Formula derivation connecting vertical water flowrate & horizontal distance moved by a suspended sphere

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$W-T\cos(\beta )=\dot{m}v_{0}\cos\left ( \theta \right)- \dot{m}v_{i}$$

Good.

Just to practice this whole convention issue you were(hopfully) having rewrite your equations for "up is positive" and "right is positive".

 

[printereater]

Good.

Just to practice this whole convention issue you were(hopfully) having rewrite your equations for "up is positive" and "right is positive".

HAHAHA yea. btw for my x-coordinate direction it should be $-\dot{m}v_{0}\sin\left ( \theta \right )$ and not $\dot{m}v_{0}\sin\left ( \theta \right )$ right

 

[erobz]

wouldn't it just be $\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right)$ then because both vectors are facing downwards?

no, becuase it is outflow minus inflow vectorally

$$ \sum \vec{F} = \dot m \vec {v_o} - \dot m \vec {v_i}$$

 

[printereater]

Oh so, I am not literally adding up the vectors, I am using the formula that was derived from Reynold's transport theorem?

 

[erobz]

Oh so, I am not literally adding up the vectors, I am using the formula that was derived from Reynold's transport theorem?

yes, it is the simplified formula of RTT. The minus sign comes from the dot products in the full treatment.

 

[printereater]

Got it. Alright, I need to go and sleep now. Thank you so much for your help, I really appreciate it so much. Can we continue tomorrow please

 

[erobz]

Got it. Alright, I need to go and sleep now. Thank you so much for your help, I really appreciate it so much. Can we continue tomorrow please

Sounds good.

 

[printereater]

I suspect it's not a trivial problem to solve theoretically. But experimentally, you should be able to find some functional relationship between the outflow angle and the mass flowrate. you have to measure ( control) incoming flowrate, and measure the angle of deflection.

You are going to be working with a control volume that is something like this:

View attachment 339090

And you apply Reynolds Transport Theorem - "The Momentum Equation" in fluid mechanics.

Which says for steady flow with uniform velocity distribution (one inlet-one outlet):

$$ \sum \vec{F} = \dot m \vec{v_o} - \dot m \vec{v_i} $$

Also, can you please show me how you simplify Reynold's transport theorem to this formula please or provide a website that shows it. I tried to look for it but I could not find anything

 

[erobz]

Also, can you please show me how you simplify Reynold's transport theorem to this formula please or provide a website that shows it. I tried to look for it but I could not find anything

We can go over that tomorrow. But tomorrow as a favor to both of us(you and I) you can open with rewriting the equations for "up positive, right positive" just to be sure you aren't still confused(please). I'm hoping to see the step where you take the sign of $v$ relative to coordinate into consideration before simplifying. It's always trivial until we get into uncharted territory, then everyone forgets which way is up, and why that minus sign is there...

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=-\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$W-T\cos(\beta )=\dot{m}v_{0}\cos\left ( \theta \right)- \dot{m}v_{i}$$

Change the sign back in the $x$ direction when you wake up...its not negative on the RHS.

 

[printereater]

We can go over that tomorrow. But tomorrow as a favor to both of us(you and I) you can open with rewriting the equations for "up positive, right positive" just to be sure you aren't still confused(please). I'm hoping to see the step where you take the sign of $v$ relative to coordinate into consideration before simplifying. It's always trivial until we get into uncharted territory, then everyone forgets which way is up, and why that minus sign is there...

Hello! Yes, definitely. I am used to taking up and right as positive all the time but I took left and down as positive to avoid dealing with negative values. i ended up confusing my self so much though.

 

[printereater]

In x-coordinate direction,
$$-T\sin(\beta )=-\dot{m}v_{0}\sin(\theta))$$
$$\therefore T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))+mv_{i}$$

 

[erobz]

In x-coordinate direction,
$$-T\sin(\beta )=-\dot{m}v_{0}\sin(\theta))$$
$$\therefore T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))+mv_{i}$$

Alright, it seems like you get it.

The next step is to use continuity to eliminate the inflow/outflow velocities in terms of variables you plan to control $\dot m$, and variables you can measure/estimate $A_i,A_o$.

