(Original post by lawlieto)
The following question is from an American physics textbook:
A glass ball of radius 2 cm sits at the bottom of a container of milk that has a density of 1.03 g/cm^3. The normal force on the ball from the container's lower surface has magnitude 9.48*10^2 N. What is the mass of the ball?
Initially I thought normal force = weight because it sits at the bottom but the mark scheme says:
buoyant force + normal force = weight, ie:
FN + ρmilk g V − mglass g = 0 (copied from mark scheme)
This is why I asked the above and it made me confused, so I wondered if it's a sphere, then there is buoyant force because the sphere touches the bottom only at 1 point and the milk is actually exerting pressure on the rest of the sphere from the bottom of it?
Fluid exerts buoyant force on all floating and sinking objects whose magnitude is given by Volume of object inside water X density of fluid X g. If you draw a freebody diagram then you have force acting upwards = Normal force + Buoyant force and force acting downwards = weight of glass.
As the glass is stationary if you equate upward forces to downward force(s) then you get the required result. The mathematical bit is very simple but the conceptual bit:
1) Fluid exerts buoyant force on any floating or sinking object whose magnitude is equal to the weight of the fluid displaced. (Google Archimedes's Principle).
2) The volume of fluid displaced is equal to the volume of object immersed (in this case volume of glass ball). So upthrust (buoyant force) = density of the fluid X mass of the ball.
I was referring to container before not object immersed inside fluid. Buoyant force acts on object inside fluid not on walls of the container. I am sorry if this added confusions.
Note: You don't care about the surface area of contact. The buoyant force is due to fluid not due to base of the container.
Feel free to ask if you have any questions. I will try my best to answer all of those.
Oh and I forgot to answer the following. just edited :d
Normal force is the reaction force due to ground/surface/base (Newton's third law). This force is not equal to weight here because not all weight acts on the base. Some of the weight is already counterbalanced by the buoyant force. So, (weight  buoyant) force acts at the bottom of the container and equal magnitude of force (normal force) acts from ground to the object.
Hope this clarifies.
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tangotangopapa2
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Last edited by tangotangopapa2; 09082016 at 18:43. 
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(Original post by lawlieto)
It makes sense.. thanks
A quick question: is there a buoyant force on a rectangular block if it's at the bottom of a container that contains some fluid?
Yes, there is buoyant force on any object that is either floating or sinking. I heartily apologise for any inconvenience this might have caused. I am having hard time concentrating today.
Posted from TSR MobileLast edited by tangotangopapa2; 09082016 at 21:39. 
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 09082016 22:53
Soft Question. Are BPHO round 1 papers set to be solved using calculators? It says nothing about it in the question paper or instructions sheet.
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 10082016 12:09
(Original post by tangotangopapa2)
Fluid exerts buoyant force on all floating and sinking objects whose magnitude is given by Volume of object inside water X density of fluid X g. If you draw a freebody diagram then you have force acting upwards = Normal force + Buoyant force and force acting downwards = weight of glass.
As the glass is stationary if you equate upward forces to downward force(s) then you get the required result. The mathematical bit is very simple but the conceptual bit:
1) Fluid exerts buoyant force on any floating or sinking object whose magnitude is equal to the weight of the fluid displaced. (Google Archimedes's Principle).
2) The volume of fluid displaced is equal to the volume of object immersed (in this case volume of glass ball). So upthrust (buoyant force) = density of the fluid X mass of the ball.
I was referring to container before not object immersed inside fluid. Buoyant force acts on object inside fluid not on walls of the container. I am sorry if this added confusions.
Note: You don't care about the surface area of contact. The buoyant force is due to fluid not due to base of the container.
Feel free to ask if you have any questions. I will try my best to answer all of those.
Oh and I forgot to answer the following. just edited :d
Normal force is the reaction force due to ground/surface/base (Newton's third law). This force is not equal to weight here because not all weight acts on the base. Some of the weight is already counterbalanced by the buoyant force. So, (weight  buoyant) force acts at the bottom of the container and equal magnitude of force (normal force) acts from ground to the object.
Hope this clarifies.
Buoyant force is due to the fact that pressure at the bottom of the object is higher than on top, as hydrostatic pressure depends on depth. If the rectangular block is at the bottom of the container and there is no fluid below it, the fluid doesn't exert pressure on the rectangular object from the bottom of the object, so I thought there would be no buoyant force on a perfect rectangular object if it's at the bottom of a container that contains a fluid? Whereas if it's a sphere, only 1 tiny point of the sphere is touching the bottom, so there is fluid pushing the sphere from the bottom, so there will be a buoyant force. 
