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Calc Woes

Soo i have my final for Calc coming and i r scared shitless. I was wondering if i can get some help on these 2 word problems i have.

1)There is a cable that is 40feet long and weighs 30lbs. It is hanging over the side of a building. How much work is Req’d to pull 10feet of it up?

2)A Chain lying on the ground is 10m long and its mass is 80KG. How much work is required to raise ONE END of the chain to a height of 6m.

First problem i am getting like 525ftlbs but im not sure if that is correct.

I think your first one is wrong since pounds are a unit of force. So esentially each foot is 3/4 a pound the work would be the integral from 0 to 10 of (3/4)x dx which is 37.5

#2 each meter is 8 kg’s. So the force required to lift one foot would be 8*9.8 = 78.4 newtons So I’m thinking the work would be the integral from 0 to 6 of 78.4x dx

[quote]stumpy wrote:
I think your first one is wrong since pounds are a unit of force. So esentially each foot is 3/4 a pound the work would be the integral from 0 to 10 of (3/4)x dx which is 37.5[/quote]

So then it would be 525/2 right? 262.5 ftlbs

[quote]stumpy wrote:
#2 each meter is 8 kg’s. So the force required to lift one foot would be 8*9.8 = 78.4 newtons So I’m thinking the work would be the integral from 0 to 6 of 78.4x dx[/quote]

hrmm fuck these KG’s and shit make it so confusing lol…

For the first one, our variable is the mass, however, as we pull the rope up over the edge we decrease it’s mass. This means we want to express the mass of the rope hanging over the edge in terms of the length we have pulled.

Let a = linear density of the rope (assumed to be uniform) = 30 lbs / 40 feet = 0.75 lbs/ft
Let x = length of rope we have pulled up over the edge.
Let m = the mass of the rope hanging over the edge = 40 lbs - 0.75x (since as we pull the rope up, the mass hanging over decreases).

Now, since W = Fx = mgx = (40 - 0.75x)g*x = 40gx-0.75gx^2

Therefore dW/dx = 40g-0.75gx and dW = (40g-0.75gx)dx and we can now integrate this expression from 0 to 10 and solve for W.

I’m sure I made some error as I’m just about to head to bed, but if nobody else has posted a solution for the second one I’ll give it a shot tomorrow morning.

EDIT: MAKE SURE YOU CONVERT YOUR UNITS. I was just too lazy and was only concerned with the process.

Mathematica is your friend.

sounds like a challenge. but i’ll have to convert the units first. where i’m at, we use the metric system yo.

this should be the answers. hope its clear to you but my hand writing is pretty much a mess.

[quote]malty_goodness wrote:
this should be the answers. hope its clear to you but my hand writing is pretty much a mess.[/quote]

Wait wat are you even using Calculus?

Like I tell my calc students: Math has a lot in common with viagra…you know, harder and harder. :slight_smile:

if by calculus u mean using differentiations and integrations?
in my own calculations no. but it can be inferred from the graphs no? find the equation of the line in the form y = mx + c, where m is the gradient and c = 0 since the line cuts the y axis at zero. integrate the equations with the relevant limits in either case and then mulitply that answer with the value of gravity.

Hey malty sorry earlyer i just wasnt paying attention lol :p. for the problem 2) im getting 1411.2 J any ideas as to why the discrepancies? Also talking to some people they are also saying it is 144J why did you change it?

for question 2, to get the value of 144, you would have multiplied the weight of the chain with the distance moved. so based on the graph, this value would be obtained with the equation: 0.5 * 6 metres * 48 kilograms.
however, this value would not be the value of work done.

work done = force * distance

also,
force = mass * acceleration

therefore: work done = mass * acceleration * distance

hence the reason we multiply 144 by the value of gravity, for which i used the value of 9.81. i suppose the difference between your answer and mine lies in the number of significant numbers used in calculations, in particular the value of gravity.

For number 2,

Work = int(Force*dh) where h is height.

So the force at any instance is 8 (kg/m) * 9.81 (m/s/s) * h (m)

Work = int(89.81hdh)
Work = 0.5
89.81h^2|0-6
Work = 0.589.816^2-0
Work = 1412.64 (J) (N
m)

This is likely the way your teacher would want you to do the problem for class.
Of course there are other ways like malty’s, but that wouldn’t probably be accepted.

Also, work=change in potential energy
so it would also work to say potential energy = 6(m)*8(kg/m)*3(m)9.81(m/s/s) = 1412.64 J
Which is just (height)
(force) applied at the center of gravity

My work is NOT 100% correct BTW. I fudged through the negative signs which are only important if your teacher incorporates dot products into the definition of work…

In that case, force is dotted with dh instead of multiplied, but the Force is negative because it is applied in the negative h direction. But the dot product also includes a cos(omega) term where omega is the angle between the h and Force. So it is cos(omega)=cos(pi)=-1, so the negatives just cancel out. So whether or not you have to include that depends on your teacher and how you were taught the subject. I think in my calc class I had to include all of that stuff.

Likewise for number 1,

Weight at any instance is going to be 30/40 (lbs/ft)* h (ft)

so work = int((30h/40)-dh) definite integral from 40-30 ft (dh is negative because it is in the negative h direction)
work = -30/40 * h^2 | 40-30ft
work = (30/40) * 40^2 - (30/40) * 30^2
work = 525 ft-lbs

So your answer is correct.