Calculus problemwork needed to produce a pile of sand

[kagaku]

How much work must be done in producing a conical heap of sand of base radius 1.2 meters and height 1 meter? The specific weight of sand is 2 grams per cubic centimeter. You may assume that all the sand is taken from the surface of the Earth (that is, from height 0 meters).

I have no idea how to set up the problem...once it's set up i'll be able to do the math, but I'm just stuck right now. Any suggeestions/help is greatly appreciated. Thanks.

 

[whozum]

You'll want to show some work before getting some help.

My suggestion is to build a reimann sum taat describes the work required to lift up one layer of width $\Delta x$ to set up your integral.

 

[lurflurf]

How much work must be done in producing a conical heap of sand of base radius 1.2 meters and height 1 meter? The specific weight of sand is 2 grams per cubic centimeter. You may assume that all the sand is taken from the surface of the Earth (that is, from height 0 meters).

I have no idea how to set up the problem...once it's set up i'll be able to do the math, but I'm just stuck right now. Any suggeestions/help is greatly appreciated. Thanks.

Consider a thin cylindrical slice.
V=pi R^2*h
You want to approximate the cone with a number of such slices.
If you place a slice at height y 0<y<1m
The radius should be 1.2(1-y)
Find the work needed to lift a slice into position.
Make a cone approximation with n slices and express the work as a sum of the work for each individual slice.
Take the limit as the thickness of slices becomes small.
This will be an integral which gives the work.
Recall
work=force*distance
force=mass*acceleration
in this problem acceleration=g (due to gravity)~9.81 m/s^2 most places on the surface of earth.
mass=density*volume

 

[kagaku]

using those equations..the work would be the integral of density*volume*acceleration with respect to distance/heighth?
density = 2g/cm^3 = 2000000g/m^3
r = 1.2(1-h)
volume = pi(r^2)(h) = 1.44pi(h-2h^2+h^3)
acceleration = -9.81m/s^2

so it's...-28252800 * the integral of (h-2h^2+h^3) from 0 to 1

which is -28252800[(1/2)h^2 - (2/3)h^3 + (1/4)h^4] from 0 to 1

which equals -28252800[1/2 - 2/3 + 1/4] = -2354400

that means I'm completely wrong because there's no way the work required to overcome gravity is going to be negative...please help

 

[lurflurf]

using those equations..the work would be the integral of density*volume*acceleration with respect to distance/heighth?
density = 2g/cm^3 = 2000000g/m^3
r = 1.2(1-h)
volume = pi(r^2)(h) = 1.44pi(h-2h^2+h^3)
acceleration = -9.81m/s^2

so it's...-28252800 * the integral of (h-2h^2+h^3) from 0 to 1

which is -28252800[(1/2)h^2 - (2/3)h^3 + (1/4)h^4] from 0 to 1

which equals -28252800[1/2 - 2/3 + 1/4] = -2354400

that means I'm completely wrong because there's no way the work required to overcome gravity is going to be negative...please help

O.k. you are most of the way there.
lets review and see where things went wrong.
Whether g=-9.81 m/s^2 or 9.81 m/s^2 depends on if we call up positive or negative. Just reason that we are resisting gravity so our work is positive so is g.
on units we measure work in joules=J=kg*m^2/s^2
so take all distances in m and masses in kg
you might consider using 1 KJ=1000 J
also you had pi in at the beginning and lost it along the way
-28252800[1/2 - 2/3 + 1/4] = -2354400
should be
28252.8*pi[1/2 - 2/3 + 1/4]J = 2354.400pi J=2.354pi kJ=7.397 kJ
also see what you book uses for g some use 9.8 or even 10 to make the numbers nicer.

 

[kagaku]

ok, thanks for the help...i guess if i had taken physics this whole problem would have made a lot more since.

but one last question...

isn't the volume of a cone 1/3 pi r^2 h?
this would make it 1/3 of 7.397 or 2.466 J

 

[lurflurf]

1/3[/B] pi r^2 h?
this would make it 1/3 of 7.397 or 2.466 J

The volume of the cone is not directly relevant since the work varies with the height, the work you found found is in effect the volume weighted with the height.
work=density*radius^2*g*pi*integral(0<y<1)(y(1-y)^2 dy)
but
volume=pi*radius^2*integral(0<y<1)((1-y)^2 dy)=pi*radius^2/3
So the integral that gives the volume is not the same as the one that gives work. The one for work has an extra y factor.

 

[kagaku]

ok. thanks for helping me understand the problem and working me through it.