In the last video, we calculated BF of the item as being W(in air) - W(in liquid). Isn't the W in liquid just 2N since the liquid weighed 100 N prior to inserting the block? Thus, wouldn't BF be 25-2?
A bucket of water is hanging from a spring balance. An iron piece is suspended into water without touching sides of bucket from another support. The spring balance reading will- a) Remain same b) Decrease c) Increase d) Increase/decrease depending on depth of immersion
Beautiful problem and very clearly explained. Thank you (: I do have a question that may sound silly - so the block counters the buoyant force by pushing back on the water with the same force *mg*. My question is about how this force gets transferred to the scale. It sems something like this : 1) block pushes the water down with a force *mg* 2) water pushes the bottom surface of the cup with the same force *mg* Since the molecules in water can move somewhat freely, my intuition tells me that the force applied by the block should end up in displacing the water molecules below the block. I'm not able to comprehend how the entire *mg* force gets transferred to the scale. Really appreciate if you/somebody clears this for me. Thanks !
Think about the concept of Newton's third law. The water pushes up against the block (BF). The water therefore pushes with an equal and opposite force against the bottom of the beaker. The beaker pushes with an equal force against the scale. The scale pushes back with an equal and opposite force.
Thank you for resoponding so fast :) I think I get the Newton's third law. This still doesn't answer my question... Here, water is between the block and bottom of the beaker, not some solid stick. If it were a solid stick, then it is easy to see how the force transfers *block->stick->beaker bottom*. I'm still finding it hard to visualize how *water* behaves like a stick in transferring the force... (I'm assuming that the *mg* force acts only over the beaker surface exactly below the block)
Thanks for the video! I have a question though. Shouldn't you have measured the cup with oil rather than the cup with water? The cup with oil would have a different weight than cup with water.
@@MichelvanBiezen Why did you use the weight of the cup with water (100N) as the baseline? If it was oil with the same volume, the weight would be lets say 80N for example. So it would be 102-80N instead of 102-100N, so wouldnt they have different bouyance forces depending on what cup and liquid you weighed? Sorry, i am confused.
@@MichelvanBiezen Ah okay. In the beginning of the video, you had stated that it was cup with water rather than cup with oil that equals the 100 N. Thank you again Professor!
Thanks for the video. I have one question though - how is this method of finding density advantageous to that of just finding out a sample's mass and volume by direct measurement?
A cube has sides of length `, and density ρ2. You drop the cube in a fluid with density ρ1, where ρ1 > ρ2. When the cube come to rest, with its top and bottom surfaces parallel to the surface of the fluid, it is submerged to a depth h. By first writing down all the surface forces and the body forces involved, show that h = L* ρ2/ρ1. Remember that air pressure also produces a force. I am stacking to proof that. can u help me ?
always start with the definition of the BF = weight of the displaced liquid = mg = d V g = d A h g (A = L^2) (d = density) Thus BF = P1 * A * h * g which is equal to the weight of the cube = mg = P2 V g = P2 A L g Set P1 A h g = P2 A L g and you get the answer.
Thank you professor with your great work... Can you add the following problem in this playlist, I see it quite special useful. A 1 kg beaker containing 2 kg of ethanol(or any fluid) rests on a scale. A 2 kg block of iron is suspended from a spring balance and completely submerged in the fluid. Determine the equilibrium readings on both scales.
Tshepo, Thanks for the input. I am adding physics videos to a number of topics that have limited coverage. I will add this when I get to Archimedes' principle.
Tshepo, You will need to find the BF value and add this value to the weight of the scale reading prior to the Iron being immersed (Original Scale Reading = 3g, see below). BF=rho(liquid)*g*v(liquid) and v(liquid) = v(iron) You have all the parameters needed except v(liquid). v(liquid) = v(iron). you need to find v(iron) weight (iron) = mass(iron) * g = rho(iron) * v(iron) * g => v(iron) = weight(iron) / ([ rho(iron) * g ]; you know rho(iron), weight, and gravity now that you know v(iron), substitute this into the BF equation above. BF = rho(liquid) * g * v(iron) new scale reading = Original Scale Reading + BF = (mass of beaker + mass of liquid) * g + BF = 3g + BF
robert The video is correct as is. I am not sure what you are claiming, but the reading on the spring should read the w of the beaker + w of the fluid + BF.
