Solving for Tension in Elevator Cable: 2300 kg, 50 m, 12 m/s

[tjbateh]

Homework Statement

An elevator and its load have a combined mass of 2300 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 50 m.

Homework Equations

The Attempt at a Solution

Any idea how to approach this problem?

 

[tiny-tim]

Hi tjbateh!

This question is in two parts, a maths part and a physics part.

The maths is to find the acceleration.

The physics is then to find the force.

What is the acceleration? ​

 

[tomcue]

To solve for the tension in the elevator cable, we can use the formula T = mg + ma, where T is the tension, m is the mass of the elevator and its load, g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration of the elevator.

First, we need to calculate the acceleration of the elevator using the kinematic equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s in this case), u is the initial velocity (12 m/s downward), a is the acceleration, and s is the distance traveled (50 m). Rearranging this equation, we get a = (v^2 - u^2)/2s = (0^2 - 12^2)/2(50) = -1.44 m/s^2. This negative sign indicates that the elevator is decelerating.

Next, we can plug in the values into our formula T = mg + ma. We know that m = 2300 kg and g = 9.8 m/s^2. Substituting these values, we get T = (2300)(9.8) + (2300)(-1.44) = 22,540 N. Therefore, the tension in the elevator cable is 22,540 N.

It's important to note that this is the maximum tension in the cable, as it assumes that the elevator is brought to rest in the shortest possible distance. In reality, the tension may be slightly lower if the elevator takes a longer distance to come to a complete stop.