 

[printereater]

oh, how do I do that?

 

[erobz]

oh, how do I do that?

Mass flowrate for incompressible flow into control volume must equal mass flowrate out for steady flow ( i.e. fluid mass is not accumulating inside the control volume over time). Velocities $v_i,v_o$ assumed to be uniformly distributed over the cross-sectional inlet/outlet areas. Are you familiar with the expression that describes this?

 

[printereater]

oh so,
In x-coordinate direction,
$$ T\sin(\beta )=\rho v_{o}A_{o} v_{o}\sin(\theta))$$
$$ T\sin(\beta )=\rho (v_{o})^{2}A_{o} \sin(\theta))$$

In y-coordinate direction ,
$$ T\cos(\beta )-W=-\rho v_{o}A_{o}v_{o} \cos(\theta)+\rho v_{i}A_{i}v_{i} $$
$$ T\cos(\beta )-W=-\rho (v_{o})^{2}A_{o} \cos(\theta)+\rho (v_{i})^{2}A_{i} $$

 

[erobz]

oh so,
In x-coordinate direction,
$$ T\sin(\beta )=\rho v_{o}A_{o} v_{o}\sin(\theta))$$
$$ T\sin(\beta )=\rho (v_{o})^{2}A_{o} \sin(\theta))$$

In y-coordinate direction ,
$$ T\cos(\beta )-W=-\rho v_{o}A_{o}v_{o} \cos(\theta)+\rho v_{i}A_{i}v_{i} $$
$$ T\cos(\beta )-W=-\rho (v_{o})^{2}A_{o} \cos(\theta)+\rho (v_{i})^{2}A_{i} $$

you need to eliminate $v_i$ and $v_o$, and keep $\dot m$

 

[printereater]

ohhh nevermind, I am supposed to make $\dot{m}$ the subject of the equation and equate them

 

[printereater]

Hello! Yes, definitely. I am used to taking up and right as positive all the time but I took left and down as positive to avoid dealing with negative values. i ended up confusing my self so much though.

idk why but the same message keeps popping up on my reply box. I accidentally ended up sending it again just now

 

[erobz]

idk why but the same message keeps popping up on my reply box. I accidentally ended up sending it again just now

It happens sometimes. I don't know why.

 

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

You aren't getting what I'm saying ( or I don't understand what making $\dot m$ "the subject " is). Either way, you want each equation to not have any velocity parameters. Just the force terms, mass flowrate $\dot m$, relevant area $A_i,A_o$, and known physical properties (density). Don't join the equations yet.

Think about equating $\dot m = \dot m_i = \dot m_o$ to eliminate each velocity $v_i,v_o$ from each equation.

 

[printereater]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

 

[erobz]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

yeah for each location ( inlet/outlet) that holds.

 

[printereater]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

 

[erobz]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

Get some rest, I'll be unavailable for the weekend. If you want to continue, feel free, someone will help you. If you want to wait till Monday, that is fine too.

 

[printereater]

Oh no. If that's the case, i don't mind sacrificing my sleep HAHAHAH let's continue. Am I close to the final equation

 

[erobz]

You're, going to get tired, and learning abilities are going to decline, and mistakes are going to start to happen in your math. There is certainly no way we can get through the Reynolds Transport Theorem - "The Momentum Equation", with any level of understanding so quick.

Your goal (for the weekend- if waiting for me at least - which is not necessary ) after eliminating the velocities, is then to eliminate the Tension force from the equations by substitution, and then solve the resulting equation containing only parameters $\dot m, A_i, A_o,\rho, \theta$ for the angle the ball makes w.r.t vertical $\beta$.

I strongly suggest you just take a rest, and work at it refreshed.

 

[printereater]

yea you are right, I will just continue tomorrow. Thank you so much, once again

 

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[printereater]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

Keep factoring the y coordinate equation to clean it up. Otherwise, ok. Then keep going.

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin\(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

Your algebra looks unnecessarily complicated here. Just use the x direction equation to eliminate $T$ in the y direction.

 

[printereater]

Oh alright

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[erobz]

$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

factor the RHS a bit more and then proceed to solve for $\beta$, and you are done with that part.

