Tank is shape of right circular cylinder with radius of 3m and height of 10m axis is horizontal. tank is full of gasoline with density of 680 kg/m^3 calculate work to empty tank through a hole at the top of one of the circular ends.

Question

perl · Answer

you can integrate work here

anonymous · Answer

yeah is my F= 680(9pi)?

anonymous · Answer

is r constant at 3m?

perl · Answer

we can cut the cylinder into slices

anonymous · Answer

because when i was doing the sphere the r changes and i had to use x=sqrt(stuff)

perl · Answer

the i'th slice

perl · Answer

what is the work to move the ith slice up a distance of 10 - y

anonymous · Answer

yeah i get that, but i don't know the Force

perl · Answer

force is weight

anonymous · Answer

usually force is equal to the density times the volume correct?

perl · Answer

density = mass / volume.

volume * density = mass

perl · Answer

then 
mass * g (gravity constant) = weight

perl · Answer

the weight of the ith slice is
mass * gravitational acceleration constant (g)

but mass  = volume * density

weight = Volume * density * g

pi * r^2 * dy * density * g

anonymous · Answer

what is the function of "r" in this problem?

perl · Answer

r = 3

perl · Answer

we use r in the formula for the slice of volume (it looks like a pancake)

perl · Answer

the work to empty a cylindrical tank of a liquid of density rho is the integral

Integral {y = 0 , y = h} pi * r^2 * dy * rho * g * (h - y )

perl · Answer

here h = 10, rho = 680, r = 3 , g = 9.8  
you might be able to factor out all the constants first

anonymous · Answer

this is for calculus not physics, we don't use g

perl · Answer

yes you have to use g, since you are given kg/m^3 for density

anonymous · Answer

volume should be: integral from 0 to 10 of 680(9)(pi)(10-y) dy?

perl · Answer

use a better estimate for g

anonymous · Answer

my 9 is from r^2

perl · Answer

oh you left out g then

perl · Answer

we can try it your way

perl · Answer

yes , integrate that

perl · Answer

you should get 50 for that

perl · Answer

integral 680(9)(pi)(10-y)  =  680 * 9 * pi * 50  

(but i still think this is incorrect)

anonymous · Answer

ok, i don't know about multiplying by g i thought that since density was given we didn't need to do that

perl · Answer

usually these calculus problems are given in units of pounds and feet , so you don't need to multiply by g . this problem is given in metric system, its different

perl · Answer

we are not given the weight density, thats the problem. we are given the 'mass' density.

anonymous · Answer

i see thank you

perl · Answer

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