Search 74,106 tutors
FIND TUTORS
Ask a question
0 0

A caluclus work problem.

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?
Tutors, please sign in to answer this question.

2 Answers

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.
 
 
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 

Comments

SURENDRA K.
Thanks, you are right maybe I did not realize that the Cylinder was lying as you say and also the problem. I do not need to refer to any solved problem because i have to solve them In the best way. Thanks again.

Woodbridge calculus tutors