# Homework Help: Integration: force on submerged triangular plate

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1. Oct 15, 2014

### LBK

1. The problem statement, all variables and given/known data
Submerged (vertically) right triangle 12x9x15(hypotenuse) with 12 m leg parallel to water surface. Top of triangle is 3m below surface. Find force on triangle.

2. Relevant equations
I know mass of water is 840 k/m3. I think I should also multiply times 9.8 for gravity? One formula I found has to get Pressure=w*g*d for water mass, gravity and depth below surface respectively. Then to take pressure times area to get force. F=PA. But does this only work when submerged horizontally?

3. The attempt at a solution
I feel like I'm not "getting" how to decide what domain to integrate over and when to rotate axes. I let the surface of water be x-axis.
Assuming the short leg to be set on y-axis, I found the slope of hyp to be 3/4 so the x coordinate of dy to be L(x)=4/3y
So my integral looked like:
840* INT[-12 to -3] -y(4/3y)dy
840*[(-4y3)/9] |eval over(-12 to -3)
840*[12-768] Now this gives me a negative number, so I should just reverse their order? Is this otherwise correct? I get so confused on the origin placement. If I reverse them it becomes 840[768-12]=635,040
Then since this is metric, I express it in Newtons? 6.35*105N Should this have been multiplied times the 9.8 gravity thing or no?
Thanks for any input or suggestions!

2. Oct 15, 2014

### Staff: Mentor

Yes, because you want the weight denssity of water, not just its mass density. (The 840 number is the mass density, not the mass.) Also, use kg, not k, for kilograms.
Three things:
1. The slope of the hypotenuse is -3/4. Per your setup, the top of the triangle is at (0, -3) and the far end of the hypotenuse is at (12, -12).
2. The relationship between x and y is not x = (4/3)y. That would be a line with positive slope that passes through the origin.
3. Be careful when you write "4/3y". Some people might (incorrectly) interpret this to mean 4/(3y).
Your integral is not set up correctly. The typical pressure element is F$\Delta A$, where F is the force of the water at a given depth acting on the plate, and $\Delta A$ is the area of a thin horizontal area element.
Here F$\Delta A = mgh \Delta A = mgh \cdot x \Delta y$
If you get a negative result when the result should be positive, you've set up the integral incorrectly.

3. Oct 15, 2014

### LBK

Thank you so much. So, one last question to clarify, because I've seen examples that seem not consistent to me. If I place the X axis at the water line (3 m above the 12m side of the triangle) are my limits correct at 3 to 12, even though I"m below the x axis? My instructor did one example where we used some negative values but flipped their positions in the integral. That confused me and we're having a test over this later today.

4. Oct 15, 2014

### Staff: Mentor

You're at liberty to define where the axes are, but your work has to be consistent with that choice. "Flipping their positions" sounds a bit flaky to me. You can define the positive y-axis to point downward, then the y values would be positive, but that might be confusing when you calculate the slope. If the x-axis as at the water line, and the y-axis points up, then the limits of integration would be -12 and -3, respectively (going from low to high). In that order you should get a positive value for your result.