Dividing over zero while calculating center of gravity - Getting error

AI Thread Summary
The discussion centers on calculating the center of gravity for a given object, where participants address errors arising from dividing by zero in their calculations. The main issue involves correctly determining the signs of areas and coordinates, with a consensus that areas contributing to the body should be positive while those taken away should be negative. Participants clarify that the denominator in the centroid formula should always reflect positive contributions from the areas, regardless of their position on the coordinate axes. After several corrections and clarifications, they arrive at a more accurate calculation for the coordinates of the center of gravity, ultimately confirming that the solution manual's values were inconsistent. The conversation highlights the importance of careful consideration of signs and contributions in centroid calculations.
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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
x_c= \frac{\int x dA}{\int dA}= \frac{\int x dA}{A}
y_c= \frac{\int y dA}{\int dA}= \frac{\int x dA}{A}

The denominator is just the area of that figure which obviously is not 0! There should not be any subrtraction in your calculation.
 
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:
 
@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?
 
Should the sign between 21x12 and 21x9 be a plus, not a minus?
 
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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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?
 
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:

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)}

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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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:
 

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