Question

Consider a vertical elevator whose cabin has a total mass of \(800 \mathrm{kg}\) when fully loaded and \(150 \mathrm{kg}\) 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 \(1.2 \mathrm{m} / \mathrm{s}\) and \((b)\) the power required while the empty cabin is descending at a constant speed of \(1.2 \mathrm{m} / \mathrm{s}\) What would your answer be to ( \(a\) ) if no counterweight were used? What would your answer be to ( \(b\) ) if a friction force of \(800 \mathrm{N}\) has developed between the cabin and the guide rails?

Short answer

Answer: 1. For the fully loaded cabin rising with a constant speed, the power required is 4,708.8 W. 2. For the empty cabin descending with a constant speed, the power required is 2,943 W. 3. If no counterweight were used in the fully loaded cabin, the power required would be 9,417.6 W. 4. Considering a friction force of 800 N for the empty cabin, the power required is 1,983 W.

Step-by-step solution

Calculate the net forces acting on the cabin

First, let's find the weight of the fully loaded cabin (\(W_c\)) and the counterweight (\(W_{cw}\)) using the given masses: \(W_c\) (fully loaded) = \((800\,{\rm kg})\times(9.81\, {\rm m/s^2}) = 7,848\,{\rm N}\) \(W_c\) (empty) = \((150\,{\rm kg})\times(9.81\,{\rm m/s^2}) = 1,471.5\,{\rm N}\) \(W_{cw}\) = \((400\,{\rm kg})\times(9.81\,{\rm m/s^2})=3,924\,{\rm N}\) Next, we'll determine the net force on the cabin by subtracting the counterweight force from the weight of the cabin. For the fully loaded cabin (rising): \(F_{net} = W_c - W_{cw} = 7,848\,{\rm N} - 3,924\,{\rm N} = 3,924\,{\rm N}\) For the empty cabin (descending): \(F_{net} = W_{cw} - W_c = 3,924\,{\rm N} - 1,471.5\,{\rm N} = 2,452.5\,{\rm N}\)

Calculate the power required while the fully loaded cabin is rising and the empty cabin is descending

Now, we will calculate the power required in the two scenarios using the following power formula: \(P = F_{net} \times v\), where \(P\) is the power needed and \(v\) is the constant velocity of motion. For the fully loaded cabin (rising): \(P = (3,924\,{\rm N})\times(1.2\, {\rm m/s}) = 4,708.8\, {\rm W}\) For the empty cabin (descending): \(P = (2,452.5\,{\rm N})\times(1.2\, {\rm m/s}) = 2,943\, {\rm W}\)

Calculate the power required if no counterweight were used

In this scenario, there would be no counterweight force to balance the weight of the fully loaded cabin, and the weight would be 7,848 N. We can calculate the power required using the same formula: \(P = W_c \times v = (7,848\,{\rm N})\times(1.2\,{\rm m/s}) = 9,417.6\,{\rm W}\)

Calculate the power required considering the friction force of 800 N

In this case, we have an additional 800 N of friction force working against the motion of the empty cabin. To find the net force, we need to subtract both the weight of the empty cabin and the friction force from the counterweight force: \(F_{net} = W_{cw} - W_c - F_{friction} = 3,924\,{\rm N} - 1,471.5\,{\rm N} - 800\,{\rm N} = 1,652.5\,{\rm N}\) Next, we'll calculate the power required using the same formula as before: \(P = F_{net} \times v = (1,652.5\,{\rm N})\times(1.2\,{\rm m/s}) = 1,983\,{\rm W}\)

Key concepts

Work-Energy Principle

Understanding the work-energy principle is crucial when solving elevator physics problems. It establishes a relationship between the work done on an object and its kinetic and potential energy. In the context of an elevator, work is performed by the force that moves the elevator cabin vertically against the pull of gravity. However, if the elevator moves at constant speed, as in this problem, the kinetic energy remains unchanged, and we focus on the power needed to overcome the gravitational force.

When an elevator cabin rises, work is done against gravity to increase its potential energy; conversely, when it descends, work is done by gravity. The net work done on the cabin is equal to the change in its potential energy, which is calculated as the product of the mass, gravitational acceleration, and the change in height. If the speed is constant, then the power required to achieve this can be determined by the amount of work done per unit time.

Net Force Calculation

A correct net force calculation is pivotal to analyzing elevator dynamics. Net force is the vector sum of all the forces acting on an object. For elevators, the primary forces are the weight of the elevator cabin and the tension in the cable caused by the counterweight.

The solution to our exercise began by calculating the weights of the fully loaded elevator cabin and the counterweight using the formula for weight, which is the mass times gravitational acceleration (\( mg \)). Then, by considering the direction of motion and the balancing effect of the counterweight, the net force was found by taking the difference between these weights. Be aware that any additional forces, such as friction, will also be included in the net force calculation.

Power Computation

Finally, power computation in elevator physics involves determining the rate at which work is done or energy is transferred. Power is defined as the amount of work done divided by the time taken to do it. The unit of power is the watt (W), where one watt equals one joule per second. For our elevator, the work done is the force times the distance moved in the direction of the force.

In this example, since the elevator moves at a constant speed, the distance traveled each second is consistent, allowing us to compute the power by multiplying the net force by the velocity of the cabin. This simplification hinges on a key assumption that the speed is constant, hence acceleration is zero. If the elevator were accelerating, the calculation would be more complex, as it would need to account for changes in kinetic energy as well.