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 )

 

[rwisz]

I appreciate your attempt at finding a solution and using the correct equations. However, there are a few issues with your approach. Firstly, the density of water is given as 1000kg/m^3, not 1000kgm^3. This is an important distinction to note.

Secondly, the question asks for the amount of the rod that is underwater, not a distance. This can be solved by finding the volume of the rod that is underwater and then dividing by the total volume of the rod. Let's break down the steps to solve this problem:

1. Calculate the total volume of the rod. Since the rod is 10cm long and has a cross-sectional area of 1cm^2, the total volume is 10cm^3.

2. Calculate the volume of the metal part of the rod. This can be done by using the density of the metal (5000kg/m^3) and the volume formula (V = m/d, where m is mass and d is density). Since the metal part is 1cm long and has a cross-sectional area of 1cm^2, the volume is 1cm^3.

3. Calculate the volume of the wood part of the rod. This can be done using the same method as above, but with the density of wood (500kg/m^3) and the remaining length of the rod (9cm). The volume is 4.5cm^3.

4. Add the volumes of the metal and wood parts to get the total volume that is underwater (1cm^3 + 4.5cm^3 = 5.5cm^3).

5. Divide the total volume that is underwater by the total volume of the rod (5.5cm^3 / 10cm^3 = 0.55). This means that 55% of the rod is underwater.

In summary, the answer to the question is that 55% of the rod is underwater when the metal part is pointing downwards. I hope this helps clarify the problem and how to approach it. Keep up the good work in your physics studies!