Correction, I typed the integral wrong! Forgot to include the distance:
dF = 53.1(3)(2 - (y/3)) dy
dw = dF * Distance
W = ∫ 53.1(3)(2 - (y/3)) (y + 3) dy
W = 3106.35 ft-lb
I am trying to understand my teacher's example of a Work problem.
Cut to the chase here's a picture of the problem:
goo.gl/photos/AqZ6ENmHLg6heixD7
While I understand how to integrate fairly well, I'm still confused on how exactly my teacher set up the work problem.
Since work = force * distance
The distance would be (y + 3) [which makes since because the engine is 3 feet above the fuel tank.]
The length would be 3 [makes sense], and somehow, the width is 2 - (y/3)??? Where did that come from???
I understand that we're trying to integrate by dy, but I'm still clueless on how the teacher got that width equation. Nevertheless, the integral would lead to:
dF = 53.1(3)(2 - (y/3)) dy
W = ∫ 53.1(3)(2 - (y/3)) dy
W = 3106.35 ft-lb
So again, what exactly is "y" representing when it comes to the width? Where did that equation come from?
I think it might've been a way to flip the trapezoidal graph, but I'm not completely sure.
Any help is greatly appreciated!
2 answers
the width equation is due to the taper of the tank
as y goes from 0 to 3, the width goes from 2 to 1
the volume of the fuel increment is
... l * w * dy
... the force increment is the volume times the fuel density
as y goes from 0 to 3, the width goes from 2 to 1
the volume of the fuel increment is
... l * w * dy
... the force increment is the volume times the fuel density
1 answer