Iron block dropped into a container of water sitting on a scale

In summary: Yes, that is all correct, except that you mean "rho" (##\rho##), not "phro".When it has descended distance h into the water, what is the net force on the block?The net force on the block is zero, so it has no acceleration.
  • #1
Kate_12
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Homework Statement
consider a container filled with water on a scale. Now a block of iron of length L and cross section A has been dropped into this container. Due to the drag of the water on the block, the block drops at constant speed v. Let the depth of the water be 2L. At time t=0, the block hit the surface of the water. Describe the change of scale reading as a function of time. If there were no drag of the water, the iron block will fall freely in the water. Describe the change of scale reading as a function of time.
Relevant Equations
f=rho*V*g
Does the scale change during the block sinks? I am so confused...
Is there any difference on change of scale with the applying of drag force?
 
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  • #2
First answer the following question:
What is the relationship between the reading on the scale when the block is fully immersed and the reading before the block enters the water?
 
  • #3
Orodruin said:
First answer the following question:
What is the relationship between the reading on the scale when the block is fully immersed and the reading before the block enters the water?
Before the block enters the water, there is no change in scale (just the weight of container filled with water)
When the block is fully immersed, the change in scale would be the weight of the block, because the buoyancy force is just internal force.
 
  • #4
Kate_12 said:
When the block is fully immersed, the change in scale would be the weight of the block, because the buoyancy force is just internal force.
So long as the block is not accelerating, yes.
We are not told with what speed the block hits the water, only that it descends at constant v until it hits the bottom. So I suppose you should assume it hits at speed v. That being the case, what is the net force on the block as a function of time? What does that imply for the force the water+container system exerts on the block?
 
  • #5
haruspex said:
So long as the block is not accelerating, yes.
We are not told with what speed the block hits the water, only that it descends at constant v until it hits the bottom. So I suppose you should assume it hits at speed v. That being the case, what is the net force on the block as a function of time? What does that imply for the force the water+container system exerts on the block?
The net force on the block is zero because it drops at constant speed.
so the water+container system exerts m(the mass of the block)*g on the block.
That means, the change in scale would be the weight of the block.
Is it right?
 
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  • #6
Note that the buoyant force is the force exerted on the block by the water. What does Newton's Third Law tell you about the force exerted on the water by the block?
 
  • #7
Kate_12 said:
The net force on the block is zero because it drops at constant speed.
so the water+container system exerts m(the mass of the block)*g on the block.
That means, the change in scale would be the weight of the block.
Is it right?
Yes.
Mister T said:
Note that the buoyant force is the force exerted on the block by the water. What does Newton's Third Law tell you about the force exerted on the water by the block?
In the first part of the question, we are told there is drag, so it is not just a matter of Archimedes' principle. Instead, we must use the constant speed information.
But Archimedes is certainly the way to approach the no drag case.
 
  • #8
haruspex said:
Yes.

In the first part of the question, we are told there is drag, so it is not just a matter of Archimedes' principle. Instead, we must use the constant speed information.
But Archimedes is certainly the way to approach the no drag case.
So with drag force, the scale change would be mg(the weight of the block) regardless of time.
Without drag force, the net force on the block would be phro*V*g (V: the submerged volume of the block)
1) during submerging, the change of the scale would be phro*A*h*g(h: the submerged height of the block)
--> I can't calculate the h as a function of time...
2) after submerged, the change of the scale would be phro*A*L*g
3) after the block reaches the bottom of the container, the change of the scale would be mg(the weight of the block)
Is this a correct approach?
 
  • #9
Kate_12 said:
So with drag force, the scale change would be mg(the weight of the block) regardless of time.
Without drag force, the net force on the block would be phro*V*g (V: the submerged volume of the block)
1) during submerging, the change of the scale would be phro*A*h*g(h: the submerged height of the block)
--> I can't calculate the h as a function of time...
2) after submerged, the change of the scale would be phro*A*L*g
3) after the block reaches the bottom of the container, the change of the scale would be mg(the weight of the block)
Is this a correct approach?
Yes, that is all correct, except that you mean "rho" (##\rho##), not "phro".
To get the force as a function of time in (1), you need a differential equation.
When it has descended distance h into the water, what is the net force on the block? What is its acceleration? Can you write the equation for ##\ddot h##?
 
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  • #10
haruspex said:
Yes, that is all correct, except that you mean "rho" (##\rho##), not "phro".
To get the force as a function of time in (1), you need a differential equation.
When it has descended distance h into the water, what is the net force on the block? What is its acceleration? Can you write the equation for ##\ddot h##?
mg-##\rho##*A*h*g=m*##\ddot h##

so integral 0 to h (dh^2/(m-##\rho##*A*h))= integral 0 to t (g/m)dt^2

I think I can solve this equation and there would be log term of h as a function of time.

I really appreciate for your help.
Have a nice day!
 
  • #11
Kate_12 said:
mg-##\rho##*A*h*g=m*##\ddot h##

so integral 0 to h (dh^2/(m-##\rho##*A*h))= integral 0 to t (g/m)dt^2

I think I can solve this equation and there would be log term of h as a function of time.

I really appreciate for your help.
Have a nice day!

No, you can't integrate one side wrt h and the other wrt t.
Your ODE has the general form ##\ddot y+Cy=D##. Do you recognise that?
 
  • #12
haruspex said:
No, you can't integrate one side wrt h and the other wrt t.
Your ODE has the general form ##\ddot y+Cy=D##. Do you recognise that?
Then, is there any general form of solution to ##\ddot y+Cy=D##?
 
  • #13
Kate_12 said:
Then, is there any general form of solution to ##\ddot y+Cy=D##?
Yes. If you have studied SHM you should recognise it.
 

1. What happens to the scale when the iron block is dropped into the water?

When the iron block is dropped into the water, the scale will initially show an increase in weight due to the addition of the block. However, as the block sinks and displaces water, the scale will gradually return to its original reading.

2. Why does the scale show an increase in weight when the iron block is dropped into the water?

The scale shows an increase in weight because the iron block has mass and is being added to the system. The scale measures the force of gravity on an object, and the addition of the block increases this force, resulting in a higher weight reading.

3. What happens to the water level in the container when the iron block is dropped in?

When the iron block is dropped into the water, it displaces a volume of water equal to its own volume. This causes the water level in the container to rise, as the water is pushed out of the way to make room for the block.

4. Does the material of the block affect the scale reading?

Yes, the material of the block can affect the scale reading. Different materials have different densities, which means they have different masses per unit volume. Therefore, a block made of a denser material will have a higher weight and cause a larger increase in the scale reading when dropped into the water.

5. What other factors can affect the scale reading when the iron block is dropped into the water?

Aside from the material of the block, other factors that can affect the scale reading include the temperature and salinity of the water. Warmer water has a lower density, which can result in a smaller increase in the scale reading. Similarly, saltwater is denser than freshwater, so a block dropped into saltwater will displace less water and cause a smaller increase in the scale reading compared to freshwater.

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