Physics 33.5 Buoyancy Force (6 of 9) Apparent Weight of a Submerged Object

[@kexus4415]

great explanation...was a confusing topic to me until i saw your video👏

[@MichelvanBiezen]

Glad you found our videos! 🙂

[@tiNgk0y1]

I had nosebleed eating this knowledge yet its very fascinating..

[@MichelvanBiezen]

😂

[@tinashemukutu3597]

Thank you simplified well explained even after 5 years its very helpful in 2022 and also helpful in future generations....... Eng Mining 👷🏿‍♂️

[@MichelvanBiezen]

Glad it helped 🙂

[@DR.199]

I have a question let's assume we have a fish tank filled with water and we submerged air balloon in side it is the fish tank going to be less weight than before submerging the balloon? Please i need answer

[@MichelvanBiezen]

In order to submerge a balloon, you have to push it down (apply a force downward) or it will float on top of the water. That will cause the balloon to displace water equal to the balloon's volume, which in turn will cause a buoyancy force, which will cause the fish tank to appear heavier if it was sitting on a scale.

[@DR.199]

@@MichelvanBiezen I appreciate your reply, but what about if the force is tension, so the balloons submerged to the tank from the bottom of the tank with cables or any thing , doesn't the volume of displaced water reduces the weight of the tank since we replaced water with air? Sorry for my English, im trying my best 😅 i have a school project about an aquarium so your help is appreciate it

[@MichelvanBiezen]

In the case where the balloon is held below the water line by a cable attached to the bottom, there is no additional force placed (from above). But since the amount of water in the tank has not changed, the mass (and weight) of the tank will remain the same.

[@leonormerino1653]

what if you have the density, mass, and volume of the object and want to find the density of the liquid but you don't know its volume. You only have the terminal velocity of the object in that liquid.

[@MichelvanBiezen]

That may not be enough. The viscosity of the liquid and its temperature also play a role.

[@alexanderalbertogarciavand2830]

Imagine it is an open steel box which is 20 tonnes (outside water), what would be the weight of the same once is submerged and it is filled up with water? Depth 7 mtrs

[@MichelvanBiezen]

Let's say that the density of steel is 8000 kg/m^3 Then the volume = mass/ density = 20,000 kg / 8000 kg /m^3 = 2.5 m^3 Boyancy force = weight of water displaced = density x g x V = 1000 kg/m^3 x 9.8 m/sec^2 x 2.5 m^3 = 24,500 N Therefore 20,000 x 9.8 - 24500 = 196,000 N = 24,500 N = 171,500 N

[@koushikparameswaran1114]

Wow.. That was beautiful!

[@christianboeve281]

really great explenations btw, liked all + subscribed

[@finlaymattt]

simp

[@MJFinn]

@@finlaymattt Tf dude

[@annafoxtrot4368]

Suppose that the density and volume of the object are known. It's mass is the product of density times volume. So, if a portion of the object is submerged in a liquid with density of seawater, how would you determine the volume of displaced liquid? I'm sorry, I am confused. We really have a different equation to solve the weight of the displaced liquid and it seems it's simplified. Weight displacement = volume displacement x density of water. And to solve for volume displacement, we simply, multiply length, breadth and it's draft. BF is equal to the weight of displaced liquid right? Am I really on the right path here?

[@MichelvanBiezen]

You are correct.

[@gauravpandey7551]

thank youu sir but if appparent weight of submerge body equal to mg -buyant force + surface tension this means real case of liquid

[@MichelvanBiezen]

The forces of surface tension are negligible compared to the weight and buoyancy force.

[@gauravpandey7551]

@@MichelvanBiezen thank you sir

[@sivarishi9639]

Sir, u nailed it

[@MichelvanBiezen]

Great. Keep it going.

[@sudhibk9904]

Excellent explanation I loved the explanation and the derivation explained

[@MichelvanBiezen]

Glad it was helpful!

[@ashfakanonno3680]

@@MichelvanBiezen what is the buoyant force at the bottom of tank? or what is the normal force at the bottom of a tank

[@christianboeve281]

Cant seem to find a solution for this exercise: the density of the object (800 kg/m³) and u submerge it under a liquid (1200). the height of the object is 6cm, but now I need to find the acceleration when fully submerged. can you help with a basic equation? I'm stuck at: m(?)*a = m(liq)g - m(obj)g (currently my solution is BF = m*a > a = BF/m(obj) = 14,7 m/s²

[@christianboeve281]

pls help, im very annoyed that i cant seem to find the answer for sure :(

[@christianboeve281]

i keep thinking it has something to do with the height submerged in balance, which i found was 4 cm, so the travel distance is 2 cm if it helps? but i guess the volumes don't matter since its the same, so height isn't used. when i use Sum of F = [m(obj)*a]; a = 0,5 is this correct maybe? i urge to use m(obj-water)*a cause it makes more sense that this is the force exerted when going up, instead of the full mass maybe since its lighter (but now I'm typing this and it makes sense, as its the force of the density of the object, which is constant.

