Integration Application: Find the force on each end of the cylinder

A cylinder shaped container of tar has a diameter of 2m. If it is half full of tar with a density of 920 kg/(m^3) and is on it's side, find the force on each end.

Helpful hints:

920 kg per m^3 is the mass density not the weight density. Take g= 9.8 m/ sec^2

Please assist.

Re: Integration Application: Find the force on each end of the cylinder

Since tar is a liquid (though a very viscous liquid) it exerts the same force in every direction. In particular, a particular "point" of tar in contact with an end exerts force on that end equal to the weight of tar above it. In particular, a "point" of tar at height y above the bottom of the cylinder (which is lying on its side) has a column of tar of height 1- y (from y up to the mid line at height 1) which, taking it have have base dA, has volume (1- y)dA and so weight (920)(9.8)(1- y)dA which then exerts **pressure** (920)(9.8)(1- y) against the end. Find the total force by integrating that over the bottom half-circle of the end.

Re: Integration Application: Find the force on each end of the cylinder

Im lost. Where did dA come from? Also, what do you mean by bottom half circle of the end?

Thanks!

Re: Integration Application: Find the force on each end of the cylinder

Quote:

Originally Posted by

**HallsofIvy** Since tar is a liquid (though a very viscous liquid) it exerts the same force in every direction. In particular, a particular "point" of tar in contact with an end exerts force on that end equal to the weight of tar above it. In particular, a "point" of tar at height y above the bottom of the cylinder (which is lying on its side) has a column of tar of height 1- y (from y up to the mid line at height 1) which, taking it have have base dA, has volume (1- y)dA and so weight (920)(9.8)(1- y)dA which then exerts **pressure** (920)(9.8)(1- y) against the end. Find the total force by integrating that over the bottom half-circle of the end.

Ok. Using the given then, this is what you would evaluate for the solution, yes?

$\displaystyle \int \frac{(920)(9.8)(1-y)}{\pi } $

I assumed that you mean pi when you say bottom half-circle, as half a circle is pi right?

Re: Integration Application: Find the force on each end of the cylinder

I would go somethig like this. Since the force is F=P*A and P=q*g*h the integral should look something like this

F=∫∫q*g*y*dA over the area of the base of your cylinder.

and from my poor understanding of english i think Hall above has wrote the same thing only better(Rofl)