help!!!!!!!!!need answers please!!!!!!!!!!!! atleast try to answer it please!`?

Question

at depth in cm below the surface of a fluid in density g/cm^3
a solid circular cylinder of lenght 10cm and radius 2cm has its axis vertical, and its top end is 15cm below the surface of a liquid of relative density 1.3.calculate the thrust of
A) the upper AND lower end of the cylinder due to the fluid

from the results deduce the loss of weight of cylinder on immersion in the fluid.

Answer 1

The pressure from a column of fluid of height h is rho*h*g. The ends of your cylinder have area A=pi*r^2. Therefore the force force on the top of the cylinder (downwards) is rho*g*(15cm)*pi^r^2 and the force on bottom of the cylinder (upwards) is rho*g*(15cm+10cm)*pi^r^2. The upward force on the cylinder (loss of weight) is the difference between the two, rho*g*pi*r^2(15cm - 25cm).

You will note that this comes out rho*g*(cyl vol), or the total weight of fluid displaced by the cylinder.

rho is actual density = relative density * density of water. Since density of water = 1gm/cm^3, rho = 1.3gm/cm^3

EDITED: Somehow my first post lost the force on the top.

Answer 2

Chad;
This sounds like your homework problem, so rather than just give you the answer, let's try to figure it out. The first part is a little odd, I guess by "thrust" they mean force. So what "thrust" does the top end of the cylinder see? Well, the area (A) of the top is pi x r squared. Let's pretend the cylinder is dry and the fluid is just jelly sitting on top of it. What would be the weight of a jelly cylinder of area A that is15 cm high? 15cm x A x 1.3gm/cm cubed right? Or another way to make a more general statement is that the Force (thrust) = Depth x Area x density. You can use this formula for the thrust on the bottom of the cylinder. Now for the second part of the question: The interesting thing in this case (an object with vertical walls) is that the difference between the top thrust and the bottom thrust = the decrease in weight of the object. As a check of the answer you get, for an object immersed in a fluid, the decrease in weight of the object equals the weight of fluid displaced (Archimedes' principle) so that's easy, it's just the volume of the cylinder times the density. Hope your answers agree.
Good luck, Craig.