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Lab 1 09/22/2019 E. Jaramillo

Lab Course 141-Section 1: Physics Laboratory Report Lab number and Title: Lab 1 – Archimedes’ Principle Your Name: Emily Jaramillo Group members: Keithleen Penetrante, Luzangela Martinez Experiment Date: 09/11/2019

Introduction When an object is placed in fluid of any kind it either sinks to the bottom or floats on top of it. This phenomenon can be explained using Newton’s laws of motion which state that a net force must be pushing the object either upwards or downwards. With respect to fluids, this net force always pushes up and is therefore called the buoyant force. Said buoyant force is defined as the force of the weight of the fluid an object displaces and is known as Archimedes’ principle. Depending on the density of the fluid and object there are three possible cases which can occur, the object has a lighter density than the fluid and will float, the object has a higher density and will sink, or the object’s density is equal to that of the fluid and will remain suspended anywhere within that fluid. In this experiment, Archimedes’ principle will be studied along with buoyant forces and specific gravity to test the hypothesis that the buoyant force equals the weight of the displaced fluid. To do so four tests were performed, the first to directly measure Archimedes’ principle, the second to calculate the specific gravity of the metallic sphere to find its density to determine what kind of metal it is, the third to find the specific gravity of a wooden sphere that is known to float so as to determine what type of wood it is via its density, and the fourth to find the specific gravity of an unknown liquid in order to determine its density. For tests 1,2, and 4 there was little to no percentage of error as the values where 0%, 1.79%,

Lab 1 09/22/2019 E. Jaramillo

and 0.98% respectively. A major source of potential error is systematic as the scale used for the first three tests of the experiment only produced whole numbers. As can be seen by the percentage of error each test supported the statement that the buoyant force is equal to the displaced weight of the fluid by an object. Theory 1. Density and Pressure The density ρ of a body that occupies a volume V and has a mass m is defined as, ρ=

m V

Density is measured in units of [kg/m3]. The pressure P exerted by a force F on a surface that has an area A is defined as, P=

F A

or as force exerted on the surface per unit area. Pressure is measured in units of [N/m2], or [Pa], called Pascal.

2. Buoyant Force Consider a fluid that is at rest inside a container, as shown in Figure 1. Since the fluid is at rest, Newton’s second law of motion says that at any point inside the volume occupied by the fluid, the sum of all the forces must be zero, or F = 0. If a block of fluid of mass mf, as shown in the Figure, is taken into consideration, it must be at rest because it is just a piece of the original fluid that has been artificially separated from the rest. This means that the sum of the forces on the object must be given by,

F = -F0 - F - mfg + n = 0

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where mfg is the weight of the block, the force F0 is due to the atmospheric pressure P0, the force F is the weight of the column of fluid above the block and n is the normal reaction of the fluid that is below the block under consideration.

.

Buoyant force B is defined

Figure 1 Fluid at rest inside a container

as,

B = -F0 - F + n = mfg If the block of fluid is now replaced by a block of the same dimensions but made of any other material, the forces acting on this body will be given by, F = -mog - F0 - F + n where mog is the weight of the new body or object. Using the equation for buoyant force the last equation can be rewritten as,  F = -mog - F0 - F + n = -mog + B which shows the buoyant force B applied by the fluid on the object. This buoyant force is equal to the weight mfg of the displaced fluid, as it was shown above. The object, then, is pushed up by a force equal to the weight of the fluid it displaces. There are three possible cases for the equation above, Case A:

Lab 1 09/22/2019 E. Jaramillo

 F = -mg + B = 0 The buoyant force cancels the weight of the object and the net force is zero. The object will be at rest anywhere within the fluid. In terms of densities, this implies that mog = B  ρoVog = ρfVfg  ρo = ρf where the fact that the volume of displaced fluid Vf is equal to the volume Vo of the object was used in the equation.

