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Math Help - Integration Application: Find the force on each end of the cylinder

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    Newbie andvaka's Avatar
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    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.
    Last edited by andvaka; June 24th 2012 at 04:09 PM.
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    MHF Contributor

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    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.
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    Newbie andvaka's Avatar
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    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!
    Last edited by andvaka; June 24th 2012 at 09:33 PM.
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    Newbie andvaka's Avatar
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    Re: Integration Application: Find the force on each end of the cylinder

    Quote Originally Posted by HallsofIvy View Post
    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?

     \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?
    Last edited by andvaka; June 24th 2012 at 09:43 PM.
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  5. #5
    Imo
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    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
    Last edited by Imo; June 25th 2012 at 02:56 AM.
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