Answer to Question #108764 in Calculus for Edward

Question #108764
A right triangular plate of base 3.0 m and height 1.5 m is submerged vertically in water, with top
vertex 3.5 m below the surface. Find the force on one side of the plate.
1
Expert's answer
2020-04-11T14:58:53-0400


The total force on the plate is given by


"F=w\\displaystyle\\int_{a}^bxydy"

where

"x"  is the length (in m) of the element of area (expressed in terms of y)

"y"  is the depth (in m) of the element of area

"w"  is the density of the liquid (in N m-3)

(for water, this is w = 9800 N m-3)

"a"  is the depth at the top of the area in question (in m)

"b"  is the depth at the bottom of the area in question (in m)


Before we can proceed, we need to find "x" in terms of "y."

Now when "x=0,y=3.5," and when "x=3,y=5."

So we have: "y=mx+b"

"x=0,y=3.5: b=3.5"

"x=3,y=5:5=3m+3.5=>m=0.5"


"y=0.5x+3.5"

This gives us 


"x=2y-7"

Hence


"F=9800\\displaystyle\\int_{3.5}^{5}(2y-7)ydy=9800\\bigg[{2y^3\\over 3}-{7y^2\\over 2}\\bigg]\\begin{matrix}\n 5 \\\\\n 3.5\n\\end{matrix}=""=99225\\ N"

"F=99225\\ N"



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Comments

Assignment Expert
11.04.20, 22:01

The x ranges from 0 to 3, hence y ranges from 3.5 to 5. The definite integral is taken with respect to y, hence a=3.5, b=5.

alias
11.04.20, 20:18

how to solve for a and b? please show how. :( thanks

Marie
11.04.20, 16:59

how did you get the value of a and b?

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