Bouyant Force Confusion [closed]
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I was asked to solve the question in the picture
I naturally thought the answer was (1) because bouyant force is caused by the pressure difference on the object's top and bottom surfaces.
Turns out that the answer is (3), can someone please explain me why is it (3), or is the book wrong in this case?
homework-and-exercises newtonian-mechanics forces fluid-statics buoyancy
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
$endgroup$
closed as off-topic by Aaron Stevens, Chair, ACuriousMind♦ Feb 16 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Chair, ACuriousMind
add a comment |
$begingroup$
I was asked to solve the question in the picture
I naturally thought the answer was (1) because bouyant force is caused by the pressure difference on the object's top and bottom surfaces.
Turns out that the answer is (3), can someone please explain me why is it (3), or is the book wrong in this case?
homework-and-exercises newtonian-mechanics forces fluid-statics buoyancy
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
$endgroup$
closed as off-topic by Aaron Stevens, Chair, ACuriousMind♦ Feb 16 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Chair, ACuriousMind
1
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44
add a comment |
$begingroup$
I was asked to solve the question in the picture
I naturally thought the answer was (1) because bouyant force is caused by the pressure difference on the object's top and bottom surfaces.
Turns out that the answer is (3), can someone please explain me why is it (3), or is the book wrong in this case?
homework-and-exercises newtonian-mechanics forces fluid-statics buoyancy
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
$endgroup$
I was asked to solve the question in the picture
I naturally thought the answer was (1) because bouyant force is caused by the pressure difference on the object's top and bottom surfaces.
Turns out that the answer is (3), can someone please explain me why is it (3), or is the book wrong in this case?
homework-and-exercises newtonian-mechanics forces fluid-statics buoyancy
homework-and-exercises newtonian-mechanics forces fluid-statics buoyancy
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
edited Feb 16 at 13:28
Qmechanic♦
106k121941217
106k121941217
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
asked Feb 16 at 12:09
Kosh RaiKosh Rai
526
526
closed as off-topic by Aaron Stevens, Chair, ACuriousMind♦ Feb 16 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Chair, ACuriousMind
closed as off-topic by Aaron Stevens, Chair, ACuriousMind♦ Feb 16 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Chair, ACuriousMind
1
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44
add a comment |
1
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44
1
1
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
1
1
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
1
1
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(3) is right. The water partially supports the solid ball, exerting an upwards force equal to the weight of water displaced, which is 40 g if we take the density of water as 1 g per cm^3. So you can think of it as a case where the left side of the balance received this amount of extra weight.
It you want to think it through more fully, then, as you correctly say, this force on the ball is caused by the pressure difference on the upper and lower surfaces of the immersed ball. But notice that there is also this same amount of force acting downwards on the water! That might seem like an odd way of putting it, but it is what Newton's third law (action and reaction) says. Since the water is not accelerating, the beaker must be providing this amount of extra upwards force at the bottom surface of the water, in order to support the water. Hence the answer to the question.
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
$endgroup$
add a comment |
$begingroup$
The water is producing an upward force on the ball (upthrust) equal to the weight of water displaced by the ball - Archimedes principle.
The ball must be exerting an equal in magnitude and opposite in direction force on the water - Newton’s third law.
The water is in static equilibrium and has two downward forces acting on it, its weight and the force due to the ball.
For the ball to static equilibrium it must have a net force of zero acting on it.
So there must be an upward force on the water due to the base of the beaker/balance equal in magnitude to the weight of the water and the downward force due to the ball.
Before the ball was added the beaker/balance exerted a force equal to the weight of the water.
With the ball immersed there is now an extra force exerted by the beaker/balance which is equal in magnitude to the weight of water displaced by the ball.
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
$endgroup$
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(3) is right. The water partially supports the solid ball, exerting an upwards force equal to the weight of water displaced, which is 40 g if we take the density of water as 1 g per cm^3. So you can think of it as a case where the left side of the balance received this amount of extra weight.
It you want to think it through more fully, then, as you correctly say, this force on the ball is caused by the pressure difference on the upper and lower surfaces of the immersed ball. But notice that there is also this same amount of force acting downwards on the water! That might seem like an odd way of putting it, but it is what Newton's third law (action and reaction) says. Since the water is not accelerating, the beaker must be providing this amount of extra upwards force at the bottom surface of the water, in order to support the water. Hence the answer to the question.
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
$endgroup$
add a comment |
$begingroup$
(3) is right. The water partially supports the solid ball, exerting an upwards force equal to the weight of water displaced, which is 40 g if we take the density of water as 1 g per cm^3. So you can think of it as a case where the left side of the balance received this amount of extra weight.
It you want to think it through more fully, then, as you correctly say, this force on the ball is caused by the pressure difference on the upper and lower surfaces of the immersed ball. But notice that there is also this same amount of force acting downwards on the water! That might seem like an odd way of putting it, but it is what Newton's third law (action and reaction) says. Since the water is not accelerating, the beaker must be providing this amount of extra upwards force at the bottom surface of the water, in order to support the water. Hence the answer to the question.
