A frog is riding on the top of a cylindrical piece of wood floating in

  • Thread starter moatasim23
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A frog is riding on the top of a cylindrical piece of wood floating in still water.
Half of the wood, with a diameter of 4 cm and length 20 cm, is immersed in water.
The density of water is 1 gm/cc.
What is the mass of the wood along with the frog?
After the frog slowly goes into the water only one third of the wood remains immersed in water. Calculate the mass of the frog.
Calculate x, the distance between the water level and the center of the circular end of the wooden piece.
Briefly describe the motion of the wood after the instance the frog moves into the water. Give a rough sketch of x as a function of time.

I did the whole numerical but i didnt know how to give a rough sketch of x as a function of time.
 

HallsofIvy

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Re: Upthrust

Initially, the upward force of the water on the wood is equal to the weight (mass times g) of both wood and frog. The weight of the water displaced must be equal to that. You are given the size of the wood so can calculate its volume. Half of that is the volume of water displaced so you can calculate the volume and weight of water displaced. That is the mass of log and frog combined. When you say "I did the whole numerical", I suspect you mean that you did this. Yes, that's the easy part.

Once the frog leaves the log, it begins to move upward. If x is the distance between the surface of the water and the center of the log, the force on the log changes because the volume of water discplaced changes. Can you calculate the volume of the portiom of the log that is underwater as a function of x? That will, I believe, require an integration. Once you get that, you can calcullate the weight of water of that volume and so get the upward force on the log, again as a function of x. Now use "mass times acceleration equals force". Mass times acceleration is [itex]m(d^2x/dt^2)[/itex] where m is the mass of the entire log. The force is the force as a function of x you just calculated. So you will have a differential equation of the form [itex]m(d^2x/dt^2)= f(x)[/itex]. How you solve that depends upon your force function, f(x).

Try that and show us your work.
 
78
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Re: Upthrust

Thanks for help.Ill try this.
 

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