Fluid Statics (1)
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Title:
Fluid Statics
Group 33:
Van Tran, Elyse Gallagher, Shrenik Patel
Experiment Date:
October 31st, 2023
Goals:
In today’s lab, we will understand how to apply static equilibrium concepts to the buoyant force
and hoe pressure changes with depth in a fluid. We will also be able to find the density of an
unknown object by measuring its specific gravity.
Procedure:
Part I:
●
Suspend the brass cylinder from the force sensor and record the value as the weight of the
brass in air
●
Suspend the brass in the water and record its weight in water
●
Use your equation and the measured values to calculate the buoyant force on the brass
●
Calculate the specific gravity and density of brass from your measurements
●
Hang the brass below the wood so the wood will sink in water
●
Measure and record the weight in the air and in water, then calculate the total buoyant
force acting on both objects. Then use the known buoyant force on the brass to find the
buoyant force on the wood
●
Use equation 3 to calculate the specific gravity of the wood
●
Tabulate and report the buoyant force, specific gravity in water, and density for all the
materials you studied in this experiment
Part II:
●
Connect the absolute pressure sensor to the computer. Check the model of your pressure
sensor, open the hardware setup menu, then locate and select it from the menu
●
Fill the 1000 mL graduated cylinder with water, then slowly lower the tubing and record
the absolute pressure as a function of the depth at 5 cm intervals. Record your data in an
excel table.
●
Make a scatterplot of the pressure vs. depth with pressure on the y-axis
Precautions and Sources of Error
:
Sources of error can be human error. There was a lot of submerging and taking our best guest
when it was submerged enough. We also had to interpret our graph as best as possible.
Data:
Brass W
air
=
1.88 N
Brass W
water
= 1.67 N
W
air wood
= 2.25 N
W
water wood
= 0.65 N
W
air
wood = 0.33 N
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BF
brass
= 1.88 -1.67 = 0.21N
BF
wood
= 1.60 - 0.21 = 1.39 N
p
object
=p
F
(W
object
/F
B
)
Density of Brass: 1*1.88 N/0.21 N = 8.95 g/cm^3
Density of Wood: 0.33 N/ 0.39 = 0.85 g/cm^3
Questions:
Question 1.
Use the expressions for W
in air
and W
in water
from the figure above to show that the
buoyant force is equal to the difference between the weight in air and the weight in water. Show
your work.
F
T, air
= mg
vs
F
T, water
= mg - F
B
F
B
= W
in air
- W
in water
Question 2.
Show that one can find the density of the unknown object by dividing its weight by
the buoyant force: W
object
/F
B
=p
object
/p
fluid
To do this, use the following substitutions
W
object
= p
object
Vobject g
F
B
= W
fluid
= p
fluid
V
fluid
g
V
object
=V
fluid
. (if the object is fully submerged)
Show your work.
p
object
=p
F
(W
object
/F
B
)
p
object
=W
object
(p
Ft
*V
object
)
Question 3.
Do you think the metal is gold? To answer this, use a reliable online resource to look
up the density of gold and other common metals. If you don’t think your metal is gold, what is it
likely to be? Use your data to support your answer.
Density of Brass: 1*1.88 N/0.21 N = 8.95 g/cm^3
No, I do not think the metal is gold. The metal is not gold as the density of gold is 19.3
g/cm^3 and the density of our metal is not close to that value. The metal is likely to be
brass. Since the density of brass is 8.73 g/cm³, we got 8.95 g/cm³. Our value was closer to
the value of brass.
Question 4.
The total buoyant force acting on both objects is simply the sum of the buoyant
force on each individual object: F
B
,
total
= F
B
,
wood
+ F
B
,
slug
. Explain why this is true using the fact
that the buoyant force is due to the volume of the fluid displaced.
This is true because of the Archimedes Principle. The total buoyant force (FB, total)
acting on both objects, the wood and the slug, is the sum of the buoyant forces acting on
each individual object, FB, wood and FB, slug. This is true because the buoyant force is
due to the volume of fluid displaced by each object. Now, when we consider two objects
immersed in the same fluid, such as the wood and the slug, the total buoyant force on
both objects is the combined effect of each object displacing its own volume of fluid. So,
we can write the total buoyant force as:
FB, total = FB, wood + FB, slug
This is because each object contributes its own buoyant force based on the volume of
fluid it displaces. When you sum these two buoyant forces together, you get the total
buoyant force acting on both objects in the fluid. The principle behind this is that the
buoyant force depends on the volume of fluid displaced by each object. So, when
multiple objects are in the same fluid, you can calculate the total buoyant force by adding
up the individual buoyant forces, which are determined by the volumes of fluid displaced
by each object.
Question 5.
Is the value of the density of the wood you found greater than or less than that of
water? When fully submerged, how does the buoyant force acting on the wood compare to the
gravitational force acting on the wood? Does this explain why it floats to the surface when not
weighted down?
Density of Wood: 0.33 N/ 0.39 = 0.85 g/cm^3
The density of the wood is less than the density of water from our calculations. When an
object is fully submerged in a fluid, the buoyant force acting on the object is equal in
magnitude to the gravitational force acting on the object. The buoyant force counteracts
the gravitational force, and the object doesn't experience a net force causing it to sink or
rise; it remains in a state of equilibrium within the fluid. This upward buoyant force will
cause the object to rise to the surface of the fluid until it reaches a point where the forces
are balanced, and the object will then float. So, when the piece of wood is not weighted
down, its average density is less than that of the surrounding water, and this difference in
density results in the buoyant force pushing the wood upward and causing it to float to
the surface.
Question 6.
How does the pressure vary with depth? Support your claims with your results.
The deeper you are in a liquid the more pressure is above you because of the weight of
the liquid. After looking at our data from part 2 it is proven the deeper you are the more
pressure there is.
Question 7.
Look up the hydrostatic pressure equation. According to the equation, what physical
quantities are represented by the slope and y-intercept of the plot? What should the value of the
slope and y-intercept be?
P=P
0
+pgh, The y intercept is atmospheric pressure, the slope is the density. The value of
the slope is 0.0863 g/cm^3 and the y-intercept is 102.3 kPa.
Discussion:
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In this lab we observed buoyant force and the relation it has with objects. We found that the
buoyant force is equal to the difference of W
in air
and W
in water
. We were able to use the different
equations to find the density of
brass and wood. Our density of wood was found to be
8.95
g/cm
3
and wood to be 0.85 g/cm
3
. Our density of the slug showed that it was made of brass and
not gold since the density was not close to the gold. We found that pressure and depth are
directly related to one another. According to our graph as the depth increases, the pressure
increases as well. Some sources of error can include human error where we had to use our best
judgment when submerging the objects in water. Overall the lab allowed us to better understand
the concept of fluid dynamics.
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