Room for RyderRude and 123

Ryder Rude

15:57

so u multiplied all the equations, and u got $\frac{\Delta v_1 \Delta v_2 \Delta v_3}{\Delta v_1' \Delta v_2' \Delta v_3'}=-1$

123

Pls add recently in my sentence.

Ryder Rude

is this correct?

123

@RyderRude Yes

ACuriousMind

← 4 messages moved from The h Bar

Ryder Rude

now, u were supposed to compute k12 k23 k31. Is this correct

123

16:00

Ooops ACM kicked us from room permanently

Ryder Rude

no. u can still post there

just not this discussion. Let's finish it here

123

No i can't see that room

Ryder Rude

Click on this chat.stackexchange.com/rooms/71/the-h-bar

123

Oooh yes now i can see

Okay continue... Thanks

@RyderRude Yes this is way i found -1

Ryder Rude

but the LHS is not k12k23k31

123

16:03

@RyderRude Yes

Ryder Rude

so u proved that whatever the LHS is, is -1. But u didn't show that k12k23k31 is -1

so now we will compute k12k23k31

123

@RyderRude No this is actually +1

Ryder Rude

-1/2 x -3 x -2/3 = -1

how is this +1

123

$\frac{\Delta{V_1}}{\Delta{V_2}} = - K_{12}$

Ryder Rude

so?

123

16:06

$\frac{\Delta{V_2}}{\Delta{V_3}} = - K_{23}$

$\frac{\Delta{V_3}}{\Delta{V_1}} = - K_{31}$

Ryder Rude

let's make one thing clear. Do u think -1/2 is K12

123

Multiply all three equations together

@RyderRude Oooh i see

I need to move this minus sign to other side

@RyderRude But -1/2 given by experiment

Oooo ooooooo now i understand

Ryder Rude

are u using k12=-1/2. Yes or no?

123

Oookay.... i picked the point

Ryder Rude

ok :)

123

16:09

I was treating minus sign separate with constant

Ryder Rude

u realise that K12 is 1/2, right?

then ur problem is solved

123

Let me do it again with pen and paper

Ryder Rude

no... It is fine

123

Thanks pls give me few minutes

Ryder Rude

no... pls don't do it again. It is fine

u understood that k12 is 1/2, there is no need to do it on paper

123

16:12

What we do with the minus sign?

It will go to the other side of equation?

Ryder Rude

yeah

remember the definition of k12, the book wrote a1/a2= - k12. therefore, k12= -a1/a2

123

Pls do calculation. I think the way you wrote again give minus sign. Because there are three equations

Ryder Rude

idk what to say now...

123

If i take $K_{12} = -1/2$ along with minus sign then it gave me plus sign.

I have checked it with paper.

Ryder Rude

did u use k23= -3 and k31= -2/3?

123

16:17

Let me show you on paper. Finally i found $F_2 = F_3$

Let me share you calculation.

What ever convention i used. The final result is $F_2 = F_2$

Ryder Rude

again, k12= - a1/a2, not +a1/a2!

123

Pls calculate yourself here. At your own using my values

Ryder Rude

calculate what

123

Let me share you what you said

Oooh i found the correct result...

I was doing this other way. That's why i was getting wrong result

Ryder Rude

great

123

16:27

I was multiplying both sides of equation

Ryder Rude

oh

glad we corrected everything...

:)

123

:D

Basically i never used that way before in my life.

I always did multiplying both sides of equation

Ryder Rude

it happens. We learn new things everyday..

123

Thanks for your patience

Ryder Rude

u r welcome

123

16:31

It means we are treating constants multiplication separately as a value.

Not using mathematical way of multiplying equations

Am i correct?

Ryder Rude

what do u mean

123

Look at my page above where i created big square box

Ryder Rude

@123 the biggest square?

123

Yes

Ryder Rude

i see it

i dont see anything non mathematical...

which step do u mean exactly

123

16:33

This is what i meant. Treat constants multiplication separately

Ooookay... It means all these things are equal to each other.

Ryder Rude

the constants in the middle vertical line? 1/2, 3, and 2/3?

123

Yes

Ryder Rude

ooh so initially u have this three way equation. All these three are equal, so u can forget about the left vertical line

123

Now the correct way thinking is that. Use constants and values multiplication

Ryder Rude

yes..

maybe

123

16:36

And left side equality outside the big bracket are also equal. Then utilize that

Ryder Rude

like, u have two equations like a=b=c, and d=e=f. U can multiply to get ad=be=cf

it's not a new rule. It follows from the usual rules

@123 yes

a=b=c implies b=c, and d=e=f implies e=f

123

Yes.... Basically i was confused when i didn't write equation with values

That's why always i found minus sign

Ryder Rude

ooh

123

Book used values and constants multiplication. not the outside the big bracket

Ryder Rude

yes. u have to get used to manipulations like this. It is a part of math too

123

16:40

I am used to this type of calculations

Ryder Rude

but u were confused about this manipulation...

123

Problem was that i wasn't writing values of constants

Ryder Rude

oh

123

Now i get the correct idea of getting +1

Ryder Rude

Great

123

16:42

I was doing left and right side multiplication. not using the middle and left

By the way.... Thanks for your patience

Now i am completely satisfied with the newton's law. Thanks my friend.

How old are you?

Ryder Rude

ok :)

123

If you don't mind

Ryder Rude

I'm going back to the main room :)

123

Sure me too...

Ryder Rude

cya

123

16:46

Cya in main room

user430580

17:28

Hello

Transcript for

Room for RyderRude and 123