# Buoyancy question

## Homework Statement

Block A in the figure hangs from a spring balance D and is submerged in a liquid C contained in a beaker B. The weight of beaker is 1 kg, and the weight of the liquid is 1.5kg. The balance D reads 2.5 kg and balance E reads 7.5 kg. The volume of the block is 0.003 m^3.

What is the density of the liquid?
Ans: 1667.7kg/m^3

## The Attempt at a Solution

I know how to do the problem, and I got the right answer, but I don't understand what's going on.
So here's what I did,
Reading of E = Weight of container + Weight of liquid + Force of Buoyancy
75 = 25 + (0.003)(10)P
P= 1667.7

I don't get why reading of E isn't just (Weight of container + Weight of liquid + Weight of block A)? Why are we considering the buoyancy? If you say that the liquid exerts a buoyant force on the block, and due to Newton's third law, the block also exerts a buoyant force on the liquid (and so we consider buoyancy), aren't both these forces internal (and action- reaction pairs), so the net effect of buoyant force is zero?

#### Attachments

• IMG_20150123_101849.jpg
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Nathanael
Homework Helper
I don't get why reading of E isn't just (Weight of container + Weight of liquid + Weight of block A)?
Some of the weight of block A is being supported by the spring so that won't work.

Why are we considering the buoyancy? If you say that the liquid exerts a buoyant force on the block, and due to Newton's third law, the block also exerts a buoyant force on the liquid (and so we consider buoyancy), aren't both these forces internal (and action- reaction pairs), so the net effect of buoyant force is zero?
The net effect of the buoyant force on the entire system is zero, but the buoyant force "redistributes" the weight that is supported by D and E. The buoyant force lightens the load on the spring (the spring only has to support the weight of block A minus the buoyant force) and by doing so it exerts a little more force on balance D

When I say "the net effect of the buoyant force on the entire system is zero," I basically mean that the sum of the readings of D and E will not change if you have liquids of different density (as long as the liquid is always 1.5kg and so on). That is because, as you said, buoyancy is an internal force. But "internal" for the entire system is not necessarily "internal" for different parts of the system.

• erisedk
TSny
Homework Helper
Gold Member
Action-reaction forces can be considered as internal forces that cancel one another only if both of the two object involved in the action-reaction pair are part of your chosen system. You can choose your system in different ways. In the attachment the figure on the left shows the "system" as consisting of just the beaker and the water. The block is not part of this system. Here, the block is an external object which exerts a downward (external) force on the system, shown as BF. The weight of the block is not a force acting on this system. The force labeled N is the force which the bottom scale pushes up on the system.

In the right figure, the system is the beaker, the water, and the block. Now, the action-reaction buoyant forces between the block and the water are internal forces that cancel. So, they are not shown. The weight of the block is now an external force on the system as well as the tension force, T, pulling up on the top of the block.

#### Attachments

Action-reaction forces can be considered as internal forces that cancel one another only if both of the two object involved in the action-reaction pair are part of your chosen system. You can choose your system in different ways. In the attachment the figure on the left shows the "system" as consisting of just the beaker and the water. The block is not part of this system. Here, the block is an external object which exerts a downward (external) force on the system, shown as BF. The weight of the block is not a force acting on this system. The force labeled N is the force which the bottom scale pushes up on the system.

In the right figure, the system is the beaker, the water, and the block. Now, the action-reaction buoyant forces between the block and the water are internal forces that cancel. So, they are not shown. The weight of the block is now an external force on the system as well as the tension force, T, pulling up on the top of the block.

In the first figure, when the block is not a part of the system, why are we discriminating between the forces applied by the block that we consider? Why do we just consider the force BF and not Mg?
In the second figure, can't we calculate Tension as Mg-BF?

haruspex
Homework Helper
Gold Member
In the first figure, when the block is not a part of the system, why are we discriminating between the forces applied by the block that we consider? Why do we just consider the force BF and not Mg?
In the second figure, can't we calculate Tension as Mg-BF?
You have a choice.
You can say, I know the interaction between the block and the fluid is the buoyancy force, so I can split the system at that boundary. When considering the fluid, beaker and items below I can represent the upper items by a downward force with the magnitude of BF. When considering the upper parts, I can replace all the lower parts by the BF.
Or you split the system just below balance D, say. Now you don't care about the BF because that's internal to the lower part, and you treat the upward pull from D as an external force on that lower part.
Or you could split at both boundaries, etc.

You have a choice.
You can say, I know the interaction between the block and the fluid is the buoyancy force, so I can split the system at that boundary. When considering the fluid, beaker and items below I can represent the upper items by a downward force with the magnitude of BF. When considering the upper parts, I can replace all the lower parts by the BF.
Or you split the system just below balance D, say. Now you don't care about the BF because that's internal to the lower part, and you treat the upward pull from D as an external force on that lower part.
Or you could split at both boundaries, etc.

I don't exactly understand what you're saying, could you tell me the following
In the first figure, when the block is not a part of the system, why are we discriminating between the forces applied by the block that we consider? Why do we just consider the force BF and not Mg?
In the second figure, can't we calculate Tension as Mg-BF?

haruspex
Homework Helper
Gold Member
when the block is not a part of the system, why are we discriminating between the forces applied by the block that we consider? Why do we just consider the force BF and not Mg?
The block does not exert Mg on the fluid. Mg is a force exerted on the block, as is the tension in D. The fluid knows only of the direct forces exerted by the block at the common surface, i.e. pressure. In principle, you could integrate the pressure over the common surface to find the net force, but we can just wrap it up with the name BF and calculate it in other ways.
In the second figure, can't we calculate Tension as Mg-BF?
Which figure, which post?

Action-reaction forces can be considered as internal forces that cancel one another only if both of the two object involved in the action-reaction pair are part of your chosen system. You can choose your system in different ways. In the attachment the figure on the left shows the "system" as consisting of just the beaker and the water. The block is not part of this system. Here, the block is an external object which exerts a downward (external) force on the system, shown as BF. The weight of the block is not a force acting on this system. The force labeled N is the force which the bottom scale pushes up on the system.

In the right figure, the system is the beaker, the water, and the block. Now, the action-reaction buoyant forces between the block and the water are internal forces that cancel. So, they are not shown. The weight of the block is now an external force on the system as well as the tension force, T, pulling up on the top of the block.

The figure from this post, which is the second post.

haruspex