Help pleeeease :3
Kay, here is the last question to my homework. I need help understanding it :/
for a they even gave you the units!!
|dw:1476060141630:dw|
The rope weighs is 0.02 lbs / foot.
Hence \(W\int_0^{30} 0.02x dx\)
dat sit!! :) Or I underestimate the problem??
On the computer, you need type \("0.02*x_i^* \triangle x\)
wouldnt gravity play a role in it?
Is it not that gravity is included on calculating mass?
I don't know physics!! :)
you're using weird American units but indeed, g should be in there
If you're still wondering about the units and where gravity comes in, look at the mini-integration they first suggest on that page you attached: \(W_1 = \int \dfrac{3}{5} x ~ dx \) ....where the \(\dfrac{3}{5} \) is the \( 0.6 \dfrac{lb}{ft}\) linear density of the rope So the units of \(W_1\), shown in red below, are \(\int \dfrac{3}{5} \color{red}{(\dfrac{lb}{ft})} x \color{red}{(ft)} ~ dx \color{red}{( ft )} = lb ~ ft \qquad \triangle\) \(W_1\) is supposed to be a work calculation, which (put crudely) is the same as energy, and should have units that follow the idea that \(W = F \cdot x = m \cdot \color{blue}{a} \cdot x \) ....which has units \(lb \cdot \color{blue}{\dfrac{ft}{s^2}} \cdot ft = \dfrac{lb ~ {ft}^2}{s^2}\) You haven't got that because the middle *blue* bit is missing; but you would if you multiplied what you have with the units of acceleration (which are the units of gravity), ie: \(\dfrac{ft}{s^2}\) So whatever you get from this odd way of doing things, well to make the numbers actually useful, you need to multiply by a factor of g. which i think you know already.
and you don't actually need calculus here, you can just model it as a point mass.
kay, so i got this far, but dont know what to do next. And i need help with part a still.
@zepdrix
I'm trying to understand it, but i honestly just dont get it.
Lmao. never mind. I got it
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