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An open tube of length L = 1.8 m and cross-sectional area A = 4.6 cm^2 is fixed to the top of a cylindrical barrel of diameter D= 1.2m and height H=1.8 m

The barrel and the tube are filled with water (to the top of the tube). Calculate the ratio of hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel. Ignore atmospheric pressure for this question.

P= patm + pgh

Weight=mg=pVg

So I tried calculating the volume of the cylinder, tube then subbing it into the equation

Weight=mg=pVg

however it did not work properly as the answer is supposed to be 2.

Did i convert the Area to Radius wrong? Or is there another way im supposed to go about this

I set up my ratio as = Weight of Barrel + Tube / Weight of barrel only

V of barrel = Pi * (0.6)^2 * 1.8 = 2.03575204

I subbed this into W=mg=pVg = 1000 g/m^3 * 2.03575204 * 9.8 m/s^2

W=19950

I'm not sure how to go about the tube part as it given us the cross sectional area

http://g.imageshack.us/img255/hrw71431.gif/1/ [Broken]

An open tube of length L = 1.8 m and cross-sectional area A = 4.6 cm^2 is fixed to the top of a cylindrical barrel of diameter D= 1.2m and height H=1.8 m

The barrel and the tube are filled with water (to the top of the tube). Calculate the ratio of hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel. Ignore atmospheric pressure for this question.

## Homework Equations

P= patm + pgh

Weight=mg=pVg

## The Attempt at a Solution

So I tried calculating the volume of the cylinder, tube then subbing it into the equation

Weight=mg=pVg

however it did not work properly as the answer is supposed to be 2.

Did i convert the Area to Radius wrong? Or is there another way im supposed to go about this

I set up my ratio as = Weight of Barrel + Tube / Weight of barrel only

V of barrel = Pi * (0.6)^2 * 1.8 = 2.03575204

I subbed this into W=mg=pVg = 1000 g/m^3 * 2.03575204 * 9.8 m/s^2

W=19950

I'm not sure how to go about the tube part as it given us the cross sectional area

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