HYDRAUL-EXPT#2.docx

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EXPT.No.2 1

HYDHYDRAULICS LABORATORY

Experiment No.2Determination of the Buoyant Force by the

Archimedes Principle of Buoyancy

It was Archimedes of Syracuse, a Greek mathematician, who first discovered the law ofbuoyancy sometimes called Archimedes Principle which states that Any object, whollyor partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluiddisplaced by the object. According to the Newtons first law of motion, if the upwardforces balance the downward forces (including the weight) the object will either remainat rest or remain in motion at a constant rate. Otherwise, it will accelerate upward ordownward.

Buoyancy arises from the fact that fluid pressure increases with depth and from the factthat the increased pressure is exerted in all directions (Pascals principle) so that there

is an unbalanced upward force on the bottom of a submerged object with respect to theforces induced by water. The buoyant force of a submerge object in a body of water isequal to the weight of liquid displaced by the object. Water, with a density of one gramper cubic centimeter, provides a convenient way to determine the volume of anirregularly shaped object and then to determine its density.

Basically, objects with higher density than water will sink in it and those with lowerdensity will float on it. We can study buoyancy with the following methods;

METHOD I: Without Anchorage

If the buoyant force is equal to the weight of water displaced by the object then,

By summing up the vertical forces we will arrive to the following equation,


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EXPT.No.2 2


Where:

Bfbuoyant force

Woweight of the blockw- Width of the blockh - Draft or height of submergence

Water -specific weight of waterVsubmerged - submerged volume of the block

Equal to the product of the width, length and h

METHOD 2: With Anchorage

Sometimes there are objects or floaters that are used to suspend something in a

body of water or anchorage that are used to hold an object while afloat in water. Inengineering practice, this method is frequently used when building bridges or anystructure at sea. In this case, the computation of the existing buoyant force or forces isas follows:

By summing up the vertical forces we will arrived to the following equation:

Wo = Bf = water* Vsubmerged

Bf1+ Bf2 = Wo + Wa


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EXPT.No.2 4


1. Measure the dimensions of the wooden block.2. By using the spring balance or a weight balance, determine the weight of the

block in air ( Wb).3. Measure the dimension of the cylindrical vessel and fill it with the liquid to a

certain height.

4. Measure the original height of the liquid in the container (h1). This is the only waywe can measure the amount of displaced liquid.5. Place the block of wood in the liquid until it floats.6. Determine the final height of the liquid in the container (h2).7. Determine the volume of the submerged portion of the block using the graphical

illustration below.

Y = h2h1 (Height of the displaced liquid or height of the submerged volume)

Subtracting the base area of the wooden block from the base area of the containergives the area of the displaced liquid. Therefore, the volume of the submergedportion or the displaced liquid is given by;

Vs = (base area of containerbase area of the block) * Y

The weight of the block (Wo) which is equal to the buoyant force can then bedetermined by;

Wo = Bf = water* Vs

8. Compare the weight of the wooden block, as determined earlier using a weightbalance or spring balance, and the weight of the wooden block computed usingthe buoyancy principle.

h2h1

Side View

Top View

Wooden block

Liquid

Container


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EXPT.No.2 5


9. Compute the percent error by subtracting the weight of the wooden block frombuoyancy principle from the weight of the wooden block using a weigh balanceand dividing the difference with the weight of the wooden block using a weighbalance and multiplying by 100 percent.

% error = [(W1WO )/W1]*[100]

10.Make three trials using different wooden blocks.

METHOD 2: With Anchorage

1. Measure the dimensions of the wooden block.2. Weigh the block in air using a spring balance or a weight balance ( Wa).3. Put a string around the block in a vertical manner in such a way that there is an

excess string to tie up a weight at the bottom and the spring balance at the top.4. Get a large transparent container and fill it with water in such a way that the rise

of water in the container is negligible as in a large body of water where thevolume of displaced liquid is equal to the volume submerged of the block.

5. Submerge the anchorage in the water while holding the spring balance and whileapproximately half of the volume of the wooden block (Vb) is also submerged.6. Check the reading in the spring balance (Ww). Take note that we have to use a

spring balance so as to prevent the total immersion of the combination of thewooden block and the anchorage.

7. Remove the assembly from the water and determine the weight of theanchorage.

8. Determine the volume of the anchorage (Va) by displacement method using agraduated cylinder or beaker. This can be done by placing water in the graduatedcylinder or beaker of known volume then immersing the anchorage. The increasein volume is equal to the volume of the anchorage.

