Lizwi Posted August 18, 2019 Share Posted August 18, 2019 Why is the pressure on top of this fluid elements on the right equal to p+dp Link to comment Share on other sites More sharing options...

timo Posted August 18, 2019 Share Posted August 18, 2019 There are two approaches here, the formal one and the brain-compatible one. 1) Formally: Realize that there are hidden coordinate dependencies. You are probably looking to construct a function p(y). Since y is the coordinate at the lower side, you have p(y) at the lower edge and p(y+dy) at the upper. This is (possibly) slightly different from p(y) (because of the displacement dy). If you call the difference dp, then p(y+dy) = p(y) + dp(y) = p + dp (note that dp can and will be negative). 2) Brain compatible: Put the equation first and then define the variables: The forces on top and bottom should cancel out. There force up is the pressure force p*A from below. The force from up is the pressure force p2*A from above the small fluid element plus the weight dw of the small fluid element. Since we are talking about infinitesimal coordinates, and since p(y) should be a function, it makes sense to say that p2 = p+dp, which you can then integrate over. 1 Link to comment Share on other sites More sharing options...

studiot Posted August 18, 2019 Share Posted August 18, 2019 (edited) Timo's posh explanation is just fine but try this simplified one. Looking at the diagram I see that upwards is defined positive. Looking at your equation and the diagram pA - (p+dp) A - ρgAdy So I see that the arrow on the underside is pointing up so pA is positive. But on the top, p and dp point downwards so are negative so (p+dp) is negative on the top dy is positive and g is negative so the third term is also negative. does this help? Edited August 18, 2019 by studiot 1 Link to comment Share on other sites More sharing options...

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