Get Help Solving a Cylinder in Equilibrium Problem

In summary: 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 ...Yes on both. Do you see why analysis of torque leads to the same answer?Yes, because torque is a vector and the net force and torque on the object are the same.
  • #1
Tanya Sharma
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Homework Statement



?temp_hash=0cf574ea033b4856fe276bc9518d0c17.jpg


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 .
 

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  • #2
Tanya Sharma said:

Homework Statement



?temp_hash=0cf574ea033b4856fe276bc9518d0c17.jpg


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?
 
  • #3
Hint: Compute the horizontal component of force from the fluid on each side. (No integration needed--or at least a trivial one.)
 
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  • #4
Doc Al said:
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.
 
  • #5
haruspex said:
I was trying not to give away that force is the way to go.
Oops! Sorry about that! :s
 
  • #6
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.
 
  • #7
Doc Al said:
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 ?
 
  • #8
Tanya Sharma said:
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?
 
  • #9
haruspex said:
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) ??

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

No
 
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  • #10
Tanya Sharma said:
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.
 
  • #11
Doc Al said:
Exactly correct.

Thanks...

And what is your opinion on post#9 ?
 
  • #12
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 dFx i.e component of force in horizontal direction and for that we are multiplying P by dAx 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 .
 
  • #13
Tanya Sharma said:
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 dFx i.e component of force in horizontal direction and for that we are multiplying P by dAx 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?
 
  • #14
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
 
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  • #15
Tanya Sharma said:
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.
 
  • #16
ehild said:
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 ?
 
  • #17
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
 
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  • #18
The force due to the right liquid comes out to be [itex] 3/2*rho*g*R^2*l [/itex](from integration).So we can see the pressure =[itex] 3/2*rho*g*R[/itex] and area = R*l. so i think we can conclude that surface to be a cuboid of dimensions [itex]R*R*l[/itex]
 
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  • #19
Tanya Sharma said:
And what is your opinion on post#9 ?
Oops... forgot to respond. (But others have already responded.)

Tanya Sharma said:
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.
 
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  • #20
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.
 
  • #21
Doc Al said:
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 ?
 
  • #22
Tanya Sharma said:
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.
 
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  • #23
Doc Al said:
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.

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

Thank you very much for sharing your insight :).
 

1. How do I solve a cylinder in equilibrium problem?

Solving a cylinder in equilibrium problem involves using the principles of equilibrium, which state that the sum of all forces acting on an object must equal zero. This means that the forces acting in the vertical and horizontal directions must balance out. To solve the problem, you will need to identify all the forces acting on the cylinder and use equations such as Newton's Second Law and the torque equation to find the unknown forces or variables.

2. What information do I need to have before solving a cylinder in equilibrium problem?

Before solving a cylinder in equilibrium problem, you will need to have the dimensions and mass of the cylinder, as well as the angles at which the forces are acting. You may also need to know the coefficients of friction and the density of the fluid (if the cylinder is submerged in a fluid).

3. Can you give an example of a cylinder in equilibrium problem?

Sure, here's an example: A cylinder with a mass of 10 kg rests on a ramp inclined at 30 degrees. The cylinder is in equilibrium with the ramp and is being held in place by a rope attached to the top of the ramp. What is the tension in the rope?

4. What are the common mistakes people make when solving a cylinder in equilibrium problem?

One common mistake is not correctly identifying all the forces acting on the cylinder. It's important to consider all the forces, including friction and the normal force, in order to accurately solve the problem. Another mistake is not setting up the equations correctly, such as using the wrong trigonometric functions or not considering the direction of the forces.

5. Are there any tips for solving a cylinder in equilibrium problem more efficiently?

Yes, a helpful tip is to draw a free-body diagram to visualize all the forces acting on the cylinder. Also, try to simplify the problem by breaking it down into smaller components and solving each component separately. It's also important to double-check your calculations and make sure your units are consistent throughout the problem.

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