# Finding thickness given width, length, mass, density

• Sep 12th 2009, 06:43 PM
hemi
Finding thickness given width, length, mass, density
Here's the problem:

A sheet of metal is 29.4 mm wide and 16.1 mm long. If it weighs 4.093 g and the density of the metal is 7.23 g/cm3, what is the thickness of the sheet (in mm)?

I'm not sure how I would begin to approach a problem like this as it's asking for a variable I'm not sure how to solve down for. Any help towards this problem will be more then appreciated, thank you. :)
• Sep 12th 2009, 06:58 PM
VonNemo19
Quote:

Originally Posted by hemi
Here's the problem:

A sheet of metal is 29.4 mm wide and 16.1 mm long. If it weighs 4.093 g and the density of the metal is 7.23 g/cm3, what is the thickness of the sheet (in mm)?

I'm not sure how I would begin to approach a problem like this as it's asking for a variable I'm not sure how to solve down for. Any help towards this problem will be more then appreciated, thank you. :)

Density

$\displaystyle d=\frac{m}{V}$ by definition.

You've got $\displaystyle d=7.23g/cm^3$ and $\displaystyle m=4.093g$

Solve for $\displaystyle V$.

Then put

$\displaystyle V=(29.4mm)(16.1mm)t$ where $\displaystyle t=thickness$ in millimeters.

PS Remember to convert.

BTW: Posts:37, Thanks:0 (Wondering)
• Sep 13th 2009, 02:30 PM
hemi
Quote:

Originally Posted by VonNemo19
Density

$\displaystyle d=\frac{m}{V}$ by definition.

You've got $\displaystyle d=7.23g/cm^3$ and $\displaystyle m=4.093g$

Solve for $\displaystyle V$.

Then put

$\displaystyle V=(29.4mm)(16.1mm)t$ where $\displaystyle t=thickness$ in millimeters.

PS Remember to convert.

BTW: Posts:37, Thanks:0 (Wondering)

Thanks.

So I found $\displaystyle V=m/d$ and ended up getting the value $\displaystyle 0.5661$. Now I'm a bit confused about this part:

$\displaystyle V=(29.4mm)(16.1mm)t$ where $\displaystyle t=thickness$ in millimeters.

What is this $\displaystyle t$ variable? Am I going to rearrange for $\displaystyle t$ by plugging in the $\displaystyle V$ value I have calculated?

Yeah I haven't been thanked much I guess. Don't really know what to say about that? (Crying)
• Sep 13th 2009, 03:47 PM
VonNemo19
Quote:

Originally Posted by hemi
What is this $\displaystyle t$ variable? Am I going to rearrange for $\displaystyle t$ by plugging in the $\displaystyle V$ value I have calculated?

Yes.

Quote:

Originally Posted by hemi
Yeah I haven't been thanked much I guess. Don't really know what to say about that? (Crying)

(Rofl) I was thinking in the opposite order.
• Sep 13th 2009, 04:07 PM
hemi
So thus I can write:

$\displaystyle t = (29.4mm)(16.1mm)/0.5661$ and therefore $\displaystyle t = 836mm$ thick? Is this correct?

Also what is the unit on the value $\displaystyle 0.5661$, cubic centimeters (cc)?

This answer doesn't seem logical to me. The sheet is thicker then its length and width? (Worried)
• Sep 13th 2009, 04:42 PM
VonNemo19
Quote:

Originally Posted by hemi
So thus I can write:

$\displaystyle t = (29.4mm)(16.1mm)/0.5661$ and therefore $\displaystyle t = 836mm$ thick? Is this correct?

Also what is the unit on the value $\displaystyle 0.5661$, cubic centimeters (cc)?

This answer doesn't seem logical to me. The sheet is thicker then its length and width? (Worried)

You have to convert density to mm before you begin.
• Sep 13th 2009, 04:56 PM
hemi
Alright so let me try to rework it:

The Density Conversion:

$\displaystyle \frac{7.23g/cm^3}{10mm} = 0.723 g/mm^3$

The volume calculation:

$\displaystyle V = m/d$
$\displaystyle V = \frac{4.093g}{0.723 g/mm^3} = 5.66 mL$

The rearrangement:

$\displaystyle V = (29.4mm)(16.1mm)t$

$\displaystyle t = \frac{29.4mm)(16.1mm)}{5.66mL} = 83.62mm$

This still seems a bit strange to me. I feel like I'm doing something wrong. (Doh)
• Sep 13th 2009, 05:03 PM
VonNemo19
Quote:

Originally Posted by hemi
Alright so let me try to rework it:

The Density Conversion:

$\displaystyle \frac{7.23g/cm^3}{10mm} = 0.723 g/mm^3$

The volume calculation:

$\displaystyle V = m/d$
$\displaystyle V = \frac{4.093g}{0.723 g/mm^3} = 5.66 mL$

The rearrangement:

$\displaystyle V = (29.4mm)(16.1mm)t$

$\displaystyle t = \frac{29.4mm)(16.1mm)}{5.66mL} = 83.62mm$

This still seems a bit strange to me. I feel like I'm doing something wrong. (Doh)

This rearangement that you speak of... Here it is:

$\displaystyle V=l\cdot{w}\cdot{t}$

Therefore

$\displaystyle t=\frac{V}{l\cdot{w}}$

You've got

$\displaystyle t=\frac{l\cdot{w}}{V}$

See my point?

And where is the mL coming from? We're talking about a solid, right?
• Sep 13th 2009, 06:47 PM
hemi
Alright so right from the start:

The Density Conversion:

$\displaystyle \frac{7.23g/cm^3}{10mm} = 0.723 g/mm^3$

The volume calculation:

$\displaystyle V = m/d$
$\displaystyle V = \frac{4.093g}{0.723 g/mm^3} = 5.66 mL$

The rearrangement:

$\displaystyle V = (29.4mm)(16.1mm)t$

$\displaystyle t = \frac{5.66mm} {29.4mm)(16.1mm)} = 0.011957578mm$

That still doesn't seem correct to me?

Sincerely Von, thanks for your help and patience, I know it may seem quite obvious to you, this is relatively new stuff for me. So thanks a lot for everything. :)
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}