Dividing over zero while calculating center of gravity - Getting error

[Femme_physics]

Homework Statement

Calculate the center of gravity of this object:

http://img638.imageshack.us/img638/8412/inddddd.jpg

The Attempt at a Solution

The bottom line turns out zero, so I get something over zero therefor error!

http://img101.imageshack.us/img101/6995/theretherere.jpg

 

[HallsofIvy]

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.

 

[I like Serena]

Hi Fp! I was wondering if you'd be here today!

On the bottom line (the denominator) you have a sum of areas.
But you've taken some of those areas as being negative?

 

[Femme_physics]

@HallsofIvy - We're not using calculus in our solutions

Hi Fp! I was wondering if you'd be here today!

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?

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?

 

[sjb-2812]

Should the sign between 21x12 and 21x9 be a plus, not a minus?

 

[I like Serena]

I meant to post 8 hours ago but I had internet disconnection issues!

Aw! That must have been hard to take!

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.

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

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?

 

[I like Serena]

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).

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

Ohhhhhhhhhhhhhhhh! okay, give me a moment :)

 

[I like Serena]

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! ).

 

[Femme_physics]

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:

http://img14.imageshack.us/img14/3472/xcxcxcxg.jpg

Xc = 0.17

 

[Femme_physics]

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! ).

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!

 

[I like Serena]

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!

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?

 

[Femme_physics]

I used the wrong distances for the triangles

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

This is what I get after all the corrections

http://img862.imageshack.us/img862/3050/xcxcxcxcxcxlllllllxcxcx.jpg

 

[I like Serena]

I used the wrong distances for the triangles

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

Very good!

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?!

All the signs are correct though!

 

[Femme_physics]

Oops! I'm getting too excited lately, heh. http://img823.imageshack.us/img823/2717/cxzczxczxczx.jpg

:) Got it, right?

 

[I like Serena]

Oops! I'm getting too excited lately, heh.

:) Got it, right?

Right!

What is it you are you excited about?

 

[Femme_physics]

To solve it! :) Thanksssssssssssssssssss!

I'll add it to the corrections manual!

 

[I like Serena]

To solve it! :) Thanksssssssssssssssssss!

I'll add it to the corrections manual!

You're welcome.

But what about Yc?

 

[Femme_physics]

Same as the solution manual :)

http://img155.imageshack.us/img155/1975/24093328.jpg

 

[I like Serena]

Same as the solution manual :)

Yes. That is right!

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!