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Archimedesβ Principle
Purpose
The Purpose of this activity is to show some basic properties of the buoyant force. Namely that the
buoyant force is a function of the density of the fluid, that the buoyant force is equal to the weight of the
fluid displace by a submerged or floating object, and that the apparent weight of a submerged object is
the difference between its weight and the buoyant force acting on it.
Theory
Archimedes' Principle states that the upward buoyant force exerted on a body immersed in a fluid,
whether fully or partially submerged, is equal to the weight of the fluid that the body displaces; it is also
applicable to gases:
πΉ" = π%π
There are 2 ways to measure buoyancy: direct and displacement. Direct measurement is the difference
between the actual weight of the object (Wo) and its apparent weight (Wa) when fully submerged.
Displacement measurement utilizes the fact that the volume of fluid displaced (Vf) is equal to the volume
of the object (Vo) that is submerged. Recall that density (Ο) = m/V, such that:
πΉ" = 'π%π%*π = 'π%π+*π
From the free body diagram:
πΉ" β πΉ- = π'π%π+ β π+* = βπ+π
where, solving for acceleration we find
π = π /
π%
π+
β 11
where it can be observed that
π% > π+ β +π ππππππ‘ π€πππ πππππ‘
π% = π+ β π = 0 ππππππ‘ ππ ππ ππ¦πππππ πππ’πππππππ’π
π% < π+ β βπ ππππππ‘ π€πππ π πππ
If the object is only partially submerged, then the volume of fluid displaced is the volume of the
part of the object actually submerged. For example, a cylinder has a volume Vc = Οr2l = Acl, where Ac = Οr2
is the cross sectional area and l is the length. Assume the cylinder is submerged by a depth d, then Vsub =
Acd such that:
πΉ" = π%ππKLM = π%ππ΄Oπ
It is important to note that the buoyant force and depth of the object are directly related such
that buoyant force will increase with depth until the object is fully submerged.
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Procedure for Aluminum Block
1. In the yellow box near the top left of your screen change the material to Aluminum
a. Set the mass to 5.00 kg.
b. Record the Aluminum blockβs mass, and volume (in Liters) in Table 1.
2. On the right side, and center of your screen you will see a scale.
a. Click and drag the Aluminum block on the scale to weigh it.
b. Record its weight in Table 4.
3. At the bottom and center of your screen set the Fluid Density to Water (1.00 kg/L) if it isnβt
already.
a. Record the Fluid Density in table 4.
b. Record the initial volume, Vi, of the fluid (100.00 L) in Table 4.
4. Click and drag the Aluminum block and place it on the scale that is in the fluid near the bottom
right of the fluid βtankβ.
a. Record the new fluid volume, V, in Table 4.
b. The weight of the block in the fluid is called the apparent weight, WA. Record the blockβs
apparent weight in Table 4.
5. Click and drag the Aluminum block, and place it on βthe groundβ on the left side of your screen.
6. Change the Fluid Density to Honey (1.42 kg/L) and repeat process, entering your values in to
table 5.
7. Change the Fluid Density to Gasoline (0.70 kg/L) and repeat process, entering your values into
table 6.
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Analysis of Archimedesβ Principle Lab
Name______________________________________________
Course/Section_______________________________________
Instructor____________________________________________
Table 1
Wood Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Water Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Water (N)
Complete Table, show any calculations in the space provided (5 points)
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Table 2
Wood Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Honey Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Honey (N)
Complete Table, show any calculations in the space provided (5 points)
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Table 3
Wood Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Gasoline Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Gasoline (N)
Complete Table, show any calculations in the space provided (5 points)
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Table 4
Aluminum Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Apparent Weight (N)
Water Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Water (N)
Complete Table, show any calculations in the space provided (5 points)
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Table 5
Aluminum Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Apparent Weight (N)
Honey Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Honey (N)
Complete Table, show any calculations in the space provided (5 points)
10
Table 6
Aluminum Values
Mass (kg)
Weight (N)
Volume (L)
Volume (m3)
Density (kg/m3)
Apparent Weight (N)
Gasoline Values
Density (kg/L)
Density (kg/m3)
πP(πΏ)
πP(m3)
π (m3)
Ξπ (m3)
Weight of displaced Gasoline (N)
Complete Table, show any calculations in the space provided (5 points)
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1. Draw a free body diagram of the wood block floating in the water, or any of the fluids because
they will be identical, then write the force summation equation from that free body diagram.
(10 points)
2. Calculate the weight of fluid displaced by the wood block for all three fluids. (5 points)
3. Why are the weights of the three fluids displaced by the wood block more or less the same?
(5 points)
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4. What is the apparent weight of the wood block when it is floating in the three different fluids?
(10 points)
5. Draw a free body diagram of the Aluminum block for when it was submerged in the water, or
any of the fluids because they will be identical, and then write the force summation equation for
that free body diagram. (10 points)
6. From your force summation equation, calculate the buoyant force acting on the Aluminum block
for each of the three fluids. (5 points)
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7. Calculate the apparent weight of the Aluminum block in each of the three fluids. (10 points)
8. Why did the wood block float in the fluids while the Aluminum block sank? (5 points)

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