Michel van Biezen Earlier, I was confused by appearant weight and BF. After re-reviewing the materials I think I got it. Thanks for pointing that out and I appreciate all of your videos!
sir i have a confusion the measuring instrument shows the weight 102 N i think .If it shows 102 N then the weight of cup+oil+object is 2 N .BECAUSE NO FLUID IS FLOWN OUT it shows actual weight OF ALL is 102 N WHICH IS IMPOSSIBLE .ACTUAL WEIGHT-APPARENT WEIGHT IS equal to bouncy force is equal to loss in apparent weight .thank u sir please help me
The cup and the fluid together weigh 100 N.. When you suspend the object in the fluid the reading increases 2 N which means that the buoancy force equals 2 N.
Great videos professor TY I have a question though... shouldnt the BF = weight of the object outside the oil - weight of the object inside the liquid? BF = 25 - 2 Ty very much
Carlos. There are different ways by which you can calculate the BF 1) BF= weight of the displaced liquid 2) BF = weight in air - weight in liquid = 25 N - 23 N = 2 N 3) BF = delta F on scale = 2 N
People are confused more because this teacher is using the drawing on the board to show the weights. Why not show the weight in air and when submerged by showing the ACTUAL setup or scale FOR CLARITY.
suppose i have a 2N weight and a pot 3N and 10N water then if i drop the weight into water in the pot then the total weight of the system will be (2+3+10) or (10+3+(2-buoyancy force)) ?
The object has a weight of 25 N. The cup and fluid together have a weight of 100 N. The apparent weight of the object when immersed in the fluid is 23 N.
Just one question.... how do we find the density of an unknown object using the water and scale and we're on the INTERNATIONAL SPACE STATION?.. no gravity!! .. (Just kidding Michel.. I saw a trend here where other posters were suggesting new Situations..) Now there's an interesting problem, how to find the Density of an object WITHOUT using Archimedes Principle.. YIKES!!
Suspend the mass from a spring with a known constant. Then pull the spring away from its equilibrium point and time the oscillation. (no gravity needed) That will give you the mass. Then you can determine the volume of the object (in a number of ways) and voila.... you have the density.
EXCELLENT reply Michel!! ha ha ha.. Very GOOD.... I'll mention you to NASA for their next Flight.. lol .. now I have to go and try out your suggestion with a spring.... Thank you!..
Since the buoyancy force (BF) equals the weight of the displaced water: BF = weight (w) of the person = density * volume * g mg = rho * V * g m = rho * V m = rho * A * h h = m / (rho * A) = 70 / (1000 * 2) = 0.035 m = 3.5 cm
In the last video, we calculated BF of the item as being W(in air) - W(in liquid). Isn't the W in liquid just 2N since the liquid weighed 100 N prior to inserting the block? Thus, wouldn't BF be 25-2?
A bucket of water is hanging from a spring balance. An iron piece is suspended into water without touching sides of bucket from another support. The spring balance reading will-
a) Remain same
b) Decrease
c) Increase
d) Increase/decrease depending on depth of immersion
It will increase by the amount of the buoyancy force ( = weight of the displaced water).
Thanks for quick reply.
Also, What if the water is filled in the bucket upto the brim? Will there be any changes in reading?
If we assume that an amount of water equal to the volume of the object overflows the bucket, then the reading will remain the same.
Got it sir 👍👍
Hey Michel, you so good !!! & keep on doing these videos, they save so many lives each seconds that u can't even realize thumps up !!!
this man is amazing
WOW, If I knew Physics was THIS EASY..... I would have become a Physicist... :D .... well done Michel !!