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 10082016 13:15
(Original post by lawlieto)
But let's say, you have a rectangular object at the bottom of the container that contains the fluid (so the rectangular object is immersed completely)
Buoyant force is due to the fact that pressure at the bottom of the object is higher than on top, as hydrostatic pressure depends on depth. If the rectangular block is at the bottom of the container and there is no fluid below it, the fluid doesn't exert pressure on the rectangular object from the bottom of the object, so I thought there would be no buoyant force on a perfect rectangular object if it's at the bottom of a container that contains a fluid? Whereas if it's a sphere, only 1 tiny point of the sphere is touching the bottom, so there is fluid pushing the sphere from the bottom, so there will be a buoyant force.
1) Even if object is sinking at the bottom, there are fluid molecules below it. (This is always the case in real life. Even smoothest surfaces known like ice are not smooth in a molecular level.)
Side note: If the object molecules and base molecules are so tight that no fluid molecules can pass below them, then we say the base and object are attached and are merely parts of each other.
2) Fluid molecules exert cohesive force on each other and tend to be smooth and continuous.
3) Fluid pressure is always constant at a given depth.
Based on these assumptions, we use hydrostatic laws regardless of the shape of the object. 
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 10082016 17:06
(Original post by tangotangopapa2)
Soft Question. Are BPHO round 1 papers set to be solved using calculators? It says nothing about it in the question paper or instructions sheet.
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 10082016 21:59
(Original post by lawlieto)
But let's say, you have a rectangular object at the bottom of the container that contains the fluid (so the rectangular object is immersed completely)
Buoyant force is due to the fact that pressure at the bottom of the object is higher than on top, as hydrostatic pressure depends on depth. If the rectangular block is at the bottom of the container and there is no fluid below it, the fluid doesn't exert pressure on the rectangular object from the bottom of the object, so I thought there would be no buoyant force on a perfect rectangular object if it's at the bottom of a container that contains a fluid? Whereas if it's a sphere, only 1 tiny point of the sphere is touching the bottom, so there is fluid pushing the sphere from the bottom, so there will be a buoyant force.(Original post by tangotangopapa2)
Yes, you are correct about it. But even if we talk about sinking object we always make the following assumptions:
1) Even if object is sinking at the bottom, there are fluid molecules below it. (This is always the case in real life. Even smoothest surfaces known like ice are not smooth in a molecular level.)
Side note: If the object molecules and base molecules are so tight that no fluid molecules can pass below them, then we say the base and object are attached and are merely parts of each other.
2) Fluid molecules exert cohesive force on each other and tend to be smooth and continuous.
3) Fluid pressure is always constant at a given depth.
Based on these assumptions, we use hydrostatic laws regardless of the shape of the object.
Apparatus: Water container with smooth base filled with water, wooden or plastic object with flat base (any object that has density less than that of water).
Procedure:
Take empty container and place flatbased object at the center so that two smooth surfaces touch each other.
Very gently pour water from one side without disturbing the test object. Keep on adding until you have sufficiently enough water.
Result and Conclusion: If no water squeezes between the surface then there should be no buoyant force acting on it and the object should not rise otherwise there is buoyant force even on the sinking objects.
You could perform this experiment and find that the object always rises no matter how gently you pour water into the container.
There is still a buoyant force when the object is at rest on the bottom. The argument is even though it may seem that there is no water between the object and the floor, there actually is. Thus, there is still a buoyant force, and it's equal to the weight of the fluid the object displaces.
I know it seems strange, but you have to imagine the fluid being able to squeeze in between the object and the floor. 
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 10082016 22:15
(Original post by tangotangopapa2)
Let us perform an experiment.
Apparatus: Water container with smooth base filled with water, wooden or plastic object with flat base (any object that has density less than that of water).
Procedure:
Take empty container and place flatbased object at the center so that two smooth surfaces touch each other.
Very gently pour water from one side without disturbing the test object. Keep on adding until you have sufficiently enough water.
Result and Conclusion: If no water squeezes between the surface then there should be no buoyant force acting on it and the object should not rise otherwise there is buoyant force even on the sinking objects.
You could perform this experiment and find that the object always rises no matter how gently you pour water into the container.
There is still a buoyant force when the object is at rest on the bottom. The argument is even though it may seem that there is no water between the object and the floor, there actually is. Thus, there is still a buoyant force, and it's equal to the weight of the fluid the object displaces.
I know it seems strange, but you have to imagine the fluid being able to squeeze in between the object and the floor. 
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 12082016 18:10
As usual I'm perplexed about something
Gas is contained in a tank at a pressure of 10 atm and a temperature of 15C. If one halfof the gas is withdrawn and the temperature is raised to 65C, what is the new pressurein the tank?
I thought if one half of the gas is withdrawn then the new volume will be half the original volume, and you can use the gas laws combined, ie: p1V1/T1 = p2V2/T2, where V2 will be 0.5V1. However, the mark scheme uses the ideal gas equation in a funny way and gives a very different answer what you would get from my method.
Here's the markscheme, it's the very first question: http://www.physics.ox.ac.uk/olympiad...r2_2011_MS.pdf
Thanks 
rohan.nuck
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 12082016 21:49
(Original post by lawlieto)
As usual I'm perplexed about something
Gas is contained in a tank at a pressure of 10 atm and a temperature of 15C. If one halfof the gas is withdrawn and the temperature is raised to 65C, what is the new pressurein the tank?
