How Much of the Rod Is Submerged When Partially Made of Metal?

[elsteveo25]

Homework Statement

one cm of a 10cm long rod is made of metal and the rest wood. Th metal has a density of 5,000kgm^3 and the wood has a density of 500kg/m^3. When the metal part points down ward, how much of the rod is underwater.

Homework Equations

water density = 1000kg/m^3


3. The Attempt at a Solution [/b
i am getting a distance of 1.053meters, which cannot be right because it is only 10cm. If anyone can tell me how to properly set up this equation, that would be great.

 

[tiny-tim]

welcome to pf!

hi elsteveo25! welcome to pf!

(try using the X² icon just above the Reply box )

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[elsteveo25]

Ok, I think I may have got the correct answer, but I am not sure.
I set the density of the metal times the volume plus the density of wood times the volume equal to the density of water times volume.
Equation looks like this.

density(pi(radius^2)*height)+density(pi(radius^2)*height)=density(pi(radius^2)*height)

5000(pi (r^2)*.01) + 500(pi (r^2)*.09) = 1000(pi (r^2)*h)

This gives me the answer of .095, which equals 9.5cm.

This sounds good to me, but I just would like to make sure that I am doing this right. I did not get a lot of clarification in class on exactly how to do this.

Also, please bear with me, I am new to this thread and too physics, I did try to use the xsquared button, but it did not appear to be working right.

Thanks.

 

[tiny-tim]

hi elsteveo25!

(just got up :zzz: …)

yes, that's right …

if we call the area A , then the buoyant force is 1000Ah upward, which has to equal the total weight 5000A*.01 + 500A*.09, so h = (50+45)/1000 m = .095 m = 9.5 cm

(hmm … if you click the button, and then type "2", you should get ² )