Case B: ΣF = -mog + B > 0  B > mog In this case, when the object is completely submerged, the weight of the displaced fluid, or buoyant force, is greater than the weight of the object and the net force points upwards. The object will float, as shown in Figure 2. The object will be at rest when the weight of the fluid its submerged part displaces equals its total weight. If the total volume of the object is Vo and the submerged volume is Vf, the buoyant force is given by, B = mfg = ρfVfg

where the definition of density was used. In the same way, the weight of the object can be written as, mog = ρoVog. Since the buoyant force equals the weight of the object, B = mog ρrfVfg = ρoVog which results in, ρf V 0 = ρ0 V f

Lab 1 09/22/2019 E. Jaramillo

That is, the total volume Vo of the object is to the submerged volume Vf as the density of the fluid rf is to the density of the object ro. Whenever ρf > ρo, then Vo > Vf and the object is only partially submerged.

Figure 2 Object partially submerged Case C: F = -mog + B < 0  B < mog In this case, when the object is completely submerged, the weight of the displaced fluid, or buoyant force, is smaller than the weight of the object itself and the net force points down. The object sinks to the bottom of the container. In terms of densities, the relation is given by, B < mog  mfg < mog  mf < mo  ρfVf < ρoVo  ρf < ρo

where the fact that for a submerged body V0 = Vf was used. This shows that, whenever ρf < ρo, the object will sink.

In conclusion, Archimedes’ Principle can be stated the as, any body completely or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the body.

3. Specific Gravity

Lab 1 09/22/2019 E. Jaramillo

The specific gravity (sp. gr.) of a substance (solid or liquid), is defined as the ratio of the weight ws of a given volume of that substance to the weight ww of an equal volume of water, sp. gr. =

ws ww

It should be noticed that for a completely submerged body, the weight of water is the buoyant force exerted by the fluid on the body. Using the definition of density and taking into account that the volume of the substance is equal to the volume of water, this equation can be written as

Specific gravity has no units because it is a ratio of two densities or similar quantities. The density of water is 1 g/cm3. Specific gravity is, then, given by,

Specific gravity is numerically equal to the density of the substance. Thus, the density of a substance is found by including the proper units to the sp. gr., once this is known.

Lab 1 09/22/2019 E. Jaramillo

Experimental Procedure Materials: 

Electronic Scale



String



Metallic sphere



Wooden Sphere



Graduated Cylinder



Beaker



Hydrometer

1. Direct Measurement of Archimedes’ Principle a. Mass of metallic sphere m0 was measured. b. Graduated cylinder was filled with water until 60 mL and the mass of the water mw was recorded. c. Metallic sphere was placed inside of cylinder and the new volume of water and mass of water m’w were recorder. d. Mass of sphere submerged in water m’0 was then recorded by hanging on a hook found below the electronic scale. e. Using the measured values, the buoyant force was found using the following equations: B=( m o −m ' o ) g∧B=mdw g 2. Density of an object with ρ o > ρ w a. The specific gravity of the metallic sphere used in part 1 was calculated using the data collected in part 1 using the equation: sp . gr .=

mo m o−m'o

Lab 1 09/22/2019 E. Jaramillo

3. Density of an object with ρ o < ρ w a. Measured the mass of the wooden sphere mwood with electronic scale. b. Attached metallic sphere (sinker) to wooden sphere using string and various knots. c. The sinker was submerged in the water in the beaker while the wooden sphere was left above water. d. The mass ml of only the submerged sinker was measured. e. Then the mass of both the spheres submerged was measured (m2) f. The specific gravity was then found using the following equation: sp . gr .=

mwood m l −m 2

g. The type of wood was then identified using densities found in Table A. 4. Density of a liquid a. The mass m’o of the metallic sphere was measured while completely submerged in the unknown liquid. b. The specific gravity of the unknown liquid was then calculated using: (mo−m 'o)l sp . gr .= (mo −m ' o)w c. The right units where then given to the specific gravity thus making it the density of the liquid and the hydrometer was used to measure the density of the unknown liquid. d. The value from the hydrometer was then compared to the value for the calculated density and the percent difference was found.

