Dividing over zero while calculating center of gravity - Getting error

In summary: So the final answer is Xc = 0.17In summary, to find the center of gravity of an object, the coordinates of the centroid for a two dimensional figure are given by x_c= \frac{\int x dA}{\int dA}= \frac{\int x dA}{A} and y_c= \frac{\int y dA}{\int dA}= \frac{\int x dA}{A}. The denominator is just the area of the figure, represented by A. The areas that are part of the body have a positive sign, while areas that are taken out of the body have a negative sign. The coordinates of the centers have their respective sign. It is important to note this distinction
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  • #2
I'm not sure what that paper is showing. Since nothing is said about the density, I assume you are taking that to be constant and are actually trying to find the centroid. The coordinates of the centroid of a two dimensional figure are given by
[tex]x_c= \frac{\int x dA}{\int dA}= \frac{\int x dA}{A}[/tex]
[tex]y_c= \frac{\int y dA}{\int dA}= \frac{\int x dA}{A}[/tex]

The denominator is just the area of that figure which obviously is not 0! There should not be any subrtraction in your calculation.
 
  • #3
Hi Fp! I was wondering if you'd be here today! :smile:

On the bottom line (the denominator) you have a sum of areas.
But you've taken some of those areas as being negative? :confused:
 
  • #4
@HallsofIvy - We're not using calculus in our solutions

I like Serena said:
Hi Fp! I was wondering if you'd be here today! :smile:

I meant to post 8 hours ago but I had internet disconnection issues!
On the bottom line (the denominator) you have a sum of areas.
But you've taken some of those areas as being negative? :confused:

The two triangles don't exist, so they're in minus, and the rectangle is at the minus part of the Y axis, hence it's negative. Doesn't that make sense?
 
  • #5
Should the sign between 21x12 and 21x9 be a plus, not a minus?
 
  • #6
Femme_physics said:
I meant to post 8 hours ago but I had internet disconnection issues!

Aw! That must have been hard to take! :frown:
Femme_physics said:
The two triangles don't exist, so they're in minus, and the rectangle is at the minus part of the Y axis, hence it's negative. Doesn't that make sense?

Oh yes. My bad.

However, the rectangle at the left of the Y axis still has a positive contributing area.
It's only that the x coordinate of its center is negative.
You didn't really make that distinction in your nominator (the top line).Edit: So areas that belong to your body always have a positive contribution, while areas that are taken out of the body always have a negative contribution.
The coordinates of the centers are taken positive or negative just as they are.
 
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  • #7
It's only that the x coordinate of its center is negative.

True, that's why in my Yc I meant to do it in plus, but in Xc it's minus
You didn't really make that distinction in your nominator (the top line).

I'm not sure what distincton. Can you elaborate?
 
  • #8
Femme_physics said:
I'm not sure what distincton. Can you elaborate?

I would have written your fraction (if I'd be very explicit) as:

[tex]x_c = \frac {((+21 \cdot 12) \cdot \frac {+12} 2) + ((+21 \cdot 9) \cdot \frac {-9} 2) + ...} {(+21 \cdot 12) + (+21 \cdot 9) + (-\frac {9 \cdot 6} 2) + (-\frac {12 \cdot 6} 2)}[/tex]

That is, the areas that are part of the body have a plus sign in front.
The areas that are taken out of the body have a minus sign in front.
The x coordinates of the centers have their sign as applicable.
 
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  • #9
But the 21x9 rectangle shape is at the negative side of the X axis, doesn't that mean that while calculating its distance to the Y axis I need to take a negative value?
 
  • #10
Femme_physics said:
But the 21x9 rectangle shape is at the negative side of the X axis, doesn't that mean that while calculating its distance to the Y axis I need to take a negative value?

For its distance to the Y axis - yes, take a negative value (-9/2).
For its contribution to the area - no, take a positive value (+21x9).

Sorry, out of ideas how to explain this better (for now).
 
  • #11
For its distance to the Y axis - yes, take a negative value (-9/2).
For its contribution to the area - no, take a positive value (21x9).

But regardless minus times a plus is a minus, so wasn't I right in minusing it?
 
  • #12
Femme_physics said:
But regardless minus times a plus is a minus, so wasn't I right in minusing it?

