Procedure:

Getting Familiar

1. On the Intro screen, mess with the apparatus, changing the blocks and fluids.

Lab Setup

1. There are 5 different fluids to choose from in the lab and five different types of materials. (Styrofoam, wood, ice, brick and aluminum)

2. Use the table supplied to organize your work.

Lab Procedure: Part 1

1. In each of the scenarios below, determine first, by predicting, whether the object will sink or float. Use a volume of 1mˆ3.

2. Test each object once you have predicted and record the results.

Part 1: Write an “S” for sink or an “F” for float. Predictions first!!

Air Gasoline Olive Oil Water Honey

Pred. Act. Pred. Act. Pred. Act. Pred. Act. Pred. Act.

Styrofoam

Wood

Ice

Brick

Aluminum

Archimedes principle

Background

Fluids are generally thought of as liquids; however, this is a common misconception. A fluid is anything that can flow, which includes gasses as well as liquids. When an object is submerged in a fluid, it experiences an upward buoyant force Fb that opposes gravitational force Fg. This is the reason ice floats on the top of water, and a balloon filled with helium rises in air. If we define Fg in the negative direction, a submerged object will rise in the fluid if the net force is positive (the condition of ice rising in water, or a helium balloon rising in air), and sink if it is negative (the condition of a rock sinking in a pond).

The magnitude of the gravitational force acting on an object is proportional to its mass, but it is easily observable that the buoyant force acting on a submerged object is not proportional to the object’s mass: a small rock may have the same mass as a tennis ball, but a tennis ball floats in water and the rock does not. So, what is different between these two objects? Their masses may be the same but their volumes are different, and so is the volume of water displaced by each once submerged.

In this activity you will explore the relationship between the buoyant force acting on an object and the volume of fluid displaced by the object, and draw conclusions that help establish the mathematical relationship between buoyant force and a) the volume of the submerged object, and b) the weight of the fluid displaced by the submerged object.

Data Analysis

Watch the video in this link (if that doesn’t work, here) and collect the following data:

Part 1 – Brass Cylinder

Brass cylinder length l (cm):

Brass cylinder radius r (cm):

Brass cylinder area Acyl (cm2):

Table 1: Buoyant force and displacement values for a brass cylinder submerged in a fluid

Depth

(cm) Vsubm

(cm3) Tension

(N) mdisp

(g) Fb

(N) wdisp

(N)

0 0.00 0.00 FT1 0.00 0.00 0.00

1/4l FT2

1/2l FT3

3/4l FT4

l FT5

Part 2 – Aluminum Cylinder

Aluminum cylinder length l (cm):

Aluminum cylinder radius r (cm):

Aluminum cylinder area Acyl (cm2):

Table 2: Buoyant force and displacement values for an aluminum cylinder submerged in a fluid

Depth

(cm) Vsubm

(cm3) Tension

(N) mdisp

(g) Fb

(N) wdisp

(N)

0 0.00 0.00 FT1 0.00 0.00 0.00

1/4l FT2

1/2l FT3

3/4l FT4

l FT5

• Calculate the following using the measured values for each cylinder. Record the results into or above each cylinder’s respective table.

a. The depth in centimeters in Tables 1 and 2 using cylinder length l.

b. The cross-sectional area A, in cm2, of each cylinder using radius r.

Acyl=πr2Acyl=πr2

c. The volume of the cylinder submerged Vsubm, in cm3, at each depth.

Vsubm=depth×AcylVsubm=depth×Acyl

• If the tension FTn measured when each cylinder was submerged is equal to the difference between the gravitational force and the buoyant force:

FTn=Fg−FbFTn=Fg-Fb

and the tension FT1 measured when each cylinder was suspended above the water is equal to the gravitational force:

FTl=FgFTl=Fg

then the buoyant force on each cylinder is equal to:

Fb=FTl−FTnFb=FTl-FTn (1)

For both cylinders, use Equation 1 to calculate the buoyant force Fb at each depth. Record the results into each cylinder’s respective table.

Analysis Questions

• What type of mathematical relationship (proportional, squared, inverse, inverse squared, et cetera) between buoyant force and submerged volume is implied by your data?

• The buoyant force Fb acting on an object that is partially or completely submerged in a fluid is described by the equation:

Fb=ρVgFb=ρVg (2)

where V is the submerged volume of the object and ρ is the density of the fluid in which the object is submerged.

From your data, knowing the volume, the force, and gravity (9.8m/sˆ2) what is the value of density (ρρ) that you obtained? How does this value compare to the real value of the density of those materials?