Results and Discussion 1. Direct Measurement of Archimede’s principle In this part of the experiment the mass of a metallic sphere in both air and submerged in water was measured using an electronic scale and compared the mass of the water in a

Lab 1 09/22/2019 E. Jaramillo

graduated cylinder with an original volume of 60mL and the mass once 8.5mL of water where added. This comparison allowed one to compare if the mass of the displaced fluid is equal to that of the metallic sphere. The measured masses were recorded as follows: mo = 66 g = 0.066 kg Type of Metal = _____? _____ Mark = 60mL mw = 60g = 0.060kg New mark = 68.5 mL m’w = 68 g= 0.068kg mdw = m’w - mw = 8g =0.008 kg m’o = 58g = 0.058 kg

Archimedes’ principle was then further supported by calculating the buoyant forces with two equations, one relating the mass of the object in air and water and the other relating the mass of the displaced water. Both calculated buoyant forces where equal and therefore resulted in a 0% difference as can be shown by the reconciled data below. Therefore, showing that the buoyant force is equal to the mass of fluid displaced by an object.

( ms )n=0.0784 N

' B mo =(m o −m o ) g= ( 0.068− 0.050 ) kg 9.8

2

( sm )=0.0784 N

B mdw =mdw g=0.008 kg 9.8

%diff =

2

|0 .0784 N−0.0784 N| |B mo −B mdw| ∗100= ∗100=0 % 0.0784 N +0.0784 N |B mo −B mdw|

(

2. Density of an object with ρo > ρw

2

)

(

2

)

Lab 1 09/22/2019 E. Jaramillo

In this part of the experiment the data recorded for the metallic sphere in part 1 was used to calculate the specific gravity and then the density of the metallic sphere. sp . gr .=

mo 0.066 kg =8.25 = m o−m'o 0.066 kg−0.058 kg

It is important to note that specific gravity and density are numerically equivalent, however, density requires the proper units. The density found was similar to that of Brass found in Table A. Density (ρo) = 8.25 g/cm3 Closest Density as per Table A: Brass = 8.4 g/cm3

%error =

|

|ρmetallic sphere −ρbrass| ∗100= ρ brass

|

g 8.4 g ¿− c m c m3 ∗100=1.79 % g 8.4 c m3

8.25

3. Density of an object with ρo < ρw This part of the experiment was employed to calculate the density of a wooden sphere that would float on water or is less dense than water via a metallic sphere or sinker. The sinker would weigh the wooden sphere down in order to submerge it completely in the water. At first the mass of the wood in air was measured and then the mass of only the submerged sinker as ml and then both spheres submerged as m2. See reconciled data below: mwood = 6g = 0.060 kg ml = 65g = 0.065 kg m2 = 56 g = 0.056 kg Using the measured values above the specific gravity of the wooden sphere was calculated, changed to density and then compared to the values in Table A. from which it was deduced that the wooden sphere was made of oak. Calculations are shown below:

Lab 1 09/22/2019 E. Jaramillo

sp . gr .=

mwood 0.060 kg =0.67 = ml−m2 0.065 kg−0.056 kg

ρ wooden sphere = 0.67 g/cm3 Type of Wood as per Table A = Oak 4. Density of a liquid The metallic sphere used in part 1 was utilized in this part of the experiment to calculate the density of an unknown liquid. By using an object for which the buoyant force is known in water it is easier to find the specific gravity of the unknown liquid as it becomes necessary to only find the buoyant force of the metallic sphere in the unknown liquid. To do so, the same procedure of measuring the mass both in and then submerged in the unknown liquid was followed. Using the reconciled data from part one the specific gravity of the unknown liquid was found as follows.

m’o = 58.4 g = 0.0584 kg sp . gr .=

(mo−m ' o)l ' o w

(mo −m )

