# How do I find the tension in a cable in an elevator cab?

1. Jan 31, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42 m."

$m = 1600 kg$
$v = 12 m/s$
$d = 42 m$

2. Relevant equations
$t = d/v$
Answer: $T = 1.8 * 10^4 N$

3. The attempt at a solution
$t = (42 m)/(12 m/s) = 3.5 s$
$v/s = (12 m/s)/(3.5 s) = 3.43 m/s^2$
$T = (1600 kg)(9.8 m/s^2 - 3.43 m/s^2) = 10192 N ≠ 18000 N$

I honestly have no idea how to tackle this problem, am just making guesses, and find it difficult to be doing this everyday as a student engineer. It's supposed to be so simple, yet I can't bloody do this. I'm doubting my ability to enter the engineering field if I can't do a simple problem such as this.

2. Jan 31, 2016

### SteamKing

Staff Emeritus
You've just thrown some calculations together at the end of your post.

There is a SUVAT equation which combines initial velocity, final velocity, distance traveled, and acceleration. Have you tried using that to find the acceleration a?

3. Jan 31, 2016

### Eclair_de_XII

The textbook did not give me any equations to work with, so I didn't even know to use them.

4. Jan 31, 2016

### SteamKing

Staff Emeritus
You've got a computer. There's a wealth of knowledge at your fingertips. Google "SUVAT equations" and you can find what you need.

P.S.: What kind of textbook doesn't have equations in it? Throw that thing away. It's defective.

5. Jan 31, 2016

### Eclair_de_XII

https://en.wikipedia.org/wiki/Equat...translational_acceleration_in_a_straight_line

$v^2 = u^2 + 2as$
$0 = (12 m/s)^2 + 2a(42 m)$
$a = -(12 m/s)^2/2(42 m)$
$a = -1.7143 m/s^2$
$F = (a + g)(1600 kg) = (-1.7143 m/s^2 - 9.8 m/s^2)(1600 kg) = -18,423 N$
$T = -F = 18,423 N$