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$
$$\frac{\rho A_{o}\tan(\beta )}{\dot{m}^{2}\sin(\theta)}=\frac{\rho}{\dot{m}^{2}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{1}{W}$$
$$\tan(\beta )=\frac{\rho\dot{m}^{2}\sin(\theta)}{\dot{m}^{2}\rho A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$\tan(\beta )=\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$
\tan(\beta )=\arctan[\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}]$$

 

[printereater]

@erobz @Delta2 Please take a look at it when you are free and let me know if I am on the right track.

 

[erobz]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

 

[printereater]

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right

 

[erobz]

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right

You are trying to solve for $\beta$ as a function of all the other variables you intend to measure. If that’s is what “making it the subject is “, then sure. You made an algebraic mistake when inverting both sides.

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\rho }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+W\rho A_{o}$$
$$\frac{1}{\tan(\beta )}=\frac{\dot{m}^{2}\rho A_{o}}{\dot{m}^{2}\rho\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}$$
$$\tan(\beta )=\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}$$
$$\beta =\arctan[\frac{1}{\frac{\ A_{o}}{\sin(\theta) }(\frac{1}{A_{i}}-\frac{\cos(\theta)}{\ A_{o}})+\frac{W\rho A_{o}}{\dot{m}^{2}\sin(\theta)}}]$$ I am not sure if there is a nicer way to express $\beta$

 

[erobz]

I am not sure if there is a nicer way to express $\beta$

Its good enough for now. Do you understand how to use it?

Latex tip. Use automatically sized parenthesizes

Instead of $$ ( 1 + \frac{1}{x} ) $$

use code

$$ \left( 1 + \frac{1}{x} \right) $$

to get

$$ \left( 1 + \frac{1}{x} \right) $$

 

[printereater]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

 

[erobz]

Hi yes I understand it. How can I use it to link the distance moved by the sphere and the flow rate though

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle $\beta_{exp}$ you experimentally measure. There is going to be error in all the variables to account for etc...

 

[printereater]

$\beta$ is the steady state angle, you need all the other variables measured (or controlled) and the equation spits out the angle the string makes with vertical (getting how far it moved is just trig), you can determine how well that fits with the angle you measure. There is going to be error in all the variables to account for etc...

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

 

[printereater]

There is going to be error in all the variables to account for etc...

Yes! Can you guide me through the calculations of uncertainties too please

 

[printereater]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

I just realised, there are 4 variables that are changing each time based on the flow rate of water $\ A_{o},\ A_{i}, \theta, \dot{m}$. So I have no choice but to plot D against $\beta$ right

 

[erobz]

I need to plot a graph using the 2 variables though (Flow rate and distance moved by the sphere). If I use this formula, the distance moved by the sphere can be expressed this way, $\tan(\beta ) =\frac{D}{L} \therefore D={L}\tan(\beta )$, where D is the distance moved by the sphere and L is the length of the string. If I plot this function, I will be showing a relationship between $\beta$ and D. $\beta$ has 2 variables that are changing constantly $\theta$ and $\dot{m}$. Is there a way to overcome this and show the relationship between $\dot{m}$ and D without the $\theta$

This is the steady state solution. $\dot m$ must be controlled. The idea is that if you control the flowrate to be constant, all other experimentally measured variables $\theta, A_i,A_o$ are constant too (for each flowrate). So, the equation predicts the angle of the string $\beta$. Then, you change the flow rate(but keep it steady), and re-measure the variables, and again the equation predicts the angle. You compare how well the equation matches what you experimentally measure for $\beta$ at each flowrate.

The goal is to see how well our theory about reality, matches reality. Will it be a decent model given our assumptions, and our inability to precisely measure the variables that contrive it? That is the purpose of an experiment.

 

[printereater]

Right, I get what you mean but for the graph I plot, am I going to plot $\beta$ against $\dot{m}$ or do I just plot $D=L\tan (\beta )$ to show the relationship between flowrate of water and sphere. With this formula I can show how $\beta$ changes with D to indirectly show the relationship between flowrate and D because what's affecting $\beta$ is $\dot{m}$ but I am wondering if there is a direct way to show the relationship flowrate and D.