How Do You Find the Centroid and Hydrostatic Force in Given Scenarios?

[ineedhelpnow]

4. find the centroid of the region bounded by the given curves. $y=x^{3}$, $x+y=2$, $x=0$

5. a gate in an irrigation canal is constructed in the form of a trapezoid 3 ft wide at the bottom, 5 ft wide at the top, and 2 ft high. it is placed vertically in the canal, with the water extending to its top. find the hydrostatic force on one side of the gate.

 

[MarkFL]

5.) First, let's determine the width of the plate as a function of the depth $x$. We know the width decreases linearly, and we know two points on the line:

\(\displaystyle w(0)=5\)

\(\displaystyle w(2)=3\)

So, knowing two points on the line, can you determine $w(x)$?

 

[ineedhelpnow]

no my instructor hasnt explained any of this yet but we still have to do it and I am totally lost. i don't get what the book is saying. :/

 

[MarkFL]

no my instructor hasnt explained any of this yet but we still have to do it and I am totally lost. i don't get what the book is saying. :/

What I suggest doing is first finding the slope $m$ of the line:

\(\displaystyle m=\frac{\Delta w}{\Delta x}=\frac{3-5}{2-0}=-1\)

Now, use the point-slope formula you learned in algebra or pre-calculus to determine the equation of the line representing the width of the plate as a function of the depth $x$. What do you get?

 

[ineedhelpnow]

w(x)=5-x

 

[MarkFL]

w(x)=5-x

Good! Now, we may compute the force along one horizontal elemental rectangle having area $A(x)$ as:

\(\displaystyle dF=\rho xA(x)\)

The area of this elemental rectangle is its width at $x$ times its height $dx$, hence:

\(\displaystyle dF=\rho xw(x)\,dx\)

Note: $\rho$ is the mass density of water, usually given as \(\displaystyle 62.5\frac{\text{lb}}{\text{ft}^3}\)

So, substitute for $w(x)$, and then integrate from the surface $x=0$, to the bottom of the plate $x=2$. Can you set up the resulting integral?

 

[ineedhelpnow]

ill try but i don't think ill do it right

 

[MarkFL]

ill try but i don't think ill do it right

I bet you can...you have a constant $\rho$ times $x$ times the width function $5-x$ times the depth differential $dx$...:D

 

[ineedhelpnow]

$\int_{0}^{2} \ 62.5x(5-x),dx$ ?

 

[MarkFL]

$\int_{0}^{2} \ 62.5x(5-x),dx$ ?

Excellent! Now, pull that constant out front, distribute $x$ to the width function, and then use the power rule and the FTOC (fundamental theorem of calculus) to compute the definite integral. :D

 

[ineedhelpnow]

thanks for your help!

 

[MarkFL]

4. find the centroid of the region bounded by the given curves. $y=x^{3}$, $x+y=2$, $x=0$

First, let's get a diagram of the region:

View attachment 2716

Now, let's dig out the formulas we need:

\(\displaystyle \overline{x}=\frac{\displaystyle \int_{a}^{b} xf(x) \,dx}{\displaystyle \int_{a}^{b} f(x)\,dx}\)

\(\displaystyle \overline{y}=\frac{\displaystyle \frac{1}{2}\int_{a}^{b} \left[f(x)\right]^2\,dx}{\displaystyle \int_{a}^{b} f(x)\,dx}\)

Can you identify $a$, $b$ and $f(x)$?

 

[ineedhelpnow]

would it be from 0 to 1? I am having trouble identifying the f(x) because my book does it kinda different and it's confusing me.

 

[MarkFL]

would it be from 0 to 1? I am having trouble identifying the f(x) because my book does it kinda different and it's confusing me.

Yes, we are given the left limit, the line $x=0$, and to find the right limit, we may equate the two functions and solve for $x$:

\(\displaystyle 2-x=x^3\)

\(\displaystyle x^3+x-2=0\)

\(\displaystyle x^3-x^2+x^2-x+2x-2=0\)

\(\displaystyle x^2(x-1)+x(x-1)+2(x-1)=0\)

\(\displaystyle (x-1)\left(x^2+x+2\right)=0\)

We see that the quadratic factor has a negative discriminant, and so the only solution is:

\(\displaystyle x=1\)

Thus, we have $a=0,\,b=1$.

Now, $f(x)$ is the upper function minus the lower function...what would that be?

 

[ineedhelpnow]

$2-x-x^3$

 

[MarkFL]

...im having trouble identifying the f(x) because my book does it kinda different and it's confusing me.

Sorry for the delay...a brief summer storm knocked out my connection (satellite internet).

How does your book compute the centroid? I ask because I see in my old book that I may have given you a formula that does not apply in this case. :D

 

[ineedhelpnow]

$x=\frac{1}{A}\int_{a}^{b} \ x[f(x)-g(x)],dx$

$y=\frac{1}{A}\int_{a}^{b} \ \frac{1}{2}([f(x)]^2-[g(x)]^2),dx$

 

[MarkFL]

$x=\frac{1}{A}\int_{a}^{b} \ x[f(x)-g(x)],dx$

$y=\frac{1}{A}\int_{a}^{b} \ \frac{1}{2}([f(x)]^2-[g(x)]^2),dx$

Yes, that's what you want to use...I apologize for my mistake of posting a formula that applies only when $g(x)=0$. (Tmi)

 

[ineedhelpnow]

oh ok i think i get where to go from here. thanks :)