Get Help Solving a Cylinder in Equilibrium Problem

[Tanya Sharma]

Homework Statement

Homework Equations

The Attempt at a Solution

Since the cylinder is in equilibrium ,net force and torque on it should be zero.

I am not sure whether we are required to calculate force due to the two fluids as it seems to be a simple conceptual problem . I think I am missing some trick .

I would be grateful if somebody could help me with the problem .

 

[haruspex]

Homework Statement

Homework Equations

The Attempt at a Solution

Since the cylinder is in equilibrium ,net force and torque on it should be zero.

I am not sure whether we are required to calculate force due to the two fluids as it seems to be a simple conceptual problem . I think I am missing some trick .

I would be grateful if somebody could help me with the problem .

I don't think there is any trick - just do the integration. But which will you use, net zero force or net zero torque?

 

[Doc Al]

Hint: Compute the horizontal component of force from the fluid on each side. (No integration needed--or at least a trivial one.)

 

[haruspex]

Hint: Compute the horizontal component of force from the fluid on each side. (No integration needed--or at least a trivial one.)

I was trying not to give away that force is the way to go.

 

[Doc Al]

I was trying not to give away that force is the way to go.

Oops! Sorry about that! :s

 

[Orodruin]

So the question I would like to ask OP now is if (s)he can argue for why force is the way to go and momentum equilibrium is automatic if force equilibrium is fulfilled.

 

[Tanya Sharma]

Hint: Compute the horizontal component of force from the fluid on each side. (No integration needed--or at least a trivial one.)

Thank you very much for your response :) .

You are right that a trivial integration is required . This is precisely why I said in OP that I was missing some trick . Well... you were spot on as usual :D . Anyways I would still like to know how we can deal with no integration oo).

I would like to discuss a couple of points .

1) Suppose instead of having the convex surface towards the liquid ,it were some other arbitrary shape of same height h ,say a similar concave surface ,or an inclined surface(instead of a cylinder there was a wedge) , a straight vertical plane ( instead of a cylinder there was a cuboid ) , then too , the horizontal force of liquid would remain same in all these cases ??

2) Another way to think about this is to find the projection of area exposed to liquid on a vertical plane and then to calculate the force due to liquid on this projected vertical area ??

Do you agree ?

 

[haruspex]

Thank you very much for your response :) .

You are right that a trivial integration is required . This is precisely why I said in OP that I was missing some trick . Well... you were spot on as usual :D . Anyways I would still like to know how we can deal with no integration oo).

I would like to discuss a couple of points .

1) Suppose instead of having the convex surface towards the liquid ,it were some other arbitrary shape of same height h ,say a similar concave surface ,or an inclined surface(instead of a cylinder there was a wedge) , a straight vertical plane ( instead of a cylinder there was a cuboid ) , then too , the horizontal force of liquid would remain same in all these cases ??

2) Another way to think about this is to find the projection of area exposed to liquid on a vertical plane and then to calculate the force due to liquid on this projected vertical area ??

Do you agree ?

Yes on both.
Do you see why analysis of torque leads to the same answer?

 

[Tanya Sharma]

Yes on both.

Do you agree that apart from horizontal force ,vertical force by the fluid on all different shapes also remain same ?

If yes ,then does that mean instead of a cylinder of radius 'r' and length 'l' ,it were a cuboidal shaped object having width '2r' , height '2r' and length 'l' , the force balance equations would remain same (i.e Pressure by the fluid would remain same) ??

Do you see why analysis of torque leads to the same answer?

No

 

[Doc Al]

I would like to discuss a couple of points .

1) Suppose instead of having the convex surface towards the liquid ,it were some other arbitrary shape of same height h ,say a similar concave surface ,or an inclined surface(instead of a cylinder there was a wedge) , a straight vertical plane ( instead of a cylinder there was a cuboid ) , then too , the horizontal force of liquid would remain same in all these cases ??

2) Another way to think about this is to find the projection of area exposed to liquid on a vertical plane and then to calculate the force due to liquid on this projected vertical area ??

Do you agree ?

Exactly correct.

 

[Tanya Sharma]

Exactly correct.

Thanks...

And what is your opinion on post#9 ?

 

[Tanya Sharma]

Despite solving the problem I am a little unsure . The force dF exerted by fluid on a tiny surface element dA is radially towards the center (on the axis ) ,given by dF = PdA . Now we are looking for dF_x i.e component of force in horizontal direction and for that we are multiplying P by dA_x i.e component of area along YZ plane . Right ??

Now when I compare this situation with that of calculating flux of a constant electric field across a hemispherical cap , the flux is given by E($\pi r^2$). Here ($\pi r^2$) is the projection of the hemispherical surface along a plane perpendicular to the electric field .

But in the former case P is a scalar quantity whereas E in the later case is vector .

Are the two concepts similar or am I mixing up two unrelated concepts ?

Sorry if I am not clear . But this is how I came up with the reasoning in this question .

Kindly give your views .

 

[haruspex]

Despite solving the problem I am a little unsure . The force dF exerted by fluid on a tiny surface element dA is radially towards the center (on the axis ) ,given by dF = PdA . Now we are looking for dF_x i.e component of force in horizontal direction and for that we are multiplying P by dA_x i.e component of area along YZ plane . Right ??

Now when I compare this situation with that of calculating flux of a constant electric field across a hemispherical cap , the flux is given by E($\pi r^2$). Here ($\pi r^2$) is the projection of the hemispherical surface along a plane perpendicular to the electric field .

