I/P convertor equation finding 2 unknowns

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In summary: You can use units in algebra just like they were numbers, so that N A^(-1) is the same as N/A, so that the equation (Kc y)/(Ax) = 1/20 with Kc = 2500 N/A and A = 5 cm^2 and x = 50 mm is the same as the equation (2500 N/A y)/((5 cm^2)(50 mm)) = (1/20) mA, which you can solve for y.In summary, the conversation involves a discussion of finding two distances x and y, with the constraint that x + y = 100 mm. The conversation also mentions an equation involving Po, Kc, y, A, x,
  • #1
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Homework Statement


Basically its all on the attached sheet as i don't know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.








But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA



We know the values of Kc and A and hence can find x and y. Sorry but I can't fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?


(N.B. be sure to have all measurements in S.I units.)



Homework Equations



Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x


A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar


The Attempt at a Solution



i can't attach any files I am afraid as it won't work?


any help greatly appreciated
 
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  • #2
tommoturbo said:

Homework Statement


Basically its all on the attached sheet as i don't know how to stick images in the post.

im having trouble finding two distances x and y (x+y= 100mm) just need a little help rearranging the equation on the attached word doc.








But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA



We know the values of Kc and A and hence can find x and y. Sorry but I can't fathom how to rearrange it as A=5cm^2 this multiplied by an unknown number in mm means it is tiny and hence once divided by the top figure will give a massive value for Po. When we know it is 0.05?


(N.B. be sure to have all measurements in S.I units.)



Homework Equations



Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x


A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar


The Attempt at a Solution



i can't attach any files I am afraid as it won't work?


any help greatly appreciated

Attaching files should work. Is it real big or something? Also, it's better to post as a PDF -- less chance of transfering worms etc. You can get a free PDF converter from PrimoPDF if you don't have one already.
 
  • #3
Since x + y = 100, you can replace y with 100 - x so that your equation for Po involves only one variable.

You haven't provided enough information for me to figure out what you're doing. E.g., what units is Kc in? I don't understand N/A. A usually represents amperes, but that doesn't seem to be what you're doing.
 
  • #4
Hi guys thanks for the replies sorry its so vague i did it all on a word doc importing graphs and the equation etc thinking i could post it but the uploader won't work.

the units for the constant Kc are in Newtons but it has an A to the minus 1 after it.


ill try and post it from home it might not work on the pc here.

thanks again for replying.
 
  • #5
HI guys here is the notes i have which should clear things up i hope
 

Attachments

  • image003.bmp
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  • ICP TMA2 Q9[1].doc
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  • #6
tommoturbo said:
But ‘A’ passes through origin so c = 0.

m = 1/20 bar per mA

Po= (Kc*y)/A*x *Iin + (Fs*z)/A*x
This formula is different from the one in the attachments. This is the one in your attached files:
[tex]P_o = \left( \frac{K_c y}{A x}\right)I_{in} + \frac{F_s z}{Ax}[/tex]

The way you wrote it above indicates that only A is in the denominators, not Ax.

This problem is a lot more confusing that it should be. For one thing, A is being used for three different purposes, making it difficult to figure out what A represents in a given situtation. One use for A is identifying one of the two straight lines (A and B). Another use is area. The third use is for amperes.

The equation above gives Po as a linear function of Iin. Since line "A" (better, line 1) goes through the origin, (Fs z)/(Ax) must be zero.

You are given that m = (1/20) mA, which means that the expression that multiplies Iin must be equal to 1/20 mA, so solve the equation (Kc y)/(Ax) = 1/20.

tommoturbo said:
A= 5cm^2
Z= 75mm
x+y = 100mm
Kc= 2500N A^-1
Po is in bar and is 0.2 at 4mA
Iin is mA and is 4 at 0.2Bar

The notation N A^(-1) is the same as N/A, or Newtons per ampere.
 

1. What is an I/P convertor equation?

An I/P convertor equation is a mathematical equation that helps convert input values into output values. It is commonly used in scientific research and engineering to analyze and predict the relationship between two variables.

2. How do I use the I/P convertor equation to find two unknowns?

To use the I/P convertor equation to find two unknowns, you will need to have two input values and their corresponding output values. Then, plug these values into the equation and solve for the two unknown variables. This will give you the relationship between the two variables.

3. Can the I/P convertor equation be used for any type of data?

Yes, the I/P convertor equation can be used for any type of data as long as there is a clear relationship between the input and output values. It is commonly used in fields such as physics, chemistry, and economics.

4. What is the benefit of using the I/P convertor equation?

The I/P convertor equation allows scientists to analyze and predict the relationship between two variables. This can help in understanding complex systems and making informed decisions in research and development.

5. Are there any limitations to using the I/P convertor equation?

One limitation of using the I/P convertor equation is that it assumes a linear relationship between the two variables. This may not always be the case in real-world situations, so it is important to carefully evaluate the data and the equation's applicability before using it.

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