# Fluids and buoyance, determine how much of the cylinder is in the oil

1. Jan 30, 2005

### Rachel C

I have two physics problems on fluids that I'm stuck on and would appreciate any help! This is an algebra based physics class... (no calculus).

1. A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 7.00 kg. This cylinder is floating in water. Then oil (rho = 725 kg/m^3) is poured on top of the water. How much of the height of the cylinder is in the oil?

I believe this problem involves Archimedes' Principle, which states that the magnitude of the buoyant force (F_b_) equals the weight of displaced fluid(W _fluid_).
F_b_ = P2*A - P1*A = rho*g*h*A
A = (pi*r^2)*h = 0.00848 m^3
rho for water = 1000 kg/m^3

Thats about as far as I can go... I don't know how to determine the height of the cylinder that is in the oil. ​

2. A pump and its horizontal intake pipe are located 12 m beneath the surface of a reservoir. The speed of the water in the intake pipe causes the pressure there to decrease, in accord with Bernoulli's principle. Assuming nonviscous flow, what is the maximum speed with which water can flow through the intake pipe?

Bernoulli's equation is....
P1 + 1/2*rho*v1^2 + rho*g*y1 = P2 + 1/2*rho*v2^2 + rho*g*y2

I think y1 = 0 m and y2 = 12 m. Then I am trying to solve for v1. I know that rho = 1000 kg/m^3, and g = 9.8 m/s^2. I'm not exactly sure what a reservoir is... but I think the water does not move, so you can assume v2= 0 m? But then I don't know P2 or P1....so how am I suppose to solve for this?​

2. Jan 31, 2005

### Andrew Mason

Find density of the cylinder. According to Archimedes principle, if the density is less than the density of the medium (water), it floats and displaces a volume of water whose mass is equal to the mass of the floating object. It floats, so it displaces 7 kg of water, or .007m^3 of water. Therefore, the cylinder displaces .007/.00848 of its volume of water, which (if the cylinder is floating with its axis pointing up) is the ratio of its height below water to total height.

The issue here is: how fast can the pressure on the intake pipe 'feed' the pump. The pressure in the reservoir at the intake is: P = atmospheric pressure + pressure due to water height. The pressure inside the pump can be no less than 0. Assume there is no change in height in moving from intake through the pump so ignore the potential energy terms ie. y1=y2

In the reservoir, water is static so v = 0. In the pump, lets assume a perfect vaccum: (P = 0). So with maximum flow all of the pressure energy is converted to flow (kinetic) energy:

$$P_{res} + 0 = 0 + \frac{1}{2}\rho v^2$$

$$P_{res} = \frac{1}{2}\rho v^2$$

$$v = \sqrt{\frac{2P_{res}}{\rho}}$$

AM

Last edited: Jan 31, 2005