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By the end of this section, you will be able to:
  • Define buoyant force
  • State Archimedes’ principle
  • Describe the relationship between density and Archimedes’ principle

When placed in a fluid, some objects float due to a buoyant force. Where does this buoyant force come from? Why is it that some things float and others do not? Do objects that sink get any support at all from the fluid? Is your body buoyed by the atmosphere, or are only helium balloons affected ( [link] )?

Figure A is a drawing of a ship anchor submerged underwater next to some sea shrubs. Figure B is a photo of a floating submarine with a wake on 3 sides. Figure C is a photo of many colored balloons floating in air.
(a) Even objects that sink, like this anchor, are partly supported by water when submerged. (b) Submarines have adjustable density (ballast tanks) so that they may float or sink as desired. (c) Helium-filled balloons tug upward on their strings, demonstrating air’s buoyant effect. (credit b: modification of work by Allied Navy; credit c: modification of work by “Crystl”/Flickr)

Answers to all these questions, and many others, are based on the fact that pressure increases with depth in a fluid. This means that the upward force on the bottom of an object in a fluid is greater than the downward force on top of the object. There is an upward force, or buoyant force    , on any object in any fluid ( [link] ). If the buoyant force is greater than the object’s weight, the object rises to the surface and floats. If the buoyant force is less than the object’s weight, the object sinks. If the buoyant force equals the object’s weight, the object can remain suspended at its present depth. The buoyant force is always present, whether the object floats, sinks, or is suspended in a fluid.

Buoyant force

The buoyant force is the upward force on any object in any fluid.

Figure is a schematic drawing of the cylinder filled with fluid and opened to the atmosphere on one side. An imaginary object with the surface area A, that is smaller than the surface area of the cylinder, is submerged into the fluid. Distance between the top of the fluid and the top of the object is h1. Distance between the top of the fluid and the bottom of the object is h2. Forces F1 and F2 are applied to the top and the bottom of the object, respectively.
Pressure due to the weight of a fluid increases with depth because p = h p g . This change in pressure and associated upward force on the bottom of the cylinder are greater than the downward force on the top of the cylinder. The differences in the force results in the buoyant force F B . (Horizontal forces cancel.)

Archimedes’ principle

Just how large a force is buoyant force? To answer this question, think about what happens when a submerged object is removed from a fluid, as in [link] . If the object were not in the fluid, the space the object occupied would be filled by fluid having a weight w fl . This weight is supported by the surrounding fluid, so the buoyant force must equal w fl , the weight of the fluid displaced by the object.

Archimedes’ principle

The buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle    is

F B = w fl ,

where F B is the buoyant force and w fl is the weight of the fluid displaced by the object.

This principle is named after the Greek mathematician and inventor Archimedes (ca. 287–212 BCE), who stated this principle long before concepts of force were well established.

Figure A is a drawing of a person submerged in water. Force wobj is expressed by the person, force Fb is applied by the water to the person. Figure B is a drawing in which the person is replaced by water. Now Force wfl is expressed by the water that replaced the person, force Fb remains the same.
(a) An object submerged in a fluid experiences a buoyant force F B . If F B is greater than the weight of the object, the object rises. If F B is less than the weight of the object, the object sinks. (b) If the object is removed, it is replaced by fluid having weight w fl . Since this weight is supported by surrounding fluid, the buoyant force must equal the weight of the fluid displaced.

Questions & Answers

state Hooke's law of elasticity
Aarti Reply
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian Reply
please help me solve this 👆👆👆
Ian
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
how please
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
Duncan Reply
I do not get the question can you make it clearer
Ark
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Liteboho Reply
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
Duncan
what is the definition of model
matthew Reply
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
matthew
yes
Duncan
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
Baijnath Reply
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Roshani Reply
0
Keshav
what is soln..
Keshav
o
Duncan
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta) therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆) on moon, the only difference is the gravity. Gravity on moon = 0.166*g substituting that value in R, we get the new R
Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
Sharath
state Newton's first law of motion
Awal Reply
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
Emmanuel Reply
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result. V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr. the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
Kumar
103.077640640442km/h
Peter
state Newton's first law of motion
Kansiime Reply
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Govindsingh
how would you measure displacement in your car?
Grace Reply
what is constellation
Charles Reply
The product of a. (vector b× vector a)
Umesh Reply
I want to join the conversation
Kumaga Reply
ok
Kamal
Ok
Bishal
ok
Rohit
Two charges 1uc and 3uc are separated 4m apart. find the point on the line connecting them at which their electric field intensity balances each other
Chukwurah
hmmm
JMPSCL
a particle projected from origion moving on x-y plain passing through p&q point (9,7)(18,4)find the equation of trajectory
ali Reply
what is the equation of trajectory
Kumaga
Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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