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Chapter 4: Problem 26

An elevator cabin has a mass of \(358.1 \mathrm{~kg},\) and the combined mass of the people inside the cabin is \(169.2 \mathrm{~kg} .\) The cabin is pulled upward by a cable, with a constant acceleration of \(4.11 \mathrm{~m} / \mathrm{s}^{2}\). What is the tension in the cable?

Short Answer

Expert verified
Answer: The tension in the cable is approximately \(7337.9\,\mathrm{N}\).

Step by step solution

01

1. Identify the given information and unknown

We are given: - Mass of the elevator cabin: \(m_{1}=358.1 \mathrm{~kg}\) - Mass of people inside the cabin: \(m_{2}=169.2 \mathrm{~kg}\) - Acceleration of the elevator: \(a= 4.11 \mathrm{~m/s^2}\) Unknown: - Tension in the cable: \(T\)
02

2. Calculate the total mass of the elevator system

To find the total mass of the elevator system (cabin and people inside), we add the masses of the cabin and the people: \(m_{total} = m_{1} + m_{2} = 358.1\,\mathrm{kg} + 169.2\,\mathrm{kg} = 527.3\,\mathrm{kg}\)
03

3. Calculate the weight of the elevator system

The weight of the elevator system (cabin and people inside) can be calculated using the equation: \(W = m_{total} \cdot g\) Where \(g = 9.81\,\mathrm{m/s^2}\) is the acceleration due to gravity. So, \(W = 527.3\,\mathrm{kg} \cdot 9.81\,\mathrm{m/s^2} = 5170.79\,\mathrm{N}\)
04

4. Apply Newton's second law

According to Newton's second law of motion, the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = m_{total} \cdot a\) We can calculate the net force acting on the elevator system: \(F_{net} = 527.3\,\mathrm{kg} \cdot 4.11\,\mathrm{m/s^2} = 2167.113\,\mathrm{N}\)
05

5. Calculate tension in the cable

Now, we need to find the tension in the cable (\(T\)) supporting the elevator system. The upward force (tension) must be greater than the downward force (the weight) to make the elevator move upward with acceleration. Therefore, \(T - W = F_{net}\) Solving for \(T\): \(T = F_{net} + W = 2167.113\,\mathrm{N} + 5170.79\,\mathrm{N} = 7337.903\,\mathrm{N}\) So, the tension in the cable is approximately \(7337.9\,\mathrm{N}\).

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Most popular questions from this chapter

A person stands on the surface of the Farth. The mass of the person is \(m\) and the mass of the Earth is \(M\). The person jumps upward, reaching a maximum height \(h\) above the Earth. When the person is at this height \(h,\) the magnitude of the force exerted on the Earth by the person is a) \(m \mathrm{~g}\). c) \(M^{2} g / m\) e) zero. b) \(\mathrm{Mg}\) d) \(m^{2} g / M\).

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An elevator cabin has a mass of \(358.1 \mathrm{~kg}\), and the combined mass of the people inside the cabin is \(169.2 \mathrm{~kg} .\) The cabin is pulled upward by a cable, with a constant acceleration of \(4.11 \mathrm{~m} / \mathrm{s}^{2}\). What is the tension in the cable?
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