9. Determine the weight of the block from the following formula derived fromsumming up vertical forces as shown below;

Bf1+ Bf2+ Ww = Wo + Wa

Water Surface

Ww

Wo

Wa

Bf1

Bf2


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EXPT.No.2 7


Sample Liquid: OIL

UNITT R I A L S

1 2 3

Weight of wooden block ( W1) N 0.78 0.74 0.76

Original ht. of liquid (h1) m 0.054 0.054 0.054

Width of block ( w) m 0.05 0.05 0.05

Length of block (L) m 0.05 0.05 0.05

Base area of block m2 0.0025 0.0025 0.0025

Base area of container m2 0.00785 0.00785 0.00785

Final ht. of liquid ( h2) m 0.064 0.063 0.062

Ht. of displaced liquid (Y) m 0.01 0.009 0.008

Base area of displaced liquid m2 0.00785 0.00785 0.00785

Specific weight N/m3 9810 9810 9810

Wt. of block from buoyancyprinciple ( Wo)

N 0.77 0.69 0.62

% ERROR % 9% 7% 18%


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EXPT.No.2 8


Method # 2: WITH ANCHORAGE

Sample Liquid: WATER

UNITT R I A L S

1 2 3

Weight of wooden block ( W1) N 0.78 0.74 0.76

Weight of anchorage (Wa) N 0.8 0.8 0.8

Width of block ( w) m 0.05 0.05 0.05

Length of block (L) m 0.05 0.05 0.05

Thickness of the block m 0.05 0.05 0.05

Spring balance reading (Ww) N 0.78 0.78 0.76

Volume of the submergedportion of the wooden block

m3 62.5x10

-6 62.5x10

-6 62.5x10

-6

Volume of the anchorage m3 3.11x10-7 3.11x10-7 3.11x10-7

Unit weight of water N/m3

9810 9810 9810Bouyant force # 1 N 0.58 0.58 0.58

Buoyant force # 2 N 0.2 0.2 0.2


N 0.63 0.67 0.68

% ERROR % 19% 9% 10%

Sample Liquid: OIL

UNITT R I A L S

1 2 3Weight of wooden block ( W1) N

Weight of anchorage (Wa) N

Width of block ( w) m

Length of block (L) m

Thickness of the block m

Spring balance reading (Ww) N

Volume of the submergedportion of the wooden block

m3

Volume of the anchorage m

3

Unit weight of water N/m

3 9810 9810 9810

Bouyant force # 1 N

Buoyant force # 2 N


N

% ERROR %


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EXPT.No.2 9


Observations

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VI. Conclusions

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EXPT.No.2 10


VII. Review Questions

I. Encircle the Letter of the best answer.

1. The buoyant force theory was first discovered by __________.

a.) Galileob.) Newtonc.) Archimedesd.) Pascal

2. The buoyant force for a floating body is equivalent to _________.

a.) The weight of water in the container.b.) The volume of the body multiplied by the unit weight of water.c.) The weight of the body.

d.) The volume submerged multiplied by the unit weight of the body.

3. If a block of wood has a unit weight less than that of water which of the followingstatement is correct?

a.) The block will sink all the way down.b.) The block will be suspended somewhere under water.c.) Its behavior cannot be determined when in water.d.) Part of the block will be submerged and part will stay above water.

4. If a floater shall hold an anchorage, which of the following criteria should be true andprecise so that they would be suspended in a body of water?

a.) The weight of the floater should be less than the anchorage.b.) The combined unit weight of the floater and anchorage should be less than

that of waterc.) The unit weight of both should be less than that of waterd.) The volume of the floater should be greater than that of the anchorage.

5. How deep along its thickness will a box is immersed in water if its dimension is 2 m x1.5 m x 0.5 m and weighs 11,000 N?

a) 0.374 mb) 1.495 mc) 1.122 md) None of the above

6. What should be the weight of a cube with sides measuring 1 m if its draws 0.5m of oilof specific weight equal to 0.8 (It is submerged halfway in oil).


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EXPT.No.2 11


a) 7845.6 Nb) 3922.8 Nc) 1961.4 Nd) 4903.5 N

7. What should be the weight of a body measuring 2m in length x 1.5 m in width x 0.5 min thickness if an anchor weighing 15,000 N with volume equal to 0.5 cubic meter is justenough to submerge its entire thickness in water?

a) 5193 Nb) 289.5 Nc) 34,614 Nd) 4,614 N

8. What should be the weight of a boy if a barge (a floater) weighing 12,000 N with avolume of 2 cubic meter is just enough to carry him in water?

a) 2,193 Nb) 19,614 Nc) 7,614 Nd) None of these

9. What will happen if two balls are put in water with the following descriptions; Ball #1weighs 2,500 N with volume equal to 0.45 cubic meter, Ball #2 weighs 1,800 N withvolume equal to 0.15 cubic meter?

a) Both of the ball will sink in waterb) Ball #1 will float while ball #2 will sink in waterc) Both of the ball will float in waterd) Ball #2 will float while ball #1 will sink in water

10. What should be the volume of the barge so that it will draw salt water (s = 1.04)which is equal to its vertical height? The barge weighs 1,500 N.

a) 0.1529 cubic meterb) 0.1471 cubic meterc) 0.9807 cubic meterd) 0.9615 cubic meter

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