Chapter 2.8, Problem 119RP

Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

To determine

The power required to raise the cabin at a constant speed of 1.2 m/s.

The power required if no counterweight were used.

Answer to Problem 119RP

The power required to raise the cabin at a constant speed of 1.2 m/s is $\underset{\_}{4.71\text{kW}}$.

The power if no counterweight were used is $\underset{\_}{9.42\text{kW}}$.

Explanation of Solution

Calculate the power required to raise the cabin at a constant speed of 1.2 m/s.

${\dot{W}}_{\text{req}}=mgV$ (I)

Here, the weight of the elevator cabin is m, acceleration due to gravity is g, and constant speed is V.

Since there is no usage of counterweight, the mass is double to 800 kg.

Calculate the power if no counterweight were used.

$\dot{W}=2\times {\dot{W}}_{\text{req}}$ (II)

Conclusion:

Substitute 400 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, and $1.2\text{m/s}$ for V in Equation (I).

$\begin{array}{c}{\dot{W}}_{\text{req}}=\left(400\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(\text{1}\text{.2}\text{m/s}\right)\\ =\left(400\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(\text{1}\text{.2}\text{m/s}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m/s}}\right)\\ =4.71\text{kW}\end{array}$

Thus, the power required to raise the cabin at a constant speed of 1.2 m/s is $\underset{\_}{4.71\text{kW}}$.

Substitute 4.71 kW for ${\dot{W}}_{\text{req}}$ in Equation (II).

$\begin{array}{c}\dot{W}=2\times 4.71\text{kW}\\ =9.42\text{kW}\end{array}$

Thus, the power if no counterweight were used is $\underset{\_}{9.42\text{kW}}$.

To determine

The power required to raise the mass of 250 kg at a constant speed of 1.2 m/s if the empty cabon is descending.

The total power needed if a friction force of 800 N has developed between the cabin and the guide rails.

Answer to Problem 119RP

The power required to raise the mass of 250 kg at a constant speed of 1.2 m/s if the empty cabon is descending is $\underset{\_}{2.94\text{kW}}$.

The total power needed if a friction force of 800 N has developed between the cabin and the guide rails is $\underset{\_}{\text{3}\text{.90}\text{kW}}$.

Explanation of Solution

Calculate the power required to raise the cabin at a constant speed of 1.2 m/s.

${\dot{W}}_{\text{req}}=mgV$ (II)

Here, the weight of the elevator cabin is m, acceleration due to gravity is g, and constant speed is V.

Calculate the friction power if the friction force of 800 N develops between the cabin and the guide rails.

${\dot{W}}_{\text{friction}}=F_{\text{friction}}V$ (III)

Here, friction force is $F_{\text{friction}}$.

Calculate the total power needed if a friction force of 800 N has developed between the cabin and the guide rails.

${\dot{W}}_{\text{total}}=\dot{W}+{\dot{W}}_{\text{friction}}$ (IV)

Conclusion:

Calculate the mass as the counterweight is ascending.

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

Substitute 250 kg for m, $9.81{\text{m/s}}^{\text{2}}$ for g, and $1.2\text{m/s}$ for V in Equation (II).

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

Thus, the power required to raise the mass of 250 kg at a constant speed of 1.2 m/s if the empty cabon is descending is $\underset{\_}{2.94\text{kW}}$.

Substitute 800 N for $F_{\text{friction}}$ and 1.2 m/s for V in Equation (III).

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

Substitute 0.96 kW for ${\dot{W}}_{\text{friction}}$ and 2.94 kW for $\dot{W}$ in Equation (IV).

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

Thus, the total power needed if a friction force of 800 N has developed between the cabin and the guide rails is $\underset{\_}{\text{3}\text{.90}\text{kW}}$.