# Thread: Applications of Intergration (Work related)

1. ## Applications of Intergration (Work related)

1. A heavy rope, 50ft long, weights 0.5 lb/ft and hangs over the edge of a building 120 ft high.

a) How much work is done in pulling the rope to the top of the building?

b) How much work is done in pulling half the rope to the top of the building?

My solution:

Work = Force x Distance
W = $\int Fxdx$

Force = (Density) = (0.5)

So Fx = 0.5x

Then: $\int_{0}^{50} 0.5 x dx$

Is that correct?

I have no idea how to do the b part.

Thank you!

P.S. How come I can't do the upper and lower limit of defined intergration? I used \int^50_0 and it didn't work.

2. For the syntax question, you reversed the order. I fixed it for you. It should be int_{0}^{50}.

Look at your units for a guide. You know that the unit of work must be lb*ft. The force is not equal to the density. The lb is a unit of force. You must multiply $\frac{.5 lb}{ft} \times 50ft$ to get the correct unit for force. Now setup your integral.

$W=\int_{0}^{50} 25xdx$

3. Consider a length element $dx$ which is $x$ft below the top of
the building. To lift this to the top of the building will require:

$dW=\rho. dx. g. x$

work in whatever units we are using.

So the work required to lift the whole rope is:

$W = \int_0^{50} \rho. g .x\ dx$,

where $\rho$ is the mass of the rope per unit length, $g$ is
the acceleration due to gravity. Now if we are using pounds as our unit
of mass, and the foot as our unit of length, then $g\approx 32\ \mbox{ft/s^2}$,
and $W$ is in units of foot poundals, to convert it to foot pounds we
would divide by $g$, then:

$W = 0.5 \int_0^{50} x\ dx\ \mbox{ft.lb}$.

Notes:

1. This to some extent demonstrates the absurdity of customary units,
which in this context manages to confuse units of mass and force.

2. This demonstrates the absurdity of exercises in integration of this
type. This can be done in one line without integration as the work done
is the change in potential energy when a mass equal to the mass of the
rope is lifted from the position of the ropes c-of-g to the top of the building.

RonL

4. Now I'm a bit puzzled CaptainBlack. If you integrate a unit of ft-lb with respect to ft, the resulting unit would be ft^2-lb, which isn't the unit for work. What am I missing?

5. Originally Posted by Jameson
Now I'm a bit puzzled CaptainBlack. If you integrate a unit of ft-lb with respect to ft, the resulting unit would be ft^2-lb, which isn't the unit for work. What am I missing?
Sorry if the notation is confusing, the ft-lb denotes the units of W. The
integral is a density (now in slightly peculiar units of lb-force per unit length)
times an integral of something in ft wrt something in ft. So we have
dimensionally:

[W]=[lb-force]/[ft] x [ft]^2=[lb-force][ft]

RonL