I thought if one half of the gas is withdrawn then the new volume will be half the original volume, and you can use the gas laws combined, ie: p1V1/T1 = p2V2/T2, where V2 will be 0.5V1. However, the mark scheme uses the ideal gas equation in a funny way and gives a very different answer what you would get from my method.
Here's the markscheme, it's the very first question: http://www.physics.ox.ac.uk/olympiad...r2_2011_MS.pdf
Thanks 
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 12082016 23:10
(Original post by rohan.nuck)
P1V1/T1 is only applicable when there is the same number of MOLES OF GAS because PV/T=nR which is constant only when n is constant. Now in the question it says half the gas is removed, this implies that HALF THE NUMBER OF MOLES OF THE GAS is removed not half the volume since the pressure will change. Ie, half of a gas is only half of*its volume if the pressure and temperature remain constant. 
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 12082016 23:15
(Original post by lawlieto)
But if you remove half the amount of gas, does the volume remain the same? Because in the mark scheme V remains V. 
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 12082016 23:32
(Original post by tangotangopapa2)
Volume of gas is the volume of the container, so it remains same. Note that gas can ideally occupy any volume of the container. 
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 13082016 10:00
(Original post by lawlieto)
In eqn 1) it substituted "10" for pressure. but pressure is in atm, shouldn't it be converted to Pa? 
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 13082016 13:05
(Original post by tangotangopapa2)
Note the equation PV=nRT. We get: PV/nT = R = a constant. If we equate (PV/nT)1 = (PV/nT)2 we see that, as long as our units are consistent and proportional to the SI units, the equation remains same. As atm is proportional to Pa, we can use this unit here. Same thing would apply if unit of volume was not SI. But we can't use degree Celsius as unit temperature as it is not proportional to Kelvin. 
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 14082016 23:35
(Original post by lawlieto)
Suddenly it all makes sense. Thanks 
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 21082016 23:29
Question Paper 2015: https://www2.physics.ox.ac.uk/sites/..._pdf_90611.pdf
Solutions: http://www.physicsandmathstutor.com/pat/solutions2015/
Question. 6.
An unbiased coin is tossed 3 times. Each toss results in a “head” or “tail”.What is the probability(a) of two or more tails in succession?(b) that two consecutive toss results are the same?(c) that if any one of the toss results is known to be a tail, that all of thetosses resulted in tails?
In part b, in the solution only HHT HTT TTH and THH are considered. Should we not consider HHH and TTT as in those cases too, two consecutive toss results are the same?
In part c, we know that one of the result is tail and we need remaining two toss to be tail. So, probability = 1/2 times 1/2 = 1/4. But result uses conditional probability formula (which I don't understand ) and gets result 1/7.
Thanks in advance 
hellomynameisr
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 22082016 00:22
(Original post by tangotangopapa2)
Question Paper 2015: https://www2.physics.ox.ac.uk/sites/..._pdf_90611.pdf
Solutions: http://www.physicsandmathstutor.com/pat/solutions2015/
Question. 6.
An unbiased coin is tossed 3 times. Each toss results in a “head” or “tail”.What is the probability(a) of two or more tails in succession?(b) that two consecutive toss results are the same?(c) that if any one of the toss results is known to be a tail, that all of thetosses resulted in tails?
In part b, in the solution only HHT HTT TTH and THH are considered. Should we not consider HHH and TTT as in those cases too, two consecutive toss results are the same?
In part c, we know that one of the result is tail and we need remaining two toss to be tail. So, probability = 1/2 times 1/2 = 1/4. But result uses conditional probability formula (which I don't understand ) and gets result 1/7.
Thanks in advance
But for part c, he uses a conditional probability formula since it is given that a tail has already been seen. Almost always when it is stated like this, you must use this formula. 
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 22082016 00:24
Does anyone know what section of Physics the questions related to calulcating/estimating the radius (or other distances) of planets/stars?

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 22082016 01:01
(Original post by tangotangopapa2)
Question Paper 2015: https://www2.physics.ox.ac.uk/sites/..._pdf_90611.pdf
Solutions: http://www.physicsandmathstutor.com/pat/solutions2015/
Question. 6.
An unbiased coin is tossed 3 times. Each toss results in a “head” or “tail”.What is the probability(a) of two or more tails in succession?(b) that two consecutive toss results are the same?(c) that if any one of the toss results is known to be a tail, that all of thetosses resulted in tails?
In part b, in the solution only HHT HTT TTH and THH are considered. Should we not consider HHH and TTT as in those cases too, two consecutive toss results are the same?
In part c, we know that one of the result is tail and we need remaining two toss to be tail. So, probability = 1/2 times 1/2 = 1/4. But result uses conditional probability formula (which I don't understand ) and gets result 1/7.
Thanks in advance
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