dot{W}+{\dot{W}}_{\text{friction}}\\ =2.94\text{kW}+0.96\text{kW}\\ =3.90\text{kW}\end{array}$

Thus, the power required when the empty cabin is descending with the friction force is $\underset{\_}{3.90\text{kW}}$.