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

[Eclair_de_XII]

Homework Statement

"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."

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

Homework Equations

$t = d/v$
Answer: $T = 1.8 * 10^4 N$

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.

 

[SteamKing]

Homework Statement

"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."

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

Homework Equations

$t = d/v$
Answer: $T = 1.8 * 10^4 N$

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.

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?

 

[Eclair_de_XII]

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?

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

 

[SteamKing]

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

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.

 

[Eclair_de_XII]

Google "SUVAT equations" and you can find what you need.

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

There is a SUVAT equation which combines initial velocity, final velocity, distance traveled, and acceleration.

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

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

Thank you for saying this. I thought it was just me. I made a whole topic voicing my complaints on it, and I was convinced to keep using it, until now.