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
$endgroup$
add a comment |
$begingroup$
(3) is right. The water partially supports the solid ball, exerting an upwards force equal to the weight of water displaced, which is 40 g if we take the density of water as 1 g per cm^3. So you can think of it as a case where the left side of the balance received this amount of extra weight.
It you want to think it through more fully, then, as you correctly say, this force on the ball is caused by the pressure difference on the upper and lower surfaces of the immersed ball. But notice that there is also this same amount of force acting downwards on the water! That might seem like an odd way of putting it, but it is what Newton's third law (action and reaction) says. Since the water is not accelerating, the beaker must be providing this amount of extra upwards force at the bottom surface of the water, in order to support the water. Hence the answer to the question.
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
$endgroup$
(3) is right. The water partially supports the solid ball, exerting an upwards force equal to the weight of water displaced, which is 40 g if we take the density of water as 1 g per cm^3. So you can think of it as a case where the left side of the balance received this amount of extra weight.
It you want to think it through more fully, then, as you correctly say, this force on the ball is caused by the pressure difference on the upper and lower surfaces of the immersed ball. But notice that there is also this same amount of force acting downwards on the water! That might seem like an odd way of putting it, but it is what Newton's third law (action and reaction) says. Since the water is not accelerating, the beaker must be providing this amount of extra upwards force at the bottom surface of the water, in order to support the water. Hence the answer to the question.
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
edited Feb 16 at 12:53
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
answered Feb 16 at 12:29
Andrew SteaneAndrew Steane
5,269735
5,269735
add a comment |
add a comment |
$begingroup$
The water is producing an upward force on the ball (upthrust) equal to the weight of water displaced by the ball - Archimedes principle.
The ball must be exerting an equal in magnitude and opposite in direction force on the water - Newton’s third law.
The water is in static equilibrium and has two downward forces acting on it, its weight and the force due to the ball.
For the ball to static equilibrium it must have a net force of zero acting on it.
So there must be an upward force on the water due to the base of the beaker/balance equal in magnitude to the weight of the water and the downward force due to the ball.
Before the ball was added the beaker/balance exerted a force equal to the weight of the water.
With the ball immersed there is now an extra force exerted by the beaker/balance which is equal in magnitude to the weight of water displaced by the ball.
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
$endgroup$
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
add a comment |
$begingroup$
The water is producing an upward force on the ball (upthrust) equal to the weight of water displaced by the ball - Archimedes principle.
The ball must be exerting an equal in magnitude and opposite in direction force on the water - Newton’s third law.
The water is in static equilibrium and has two downward forces acting on it, its weight and the force due to the ball.
For the ball to static equilibrium it must have a net force of zero acting on it.
So there must be an upward force on the water due to the base of the beaker/balance equal in magnitude to the weight of the water and the downward force due to the ball.
Before the ball was added the beaker/balance exerted a force equal to the weight of the water.
With the ball immersed there is now an extra force exerted by the beaker/balance which is equal in magnitude to the weight of water displaced by the ball.
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
$endgroup$
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
add a comment |
$begingroup$
The water is producing an upward force on the ball (upthrust) equal to the weight of water displaced by the ball - Archimedes principle.
The ball must be exerting an equal in magnitude and opposite in direction force on the water - Newton’s third law.
The water is in static equilibrium and has two downward forces acting on it, its weight and the force due to the ball.
For the ball to static equilibrium it must have a net force of zero acting on it.
So there must be an upward force on the water due to the base of the beaker/balance equal in magnitude to the weight of the water and the downward force due to the ball.
Before the ball was added the beaker/balance exerted a force equal to the weight of the water.
With the ball immersed there is now an extra force exerted by the beaker/balance which is equal in magnitude to the weight of water displaced by the ball.
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
$endgroup$
The water is producing an upward force on the ball (upthrust) equal to the weight of water displaced by the ball - Archimedes principle.
The ball must be exerting an equal in magnitude and opposite in direction force on the water - Newton’s third law.
The water is in static equilibrium and has two downward forces acting on it, its weight and the force due to the ball.
For the ball to static equilibrium it must have a net force of zero acting on it.
So there must be an upward force on the water due to the base of the beaker/balance equal in magnitude to the weight of the water and the downward force due to the ball.
Before the ball was added the beaker/balance exerted a force equal to the weight of the water.
With the ball immersed there is now an extra force exerted by the beaker/balance which is equal in magnitude to the weight of water displaced by the ball.
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
answered Feb 16 at 12:29
FarcherFarcher
50.7k338105
50.7k338105
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
add a comment |
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
$begingroup$
All I was confused about was ball exerting the downward force. Thanks for your clarification.
$endgroup$
– Kosh Rai
Feb 16 at 12:40
add a comment |
1
$begingroup$
Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem?
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
Also, please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead.
$endgroup$
– Chair
Feb 16 at 14:15
1
$begingroup$
While it would be possible to edit this question to make it on-topic, the on-topic version would certainly be a duplicate, e.g. of physics.stackexchange.com/q/62864/50583
$endgroup$
– ACuriousMind♦
Feb 16 at 14:44