In the nominator (top line) - yes, you were right minusing it. I just wanted to point out the distinction between sign of area and sign of coordinate.

In the denominator (bottom line) - no, you were not right minusing it. The area contributes here in a positive manner.
 
  • #13
Ohhhhhhhhhhhhhhhh! okay, give me a moment :)
 
  • #14
Femme_physics said:
Ohhhhhhhhhhhhhhhh! okay, give me a moment :)

Just to make sure, before you're going to start drawing flowers, and butterflies, and bees, you do have another mistake in the nominator (I wrote "..." before on purpose! :devil:).
 
  • #15
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  • #16
I like Serena said:
Just to make sure, before you're going to start drawing flowers, and butterflies, and bees, you do have another mistake in the nominator (I wrote "..." before on purpose! :devil:).

The small triangle. It's a minus, since it doesn't exist, and since it's on the negaative side, it's arm is minus, therefor it's a plus. Is that right?

Edit: Am off back home. Talk later :) thanks!
 
  • #17
Femme_physics said:
Could the answer be

Xc = 5.654

?

The manual says something else. I don't have a scanner (not for the next two hours), but this is what the manual says through my poor quality webcam:

Xc = 0.17

Sorry, both answers are wrong!
I think you have a faulty solution manual here! :devil:
Femme_physics said:
The small triangle. It's a minus, since it doesn't exist, and since it's on the negaative side, it's arm is minus, therefor it's a plus. Is that right?

Edit: Am off back home. Talk later :) thanks!

Yes, that is right. (That's not your mistake. :P)

Edit: could you check the arm lengths?
 
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  • #19
Femme_physics said:
I used the wrong distances for the triangles

For the bigger triangle it's 8
For the smaller triangle it's 6

Very good! :smile:


Femme_physics said:
This is what I get after all the corrections

Hmm, are those areas in the denominator?
The numbers look the same as the ones in the nominator?! :confused:

All the signs are correct though! :smile:
 
  • #21
Femme_physics said:
Oops! I'm getting too excited lately, heh.

:) Got it, right?

Right! :smile:

What is it you are you excited about?
 
  • #22
To solve it! :) Thanksssssssssssssssssss!

I'll add it to the corrections manual!
 
  • #23
Femme_physics said:
To solve it! :) Thanksssssssssssssssssss!

I'll add it to the corrections manual!

You're welcome. :wink:

But what about Yc?
 
  • #25
Femme_physics said:
Same as the solution manual :)

Yes. That is right! :smile:

Funny, because I expected the solution manual to be wrong, but I didn't check.
Turns out that they have been inconsistent in their calculations! :wink:
 

FAQ: Dividing over zero while calculating center of gravity - Getting error

What does it mean to divide over zero while calculating center of gravity?

Dividing over zero while calculating center of gravity means that you are attempting to divide a value by zero while determining the center of mass or center of gravity for a system. This typically occurs when using the formula for center of gravity, which involves dividing the sum of the product of mass and distance by the total mass. If one of the distances is zero, the result is undefined.

Why does dividing over zero while calculating center of gravity result in an error?

Dividing over zero while calculating center of gravity results in an error because division by zero is undefined in mathematics. This means that there is no valid answer or solution for the calculation. The error is usually denoted as "NaN" or "undefined" to indicate that the result cannot be determined.

How can I avoid dividing over zero while calculating center of gravity?

To avoid dividing over zero while calculating center of gravity, you can check your calculations and make sure that all distances used in the formula are non-zero. This may involve double-checking your measurements or using a different formula that does not involve division.

What are the consequences of dividing over zero while calculating center of gravity?

The consequences of dividing over zero while calculating center of gravity can vary depending on the context. In some cases, it may simply result in an undefined answer and no further implications. However, in situations where precise calculations are needed, such as in engineering or physics, it can lead to inaccurate results and potentially affect the performance or safety of a system.

Is it ever acceptable to divide over zero while calculating center of gravity?

No, it is never acceptable to divide over zero while calculating center of gravity. Dividing by zero is a mathematical error and should be avoided in all calculations. If you encounter a situation where zero is a possible value for a distance, you should use a different formula or approach the problem from a different angle to avoid the error.

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