It's never too late! -Wife
Beautiful problem and very clearly explained. Thank you (:
I do have a question that may sound silly - so the block counters the buoyant force by pushing back on the water with the same force *mg*. My question is about how this force gets transferred to the scale. It sems something like this :
1) block pushes the water down with a force *mg*
2) water pushes the bottom surface of the cup with the same force *mg*
Since the molecules in water can move somewhat freely, my intuition tells me that the force applied by the block should end up in displacing the water molecules below the block. I'm not able to comprehend how the entire *mg* force gets transferred to the scale. Really appreciate if you/somebody clears this for me. Thanks !
Think about the concept of Newton's third law. The water pushes up against the block (BF). The water therefore pushes with an equal and opposite force against the bottom of the beaker. The beaker pushes with an equal force against the scale. The scale pushes back with an equal and opposite force.
Thank you for resoponding so fast :) I think I get the Newton's third law. This still doesn't answer my question... Here, water is between the block and bottom of the beaker, not some solid stick. If it were a solid stick, then it is easy to see how the force transfers *block->stick->beaker bottom*. I'm still finding it hard to visualize how *water* behaves like a stick in transferring the force... (I'm assuming that the *mg* force acts only over the beaker surface exactly below the block)
Radha great question. Now I'm wondering the same thing.....
Thank you so much for everything. I wish youwere my teacher. You are such an awesome person!
Mad respect to you. Awesome videos
Also its better to remember the submerged weight is always less than the free weight, taken in air b/c of buoyancy
Amazing videos!
Awesome videos!
Thanks go to my wife who puts in all the hard work behind the scenes.
one of the best
This absolutely amazing, i enjoyed the lessons.
Glad you enjoyed it!
Thanks for the video! I have a question though. Shouldn't you have measured the cup with oil rather than the cup with water? The cup with oil would have a different weight than cup with water.
We didn't need to know the weight of the cup and oil, since we only needed the buoyancy force to solve the problem.
@@MichelvanBiezen Why did you use the weight of the cup with water (100N) as the baseline? If it was oil with the same volume, the weight would be lets say 80N for example. So it would be 102-80N instead of 102-100N, so wouldnt they have different bouyance forces depending on what cup and liquid you weighed? Sorry, i am confused.
@@MichelvanBiezen Hi Professor. I just wanted to follow up on my last reply. Thank you!
In this example the 100 N represents the weight of the cup and the oil.
@@MichelvanBiezen Ah okay. In the beginning of the video, you had stated that it was cup with water rather than cup with oil that equals the 100 N. Thank you again Professor!
You are wonderful. Thank you
We are glad you like the videos. Thank you.
Thanks for the video. I have one question though - how is this method of finding density advantageous to that of just finding out a sample's mass and volume by direct measurement?
If the shape is irregular, using the displacement of water is the best method.
Michel van Biezen thank you! Almost feels like a celebrity answered me :D
Great content
A cube has sides of length `, and density ρ2. You drop the cube in a fluid with density ρ1, where ρ1 > ρ2. When the cube come to rest, with its top and bottom surfaces parallel to the surface of the fluid, it is submerged to a depth h. By first writing down all the surface forces and the body forces involved, show that h = L* ρ2/ρ1. Remember that air pressure also produces a force. I am stacking to proof that. can u help me ?
always start with the definition of the BF = weight of the displaced liquid = mg = d V g = d A h g (A = L^2) (d = density)
Thus BF = P1 * A * h * g which is equal to the weight of the cube = mg = P2 V g = P2 A L g
Set P1 A h g = P2 A L g and you get the answer.
Thanks so much. The way u explain is a lot better than my lecture.
Is there any other way that i can proof without using Archimedes principle ?
Thank you professor with your great work... Can you add the following problem in this playlist, I see it quite special useful.
A 1 kg beaker containing 2 kg of ethanol(or any fluid) rests on a scale. A 2 kg block of iron is suspended from a spring balance and completely submerged in the fluid.
Determine the equilibrium readings on both scales.
Tshepo,
Thanks for the input. I am adding physics videos to a number of topics that have limited coverage. I will add this when I get to Archimedes' principle.
