A caluclus work problem.

Question

A cylindrical tank of radius 4 ft and length 12 ft is lying on its side on horizontal ground. The tank initially is full of gasoline weighing 40 lb/ft³. How much work is done in pumping all this gasoline to a point 6 ft above the top of the tank?

Answer 1

The volume of the tank will be:

∏r²X12 ft³ = 603.20 ft³

Work is the act of changing the energy of a particle, body or system, in our case  is force times space so the total weight is

40X603.20 = 24128 lb

The space will be 6 ft

and the work is

24128 X 6 = 144768 ft-lbf =

196276 Joules.

 

 

Answer 2

Hello Francisco,

 

In your above question everything is ok.

But the height should be taken as (6+4)=10ft and not 6ft.

Just see, how the cylinder is lying.

Since the total weight shall be acting at the center of gravity.

So, the total height to be lifted should be 10ft and not 6 Ft

So, work done = 24128*10=241280 Ft pounds.

Kindly refer to Q no 27 on page 452 Larson calculus.

 

Surendra