Euler's equation pressure difference

[fayan77]

Homework Statement

I am after P_C - P_A
However I must do so without breaking into components. My problem has different values

L=3
H=4
SG=1.2
downward a = 1.5g
horizontal a = 0.9g
and my coordinate is conventional positive y up and positive x to the right
cos$\theta$ = 3/5
sin$\theta$ = 4/5

Homework Equations

- $\frac {\partial} {\partial l}$ (p+$\gamma$l) = $\rho$a

- $\nabla p$ = $\rho$a

The Attempt at a Solution

a = 0.9g$\hat i$ -1.5g$\hat j$
therefore,
- $\frac {dp} {dl}$-$\gamma$ = $\rho$(0.9g$\hat i$ -1.5g$\hat j$)

where dp = P_C - P_A
dl = 5 because of the 3,4,5 triangle
and in order to get rid of (i hat and j hat) i use triangle relationship and project 0.9g$\hat i$ to l direction using 0.9g(3/5) and -1.5g(4/5)
therefore,
- (P_A - P_C) / 5 = $\rho$(0.9g(3/5)-1.5g(4/5)) + $\gamma$
P_C - P_A= -5$\rho$g(.9(3/5)-1.5(4/5)+1)
P_C - P_A= not correct answer

 

[Chestermiller]

This doesn't seem to have been set up correctly. You have vectors on one side of the equations and scalars on the other. That's a no-no.

The starting equation should be:
$$-\nabla{p}-\gamma\hat{j}=\rho\hat{a}$$
To get the derivative of p in the direction from C to A, you need to dot this equation with a unit vector in that direction. What is the equation for the unit vector in the direction from C to A?

 

[fayan77]

(3/5)$\hat i$ + (4/5)$\hat j$

 

[Chestermiller]

(3/5)$\hat i$ + (4/5)$\hat j$

When you dot $\nabla p$ with this unit vector, you get the "directional derivative" $\frac{\partial p}{\partial l}$, where l is measured along the diagonal from C to A. What do you get when you dot the other terms in the equation with this unit vector?

 

[fayan77]

let (3/5)$\hat i$ + (4/5)$\hat j$ = W
where $\gamma$ = $\rho$g

-$\frac{\partial p}{\partial l}$ -$\rho$g$\hat{j}$ DOT W =$\rho\hat{a}$ DOT W
-$\frac{\partial p}{\partial l}$ = [$\rho$g$\hat{j}$ DOT W] DOT [$\rho\hat{a}$ DOT W]
-$\frac{\partial p}{\partial l}$ = $\rho$g[.9$\left( \frac 3 5 \right) \hat i$ - .5$\left( \frac 4 5 \right) \hat j$]

 

[Chestermiller]

let (3/5)$\hat i$ + (4/5)$\hat j$ = W
where $\gamma$ = $\rho$g

-$\frac{\partial p}{\partial l}$ -$\rho$g$\hat{j}$ DOT W =$\rho\hat{a}$ DOT W
-$\frac{\partial p}{\partial l}$ = [$\rho$g$\hat{j}$ DOT W] DOT [$\rho\hat{a}$ DOT W]
-$\frac{\partial p}{\partial l}$ = $\rho$g[.9$\left( \frac 3 5 \right) \hat i$ - .5$\left( \frac 4 5 \right) \hat j$]

That’s not done correctly. The dot product of two vectors is a scalar.

 

[fayan77]

got it,
-$\frac{\partial p}{\partial l}$ = $\rho$g(.14)

-(P_A - P_C) / 5 = $\rho$g(.14)

P_C - P_A = 5 (1.2)(1.94)(32.2)(.14) = 52.4

So what we did here was just projecting everything along CA?
Is there another way of doing this? I saw a solution on the internet solved differently. Do you mind explaining the other solution?

 

[Chestermiller]

got it,
-$\frac{\partial p}{\partial l}$ = $\rho$g(.14)

-(P_A - P_C) / 5 = $\rho$g(.14)

P_C - P_A = 5 (1.2)(1.94)(32.2)(.14) = 52.4

So what we did here was just projecting everything along CA?

Sure. What they wanted you to realize was that you can get the directional derivative by dotting the equation with a unit vector in the desired direction.

Is there another way of doing this? I saw a solution on the internet solved differently. Do you mind explaining the other solution?
View attachment 220408

This is just basically equivalent to resolving the equation into components, to get the partial derivatives in the x and y directions. I really don't like what they've done here. The only reason this method works is that, for this particular problem, the partial derivatives of p with respect to both x and y are constant, independent of x and y.

 

[fayan77]

okay, and thank you for your help.