An error in my texbook?

[Hitman32]

Ok, I just did a word problem in my textbook and something isn't adding up. Here is the word problem:

A freight elevator weighing 3,000 pounds is supported by a 12-foot-long cable that weighs 14 pounds per linear foot. Approximate the work required to lift the elevator 9 feet by winding the cable onto a winch.

Ok, multiplying 3,000 by 9 gives me 27,000 ft/pounds of work to lift the elevator. Then when I did the integral to find the work required to pull the cable I got 945 ft./pounds. So adding the two I got 27,945 ft./pounds of total work to lift the cable and the elevator 9 feet. The answer in the back of my book says 36,945 ft./pounds. How could this be? Is it possible the authors accidently thought the work required to lift the elevator is 36,000 ft./pounds by multiplying 4,000 pounds by 9 feet?

 

[Anubis]

yes textbooks arnt perfect

 

[Hitman32]

Originally posted by: TheEvil1
yes textbooks arnt perfect

Did you work the problem and get the same answer as me?

 

it sounds like they mult. 3000 by 12 instead of 9. 12 being the total length of the cable, but at the same time only did 9 feet when calculating the ft/pounds of the cable.

 

[Anubis]

no i jus t mentioned teh fact that the books arnt perfect

 

[Hitman32]

Originally posted by: FallenHero
it sounds like they mult. 3000 by 12 instead of 9. 12 being the total length of the cable, but at the same time only did 9 feet when calculating the ft/pounds of the cable.

That makes more sense. So did they make the mistake or did I mess up?

 

Originally posted by: Hitman32

Originally posted by: FallenHero
it sounds like they mult. 3000 by 12 instead of 9. 12 being the total length of the cable, but at the same time only did 9 feet when calculating the ft/pounds of the cable.

That makes more sense. So did they make the mistake or did I mess up?

they made the mistake...unless im missing something completely. Speaking of math, I should do my homework.

 

[brxndxn]

Since it is going into a winch, moving from start..

The first foot is moving 12ft of cable and a 3000lb elevator.
The second is moving 11ft of cable and a 3000lb elevator.

Since it is only moving 9ft, 3ft of the cable can be considered part of the elevator weight. So the elevator now weighs 3000+14*3lbs

Then, you just do 14lbs * the integral from 0 to 9 for the cable... However, since the cable is uniform weight, you can just take 1/2 the distance of movement (4.5) and multiply that times the weight of the entire cable. That's how much work should be done..

Doing that, you get 9(4.5*14+3000+14*3)lbs.. then, you get 27,945.. I think you are correct.

 

[Hitman32]

Originally posted by: brxndxn
Since it is going into a winch, moving from start..

The first foot is moving 12ft of cable and a 3000lb elevator.
The second is moving 11ft of cable and a 3000lb elevator.

Since it is only moving 9ft, 3ft of the cable can be considered part of the elevator weight. So the elevator now weighs 3000+14*3lbs

Then, you just do 14lbs * the integral from 0 to 9 for the cable... However, since the cable is uniform weight, you can just take 1/2 the distance of movement (4.5) and multiply that times the weight of the entire cable. That's how much work should be done..

Doing that, you get 9(4.5*14+3000+14*3)lbs.. then, you get 27,945.. I think you are correct.

Yeah, I did it using the integration technique. That's what this section of my book is about, thanks for confirming that the authors of the textbook messed up on that one.

 

[Yossarian]

Now you can't trust anything the authors of that book are trying to tell you. It could all be wrong, who knows!!?!?!??

 

[Hitman32]

Originally posted by: PipBoy
Now you can't trust anything the authors of that book are trying to tell you. It could all be wrong, who knows!!?!?!??

Possibly but everything else checks out.