[@tsoojbaterdene7793]

Yeah it is like this video´s problem.But dice is not fully submerged, submerged 20% of its body.How to find dice density?In two cases: glass of water with no dice,glass of water with one. What will the change in normal force of the glass of water be?Dice weight is 40 grams,Sir.

[@MichelvanBiezen]

Yes, this is how to do such a problem. Always start with BF = weight of the displaced liquid = mg = density of liquid x V submerged x g and BF = weight of the object because it floats thus mg of object = density of liquid x V submerged x g or (40) (g) = (1 g/cm^3) ( 0.2 V) (g) g cancels and V of object is 200 cm^3 density = (40) / (200) = 0.2 g/cm^3

[@tsoojbaterdene7793]

Thank you so much.😍😍😍

[@tsoojbaterdene7793]

In this case Fg-BF=0 Mg=BF. Is it true Sir?

[@MichelvanBiezen]

Yes, that is correct

[@antoinegriezmann3476]

Thnx, helped me

[@MichelvanBiezen]

Glad you found our videos and that you found them helpful. 🙂

[@johnnym500]

wouldn't it be a negative Bf over the density of liquid and gravity. if you break the weight of each down it would be mg-mg-bf and the mg cancels out leaving negative Bf.

[@MichelvanBiezen]

The Buoyancy force always acts upwards.

[@sujoyghosh8612]

How can density of the liquid= mass of the liquid/volume of the"displaced liquid" Is it not the volume of whole liquid that is inside the container ?????

[@MichelvanBiezen]

It is the same thing, the result will be the same.

[@mubashir22ful]

(mass of liquid in that volume)/(THAT VOLUME) gives you mass per unit volume

[@alshababu5973]

For an object submerged in water, which force is the force we measure with the spring scale?

[@MichelvanBiezen]

The weight minus the buoyancy force.

[@alshababu5973]

Thank You

[@Dovlo.Jessica]

Nice teaching

[@MichelvanBiezen]

Thanks and welcome

[@beckwilde]

great video!

[@thisismyoutubechannel]

amazing! thank you so much!

[@tejas9008]

hello sir, I have confusion in part 1 you have written that the BF=Weight of the object, then how could we justify apparent weight there?

[@thefelixgan]

'BF = weight of object' only applies in the case when the object Is floating, in which case the weight of the object is completely supported by the buoyancy force from the water.

[@divyarajaram7845]

@@thefelixgan so is the apparent weight zero?

[@omjadhav9066]

Thank you,Love from india.

[@MichelvanBiezen]

You're Welcome!

[@mekuriawbiadg4108]

it is nice. it help me to do my assignment .thanks so much

[@abonehhawas8957]

That nice lecture to thank you

[@MichelvanBiezen]

You are very welcome

[@joycemabunda2803]

What if you were given the mass and volume as well as the density of the water, how do you find the tension of the string?

[@MichelvanBiezen]

The tension will always be (weight of object in air) - (buoyancy force). They may give you different parameters, but you'll always try to compute those two factors.

[@someguyslastname8487]

Thank you!

[@CosmicCraw7]

thanks

[@shubhamgupta4329]

Nice

[@mayankguliani6232]

sir parts 2 3 4 and 5 are missing so how to find

[@MichelvanBiezen]

You can find those in this playlist: PHYSICS 33.5 BUOYANCY FORCE

[@hafi4777]

thank you sir , your videos are amazing, i wanted to know what if the block is hollow inside? because i have been working on a fish feeding buoy which is hollow inside and hold a silo for fish bait. i need to calculate the waterline. please give me your guidance.

[@MichelvanBiezen]

Is the density of the block (with the cavity included) less than the density of water? If yes, then we have some videos that show you how to calculate the water level on the side. (like ice or wood that floats)

[@hafi4777]

@@MichelvanBiezen thank you so much for taking time out for replying the basic problem is how to calculate the volume, should i take the volume by subtracting the void and calculate it only for solid parts? the fundamental design is a buoy ( which is hollow inside) inside have batteries , control system and silo which could ultimately hold fish-feed and the hollow part is covered by lid . i have seen your series about buoyancy which is amazing and cleared my concepts, now i want know about calculation for hollow structure like the one i mentioned above.

[@tsoojbaterdene7793]

Is that help Sir,🤗🤗😄😄

[@MichelvanBiezen]

Yes, this is how to do such a problem. Always start with BF = weight of the displaced liquid = mg = density of liquid x V submerged x g and BF = weight of the object because it floats thus mg of object = density of liquid x V submerged x g or (40) (g) = (1 g/cm^3) ( 0.2 V) (g) g cancels and V of object is 200 cm^3 density = (40) / (200) = 0.2 g/cm^3

[@yeabg2210]

short and nicee

[@gamingwithhassan486]

I could not understand even a word of the formulae.

[@johnantoniogaming7260]

Blah blah blah Haha good

[@gamingwithhassan486]

He knows a lot but he does not know how to teach.