Finding Fluid Force on Plate Submerged in Fluid

[think4432]

2. A plate shaped as in the figure [picture.jpg is attached] is submerged vertically in a fluid as indicted. Find the fluid force on the plate if the fluid has weight density 62.4 lb/ft^3

The integral I set up was the limits being -5 to -1 and the integral being (5-y)(-7/4y - 7/4) dy.


[Also on this one I was getting help on it yesterday on another thread...but the tutor seems to offline right now, and I am hoping someone else can assist me to get this problem done]

I was told that the (-7/4y - 7/4) is correct but the (5-y) is not.

After many posts on the last thread I still can't figure out the depth.

Help?

 

[HallsofIvy]

Actually, it is the other say around- the "5- y" is correct, it is the "(7/4)- (7/4)y" that is incorrect. You want (7/4)(4- y)= 7- (7/4)y.

Let "y" be the height above the base line of the triangle. Then the depth of water above that line is 5- x. Also, by "similar triangles" we have
\frac{x}{7}= \frac{4-y}{4}
where "x" is the width of the triangle at height y.
(The triangle above height y has height 4- y, width x and is similar to the entire triangle with height 4 and width 7)
x= \frac{7}{4}(4- y)= 7- (7/4)y

So a thin rectangle, at height y, with thickness dy, would have area (7- (7/4)y)dx. Taking \delta as the density of water, the weight of water above that rectangle is the area times the height above y (the volume of water above that rectangle) times the density of water: ((7-(7/4)y)dx)(5- y)\delta

Integrate that from y= 0 to y= 4.

 

[think4432]

Actually, it is the other say around- the "5- y" is correct, it is the "(7/4)- (7/4)y" that is incorrect. You want (7/4)(4- y)= 7- (7/4)y.

Let "y" be the height above the base line of the triangle. Then the depth of water above that line is 5- x. Also, by "similar triangles" we have
\frac{x}{7}= \frac{4-y}{4}
where "x" is the width of the triangle at height y.
(The triangle above height y has height 4- y, width x and is similar to the entire triangle with height 4 and width 7)
x= \frac{7}{4}(4- y)= 7- (7/4)y

So a thin rectangle, at height y, with thickness dy, would have area (7- (7/4)y)dx. Taking \delta as the density of water, the weight of water above that rectangle is the area times the height above y (the volume of water above that rectangle) times the density of water: ((7-(7/4)y)dx)(5- y)\delta

Integrate that from y= 0 to y= 4.

So wait, What is it that I am integrating? (5-y)(7-7/4y) dy

From the limits [0,4]

?