Finding Work using Calculus - The Cable/Rope Problem - Part b

Finding Work using Calculus - The Cable/Rope Problem - Part b.
In this video, I find the work required to lift up only HALF of the rope to the top of the building.

Category:

Education

Tags:

• math
• calculus
• work
• cable
• rope

[bluejimmy168]

Where's part A?



[patrickJMT]

@bluejimmy168 it was set to private for some reason, sorry. it is available now!

[choconiel]

today ill take my test for 25B and 25C your videos helped me a lot for those 2 tests :DDD watching ur videos and reading the book, i feel that i can conquer the world hahahhahaa thanks

[patrickJMT]

@choconiel ha : ) good luck on it all!

[Bob8199]

Patrick, you're the best math tutor on Youtube.

[patrickJMT]

@Bob8199 thanks! just tryin' my best here...

[espobulldog]

Absolutely awesome. Keep these coming please! There is such a lack of informational videos regarding the applications of calculus. These type of problems are great practice problems too.

[swabsDE]

why can't you do that all in one integral? integrate between 0 and 30 (60-x)2dx

[ixionike]

@farazbehrouz2002

that's unit of weight not mass. weight is a force.

[hallyo456]

Why cant you just use 30ft for delta x for the first 30ft as you did the last 30 ft? The change in distance for the first half is the same as the second half, is it not?

[TheAnthonySoriano]

What do you do if there is a block attached at the bottom of the rope? How do you set up the integral..?

[farazbehrouz2002]

the units of weight is just lb, not lb/ft.

other than that. great vid.

[Charounson]

Patrick, is it not slightly incorrect to use units of feet and pounds in a work problem? Force is defined to be measured in Newtons so shouldn't your units for force be in Newtons or some form of newtons (i.e. kiloNewtons..). Also, work is a form of energy so it's given in Joules so, the equation W=F*d is no longer homogeneous if you units at the end are really ft-lbs.. I don't believe Joules = ft-lbs.. It's possible that I'm completely wrong. :)

[turionking]

I'm using the metric system if that would make a difference?

[turionking]

@turionking And don't you have to treat the bottom section as a dead weight and not treat it as an integral?

[turionking]

Question: Why aren't you multiplying the mass by the force of gravity 9.8? :)

[newpurgatory]

@bluejimmy168 if a derivative is the slope the very instant of a point then the opposite, the antiderivative, would be like adding all these infinitesimal points from a to b (The limits of Integration). all you are doing is reversing the operation of a derivative. Thus, the opposite of the slope of an instant is the total change of something. idk it just works!