A chain, with length h = 100 feet and weighing 3 pounds per foot, hangs over the side of a building. We are going to determine the work required to lift the chain to the top of the building.

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Chapter7: Work And Kinetic Energy
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Problem 84AP: A cart is pulled a distance D on a flat, horizontal surface by a constant farce F that acts at an...
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A chain, with length h = 100 feet and weighing
3 pounds per foot, hangs over the side of a
building.
We are going to determine the work required to
lift the chain to the top of the building.
X = 0
Xi
di
x = h
This problem depends on the vertical coordinate
system to the left.
The force, F;, on the i" strip of the chain
;th
weight of the i" strip of the chain.
;th
Fi
pounds
th
The distance, di, we are moving the i" strip is
di
feet
(Must relate back to x; ).
The work, Wi, required to lift the i" strip of
Transcribed Image Text:A chain, with length h = 100 feet and weighing 3 pounds per foot, hangs over the side of a building. We are going to determine the work required to lift the chain to the top of the building. X = 0 Xi di x = h This problem depends on the vertical coordinate system to the left. The force, F;, on the i" strip of the chain ;th weight of the i" strip of the chain. ;th Fi pounds th The distance, di, we are moving the i" strip is di feet (Must relate back to x; ). The work, Wi, required to lift the i" strip of
This problem depends on the vertical coordinate
system to the left.
;th
The force, F;, on the ih strip of the chain =
weight of the itn strip of the chain.
„th
Fi
pounds
The distance, di, we are moving the
;th
strip is
di
feet
(Must relate back to xi ).
The work, Wi, required to lift the ith
the chain is Wi
strip of
F; · di.
Wi
ft-lbs
Adding up the work on all the subintervals and
allowing n 0, where n represents the
number of
subintervals, gives the integral for total work,
W, required to lift the chain to the top of the
building.
W
dx
Finally, W
foot-pounds
Transcribed Image Text:This problem depends on the vertical coordinate system to the left. ;th The force, F;, on the ih strip of the chain = weight of the itn strip of the chain. „th Fi pounds The distance, di, we are moving the ;th strip is di feet (Must relate back to xi ). The work, Wi, required to lift the ith the chain is Wi strip of F; · di. Wi ft-lbs Adding up the work on all the subintervals and allowing n 0, where n represents the number of subintervals, gives the integral for total work, W, required to lift the chain to the top of the building. W dx Finally, W foot-pounds
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