But in the former case P is a scalar quantity whereas E in the later case is vector .

Are the two concepts similar or am I mixing up two unrelated concepts ?

Sorry if I am not clear . But this is how I came up with the reasoning in this question .

Kindly give your views .

The concepts are analogous. P is a scalar, but dA is a vector (perpendicular to the surface).
If the small element length ds is at angle theta to the vertical, what horizontal force do you get from a pressure P?
What torque do you get from the total force on ds about the point of contact of the drum with the base?

 

[ehild]

The electric flux through an elementary surface is $\vec E \cdot \vec {dA}$. $\vec{dA}$ is a vector, with magnitude equal to the area dA and direction perpendicular to the surface, and pointing "outward".

In case of the cylinder in the fluid, you need the resultant force. The force exerted by the fluid on a small area on the surface of the cylinder is $\vec {dF}=-P\vec {dA}$. You need the horizontal components, $dF_x= -P\vec{dA}\cdot \hat x$ .

In both cases, you have the projection of the area onto a plane perpendicular to a certain direction.

ehild

 

[nil1996]

Thank you very much for your response :) .

You are right that a trivial integration is required . This is precisely why I said in OP that I was missing some trick . Well... you were spot on as usual :D . Anyways I would still like to know how we can deal with no integration oo).

I would like to discuss a couple of points .

1) Suppose instead of having the convex surface towards the liquid ,it were some other arbitrary shape of same height h ,say a similar concave surface ,or an inclined surface(instead of a cylinder there was a wedge) , a straight vertical plane ( instead of a cylinder there was a cuboid ) , then too , the horizontal force of liquid would remain same in all these cases ??

2) Another way to think about this is to find the projection of area exposed to liquid on a vertical plane and then to calculate the force due to liquid on this projected vertical area ??

Do you agree ?

If you try to change the surface you still have to find the relation between the force on curved surface and any other surface.So I think there is no way without integration.

 

[Tanya Sharma]

The electric flux through an elementary surface is $\vec E \cdot \vec {dA}$. $\vec{dA}$ is a vector, with magnitude equal to the area dA and direction perpendicular to the surface, and pointing "outward".

In case of the cylinder in the fluid, you need the resultant force. The force exerted by the fluid on a small area on the surface of the cylinder is $\vec {dF}=-P\vec {dA}$. You need the horizontal components, $dF_x= -P\vec{dA}\cdot \hat x$ .

In both cases, you have the projection of the area onto a plane perpendicular to a certain direction.

ehild

Hello ehild

Thank you for the reply .

Could you give response on post#9 ?

 

[ehild]

What do you mean by "vertical force by the fluid on all different shapes also remain same"?
The buoyant force is equal to the weigh of the fluid displaced, for any shapes.

ehild

 

[nil1996]

The force due to the right liquid comes out to be 3/2*rho*g*R^2*l(from integration).So we can see the pressure =3/2*rho*g*R and area = R*l. so i think we can conclude that surface to be a cuboid of dimensions R*R*l

 

[Doc Al]

And what is your opinion on post#9 ?

Oops... forgot to respond. (But others have already responded.)

Do you agree that apart from horizontal force ,vertical force by the fluid on all different shapes also remain same ?

No.

As you realize, calculating the horizontal force involves a simple integration. But directly calculating the vertical force would involve a more challenging integration. Luckily, you can apply Archimedes' principle and avoid such troubles. Then you'll see that the vertical force depends on displaced volume.

 

[Dr.D]

Just for fun, try calculating the moments on the barrier due to the fluid volumes. I'll be anxious to see how you do that.

 

[Tanya Sharma]

Oops... forgot to respond. (But others have already responded.)No.

As you realize, calculating the horizontal force involves a simple integration. But directly calculating the vertical force would involve a more challenging integration. Luckily, you can apply Archimedes' principle and avoid such troubles. Then you'll see that the vertical force depends on displaced volume.

Thanks Doc !

You mentioned in post#3 that this problem could be solved without integration . Do you mind discussing how this could be done ?

 

[Doc Al]

You mentioned in post#3 that this problem could be solved without integration . Do you mind discussing how this could be done ?

Sure. All you need care about is the horizontal force exerted by fluid pressure on each side. And that horizontal force is independent of the shape of the object. So make it easy on yourself and imagine a vertical plane in front of the body.

Start with the left side. The height of the plane is h. The pressure on the plane varies from 0 at the top to $2\rho g h$ at the bottom. (Ignoring atmospheric pressure, for simplicity. I'll leave it to you to show that including atmospheric pressure does not change the answer.) Thus the average pressure on the plane is $\rho g h$, making the force per unit length on that plane equal to $\rho g h^2$.

Do the same analysis for the right hand side and set the expressions equal. That's all there is to it.

 

[Tanya Sharma]

Sure. All you need care about is the horizontal force exerted by fluid pressure on each side. And that horizontal force is independent of the shape of the object. So make it easy on yourself and imagine a vertical plane in front of the body.

Start with the left side. The height of the plane is h. The pressure on the plane varies from 0 at the top to $2\rho g h$ at the bottom. (Ignoring atmospheric pressure, for simplicity. I'll leave it to you to show that including atmospheric pressure does not change the answer.) Thus the average pressure on the plane is $\rho g h$, making the force per unit length on that plane equal to $\rho g h^2$.

Do the same analysis for the right hand side and set the expressions equal. That's all there is to it.

. This is a very nice way of looking into the problem .

Thank you very much for sharing your insight :).