Given a floating object of density D, find the maximum number of an object put on the floating object such?
-
Given a floating object of density D, find the maximum number of an object of mass M and volume V put onto the floating object before it will sink below water. Write the answer in terms of D, M, and V. Note: Design of the object is not given. All I want is the maximum number under any conditions necessary. You may assume they are put on one time after a million years or they are put on five times within one second (but do not say to put on one billion in a nanosecond). Please do not tell me the optimal design. Currently have a contest on Oct 22 with pennies and two pieces of computer paper. I still want to know what's the best possible achievement. Thank you.
-
Answer:
Density D has to be less that that of water (call it d) or the object won't float at all. v is volume of floating object m is mass of floating object m = vD The object, when loaded is totally submerged, so buoyancy force (BF) is the weight of the water it displaced, minus it's weight. Volume of water = volume of object = v mass of water = vd weight of water = gvd Weight of object = gvD BF = gvd - gvD = gv(d-D) BF is the weight of the load it can carry, = ngM BF = gvd - gvD = gv(d-D) = ngM solving for n n = v(d-D)/M number is volume of floating object times difference in densities divided by mass of load objects dimensional check v = m³ d = kg/m³ M is kg n is m³kg/m³/kg, and it all cancels, as it should. .
UnknownD at Yahoo! Answers Visit the source
Other answers
DL, Thought I recognized your face: Couple of ideas: 1. If you just float the paper, it is simply bouancy. Which is the volume of paper * (Density water - Density Paper) 2. If you make a boat with the paper, then your displacement rules change because you are using the air for some of the volume. You just add the Volume of the air * density of water. So in designing you boat, you want maximum volume, with some stability. You also want a design that is strong enough not to deform under your proposed load. 3. If you can only float a paper, you might be able to use the surface tension effect, see: http://answers.yahoo.com/question/index;_ylt=AlolIrBKdzsh6w5sYiyhDdAjzKIX;_ylv=3?qid=20080619083318AA2zvKj (This is actually really cool. You can float a CD and load it up with pennies and see the water bulge over the top. Great science experiment.) ******** I presume you can build a boat with the paper? Can you join the two piece of paper together in a watertight manner? As I said above, it is volume, stability, and avoding deformation.
Frst Grade Rocks! Ω
Let m and m/D be the mass and volume of the floating object. Let n objects of mass M and volume V be placed on it. You are saying that the floating object alone must sink under water. Then what this means is that the total weight of all the objects must equal the mass of water occupying a volume m/D. ie m + n*M = ρ * m/D where ρ = density of water n = m/M * (1 - ρ/D)
Dr D
Related Q & A:
- How To Find Tamilnadu Mobile Number?Best solution by answers.yahoo.com
- How can I find the minimum number of open rectangles in a grid?Best solution by Stack Overflow
- How to find out the number of a missed call?Best solution by ehow.com
- How to find the serial number on a printer?Best solution by Yahoo! Answers
- How do I find the item number on a ebay item?Best solution by Yahoo! Answers
Just Added Q & A:
- How many active mobile subscribers are there in China?Best solution by Quora
- How to find the right vacation?Best solution by bookit.com
- How To Make Your Own Primer?Best solution by thekrazycouponlady.com
- How do you get the domain & range?Best solution by ChaCha
- How do you open pop up blockers?Best solution by Yahoo! Answers
For every problem there is a solution! Proved by Solucija.
-
Got an issue and looking for advice?
-
Ask Solucija to search every corner of the Web for help.
-
Get workable solutions and helpful tips in a moment.
Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.