=

(0.0665−0.0584 )l =1.01 (0.066 kg−0.058 kg)w

It is important to note that for this part of the experiment, another scale was used which showed decimals as opposed to the scale used is parts 1 through 3 which showed only whole numbers. The specific gravity found was then changed to density by adding the proper units. The density of the unknown liquid was then also found by measuring using a hydrometer. The percent difference of the measured and calculated values was then calculated. The percent difference was zero and shows the accuracy of using Archimedes’ principle to relate buoyant force of the metallic sphere in both water and the unknown liquid to find specific gravity. The results of part 4 are as follows:

Lab 1 09/22/2019 E. Jaramillo

ρfluid ( calculated)=1.01 g/ mL ρfluid ( measured) =1.02 g/ mL g −1.02 g/mL | 1.01 | mL ∗100= ∗100=0 % g | ) (1.01 mL +1.02 g/mL )

|ρcalculated − ρmeasured| %diff =

(|

ρcalculated + ρmeasured 2

2

Conclusions After having performed the experiment, our hypothesis regarding the validity of Archimedes’ principle where all supported by every part of the experiment as all the percent differences where extremely low and Parts 1 and 4 were zero. Any source of potential error if any, is systematic as the scale used for part 1 through 3 displayed measurements of mass with whole numbers. However, to combat this source of error in Part 4, a new scale which displayed decimals was utilized to measure a more accurate number for the mass of the metallic sphere in the unknown liquid. Any slight change in mass of the object could have affected the calculated specific gravity, density and in turn increasing the percent difference thus decreasing accuracy. To conclude, the results form this experiment support the fact that the relationship between specific gravity, density and buoyant force can be tied to the definition of Archimedes’ principle. .

Lab 1 09/22/2019 E. Jaramillo

References "New Jersey City University: Department of Physics." Archimedes’ Principle, September 21, 2019

Post-Lab Questions 1. Show that equations (1.4) and (1.5) are correct. (1.4) mwood - m'wood = m1 - m2 m1= mwood + m’s m2 = m'wood +m’s Substitute: mwood - m'wood = mwood + m’s – (m'wood + m’s)  All the values then cancel out as they are equivalent.

sp . gr .=

(1.5)

mwood m = wood m wood −m' wood m1 −m 2

mo ρ Vo sp . gr .= o= but V o=V f ρ f mf Vf mo mf

→

mo m 1−m 2

 Therefore equation 1.5 is correct to find the specific gravity of the wooden sphere or any other object.

2. Which would be the effects in the measurement of air bubbles attached to the bottom of a submerged object? Air is less dense than water and therefore air bubbles would cause the object to be pushed upwards as air floats on top of water.

Lab 1 09/22/2019 E. Jaramillo

3. Comment on the purity of the metal of the sphere used in section 1. of the experimental procedure. Brass is an alloy which contains a percentage of both copper and zinc and traces of other metals. Because brass is a combination of metals therefore it is very hard to know how pure the metal is. To determine the properties of metals like these it is evident that Archimedes’ principle needs to be used as it allows to find the density of the unknown metal to identify it as was done in Parts 1 and 2 of the experiment. After finding from there its composition and purity can be determined

4. Use the pressure depth relationship P = ρgh to show mathematically that the buoyant force on a wooden cube being submerged in a liquid is equal to the weight of the volume of liquid it displaces. P1=ρg h1=

F1 → F 1=ρg h1 A A

P2=ρg h2=

F2 → F 2=ρg h2 A A

F B= F 2−F 1=ρg h1 A −ρg h2 A

¿ ρg A ( h2−h1) ¿ ρgA S ¿

m gV V

¿ mg

 mg is therefore the weight of the displaced fluid

5. Was Archimedes’ Principle verified in all cases? Explain. Yes, it was verified in all cases as the percentage of error and difference for parts 1 and 4 were

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extremely low and even zero not to mention that for part 1 the buoyant forces using only the mass of the object and the mass of the displaced water found were equal. On the other hand, for parts 2 and 3 the materials could be identified using Table A thus showing just how accurate the use of Archimedes’ principle is to find the density and in turn identify the nature of an object. For more proof of verification please refer to results and discussion....