Tshepo,
You will need to find the BF value and add this value to the weight of the scale reading prior to the Iron being immersed (Original Scale Reading = 3g, see below).
BF=rho(liquid)*g*v(liquid) and v(liquid) = v(iron)
You have all the parameters needed except v(liquid).
v(liquid) = v(iron). you need to find v(iron)
weight (iron) = mass(iron) * g
= rho(iron) * v(iron) * g
=> v(iron) = weight(iron) / ([ rho(iron) * g ]; you know rho(iron), weight, and gravity
now that you know v(iron), substitute this into the BF equation above.
BF = rho(liquid) * g * v(iron)
new scale reading = Original Scale Reading + BF
= (mass of beaker + mass of liquid) * g + BF
= 3g + BF
robert
The video is correct as is.
I am not sure what you are claiming,
but the reading on the spring should read the w of the beaker + w of the fluid + BF.
Michel van Biezen
Earlier, I was confused by appearant weight and BF. After re-reviewing the materials I think I got it. Thanks for pointing that out and I appreciate all of your videos!
sir i have a confusion the measuring instrument shows the weight 102 N i think .If it shows 102 N then the weight of cup+oil+object is 2 N .BECAUSE NO FLUID IS FLOWN OUT it shows actual weight OF ALL is 102 N WHICH IS IMPOSSIBLE .ACTUAL WEIGHT-APPARENT WEIGHT IS equal to bouncy force is equal to loss in apparent weight .thank u sir please help me
The cup and the fluid together weigh 100 N.. When you suspend the object in the fluid the reading increases 2 N which means that the buoancy force equals 2 N.
ok sir.
thanks a lot ,great video
Great videos professor TY
I have a question though... shouldnt the BF = weight of the object outside the oil - weight of the object inside the liquid? BF = 25 - 2
Ty very much
Carlos.
There are different ways by which you can calculate the BF
1) BF= weight of the displaced liquid
2) BF = weight in air - weight in liquid = 25 N - 23 N = 2 N
3) BF = delta F on scale = 2 N
i am confused here.. you are talking about Newton 2nd law!! , how Newton 2nd law applicable here ??? plz explain it
People are confused more because this teacher is using the drawing on the board to show the weights. Why not show the weight in air and when submerged by showing the ACTUAL setup or scale FOR CLARITY.
suppose i have a 2N weight and a pot 3N and 10N water then if i drop the weight into water in the pot then the total weight of the system will be (2+3+10) or (10+3+(2-buoyancy force)) ?
As measured by the scale on the bottom it will be 2 + 3 + 10
Thank u teacher :)
Thank you, sir
Most welcome
Thankyou very much
Most welcome
why is the cup 25 n on one scale and 100 n on another scale
The object has a weight of 25 N. The cup and fluid together have a weight of 100 N. The apparent weight of the object when immersed in the fluid is 23 N.
0:52 Newton's second law: For every action there is an equal and opposite reaction? False. That's Newton's third law.
You are correct.
Just one question.... how do we find the density of an unknown object using the water and scale and we're on the INTERNATIONAL SPACE STATION?.. no gravity!! .. (Just kidding Michel.. I saw a trend here where other posters were suggesting new Situations..) Now there's an interesting problem, how to find the Density of an object WITHOUT using Archimedes Principle.. YIKES!!
Suspend the mass from a spring with a known constant. Then pull the spring away from its equilibrium point and time the oscillation. (no gravity needed) That will give you the mass. Then you can determine the volume of the object (in a number of ways) and voila.... you have the density.
EXCELLENT reply Michel!! ha ha ha.. Very GOOD.... I'll mention you to NASA for their next Flight.. lol .. now I have to go and try out your suggestion with a spring.... Thank you!..
how to solve this, a 70 kg man jumps off a raft 2m square moored in a fresh water lake. by how much does the raft rise. ty
Since the buoyancy force (BF) equals the weight of the displaced water: BF = weight (w) of the person = density * volume * g mg = rho * V * g m = rho * V m = rho * A * h h = m / (rho * A) = 70 / (1000 * 2) = 0.035 m = 3.5 cm