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Q4-36PE

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Found in: Page 163

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity.

The forces in the free-body diagram are tension and the weight of the elevator. The tension on the cable is calculated as $$5400 N$$.

See the step by step solution

## Step 1: Definition of tension force

The tension force is described as the force transferred through a rope, string, or wire when it is pulled by opposing forces.

## Step 2: Construct the problem

Let construct the problem with these values and the mass of the elevator. It can be done as follows:

Consider an elevator with load having a mass of $$500 kg$$ starts from rest. It moves upwards with an acceleration of $$1 m/{s^2}$$ and reached final point after $$12 s$$ with the final velocity of $$12 m/s$$. Calculate the tension on the elevator cable with the help of a free body diagram.

$$\begin{array}{c}u = 0\\a = 1 m/{s^2}\\t = 12 s\end{array}$$

$$\begin{array}{c}v = 0 + \left( {1 m/{s^2} \times 12 s} \right)\\ = 12 m/s\end{array}$$

## Step 3: Draw the free body diagram.

A free-body diagram can be drawn by considering the forces that act on the cable. There is a force, $$W$$ acting downwards that is equal to the weight of the elevator. The tension, $$T$$ on the cable is acting upwards, opposite to the weight of the elevator, and acceleration is in the upward direction as it moves upwards. Thus, the free body diagram can be drawn as follows:

The forces that act in the y direction are equal to the sum of the tension and weight of the elevator.

$$\begin{array}{l}{F_y} = ma\\T - W = ma\\T - mg = ma\\T = m\left( {a + g} \right)\end{array}$$

Here, $${F_y}$$ is the total force acting on the elevator in the y direction and $$g$$ is the acceleration due to gravity which is equal to $$9.8 m/{s^2}$$. Substitute the known values to calculate tension on the cable.

$$\begin{array}{c}T = 500 kg\left( {1 m/{s^2} + 9.8 m/{s^2}} \right)\\ = 5400 kg.m/{s^2}\\ = 5400 N\end{array}$$

Therefore, tension on the cable is $$5400 N$$.