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

[Kate_12]

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?

 

[Orodruin]

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?

 

[Kate_12]

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.

 

[haruspex]

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?

 

[Kate_12]

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?

 

[Mister T]

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?

 

[haruspex]

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.

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.

 

[Kate_12]

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?

 

[haruspex]

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 $\ddot h$?

 

[Kate_12]

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!

 

[haruspex]

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?

 

[Kate_12]

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 $\ddot y+Cy=D$?

 

[haruspex]

Then, is there any general form of solution to $\ddot y+Cy=D$?

Yes. If you have studied SHM you should recognise it.