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Undoubtedly0

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Consider a cylindrical barrel with cross-sectional area

*A*and height

*H*. Now insert a straw of height

*h*and cross-sectional area

*a*so that the total of height of the barrel and straw is

*h + H*. Assuming a fluid of uniform density

*ρ*, the total weight of the fluid is

[tex] W = m_{tot}g = \rho V g = \rho g (ha + HA)[/tex]

If the barrel is sitting on a plank, then the magnitude of the normal force

*N*of the plank on the barrel is simply

*N = W*.

However, using a different method, I come up with a different result for

*N*. By Pascal's equation, the gauge pressure at the bottom of the tank is

[tex] P_{bottom} = \rho g (h + H) [/tex]

Thus the force due to pressure acting on the bottom of the barrel is

[tex] F_P = P_{bottom}A = \rho g A (h + H) [/tex]

In order to balance this force, the normal force

*N*from the plank on the barrel must thus be equal to

*F*, yet this is a different result then from before:

_{P}*N = W ≠ F*.

_{P}= NThere must be some error in my reasoning here, but I cannot find it. Thanks a bunch for the help!