How much work is done pulling the entire 100ft length of chain up the side of a building? The chain weighs 15lb/ft.
Note: The chain weighs $\displaystyle 100 \, ft \cdot 15 \, lb/ft = 1500 lb$. The linear weight density of the chain is 15 lb/ft. I will call the linear weight density $\displaystyle \lambda = 15 \, lb/ft$ and $\displaystyle L = 100 \, ft$.
$\displaystyle W = \int \vec{F} \cdot d \vec{s}$
I am assuming at any given point that the length of chain that remains to be lifted is "x". Since we are exerting force to lift the chain in the same direction that the chain is being moved, $\displaystyle \vec{F}$ and $\displaystyle d \vec{s}$ are always in the same direction, thus
$\displaystyle W = \int \vec{F} \cdot d \vec{s} = \int Fds$
F at a given point is going to be the magnitude of the weight of the chain remaining to be lifted but in the upward direction. Thus:
$\displaystyle F = - \lambda x$
Thus
$\displaystyle W = \int_L^0 (- \lambda x)dx = - \lambda \int_L^0 xdx$
$\displaystyle W = -\lambda \left ( -\frac{1}{2}L^2 \right )$
$\displaystyle W = \frac{\lambda L^2}{2} = 75000 \, lb ft$
(What the heck is the English unit for mechanical energy? I know "ergs" but I've forgotten the other one!)
-Dan
The work done is equal to the change in potential energy of the chain, and so is equal to the mass of the chain times the height change of the centre of mass times g the accelleration due to gravity.
The only difficult thing about this problem is the ridiculous unit system it is expressed in.
I guess in these units the answer is 50.1500 =75000 foot pounds(farce).
which is also 2400000 ft poundals.
RonL
That the US education system still expresses these problems in these ludicrous units is a clear indication that there is something wrong with the intestinal fortitude of all those involved in setting the agenda for mathematics education in the States.
Oh!. I messed up my unit. I hate it when I do that (because I'm such a stickler about my students getting them right. )
Mainly it's the engineering sector in the US that is still using the English system. The explanation I've heard is that it would cost them too much money to retool into metric. Funny that it didn't seem to cost too much for the rest of the world to do it...
-Dan
Now I think it would help if schools stop attempting to teach customary units.
(any how many lost Mars probes will cover the cost of retooling anyway -
BTW we only ever hear of losses due to unit confusion on publicly funded
systems who knows how much is really lost due to inconsistent units).
RonL