A slanted cylinder full of liquid about to fall

[lorenz0]

Homework Statement

A slanted cylinder of negligible mass has radius $R=0.3m$ and an internal angle of $\alpha=70°$ lies on a flat surface and is filled up with a liquid up to a height $H$. Find the maximum value of $H$ such that the cylinder remains in equilibrium

Relevant Equations

$\vec{\tau}=\vec{r}\times\vec{F}, \sum\vec{\tau}=\vec{0}, \sum{\vec{F}}=\vec{0}$

The cylinder will cease to be in equilibrium when the sum of the torques on the cylinder calculated with respect to the rightmost point of contact of the cylinder with the plane will be unbalanced. Now, the liquid is homogeneous and the cylinder has negiglible mass so the forces (normal force of the surface and gravitational force) will act on the center of mass which is in the geometric center of the figure so the lever arm of this torque with respect to said point should be $2R-(R+\frac{L}{2}\sin(30°))=2R-(R+\frac{H}{2\sqrt{3}})=R-\frac{H}{2\sqrt{3}}$.

Now, the thing that confuses me is that these two torques have the same $F$ and the same $r$ so their algebraic sum is always $0$.

There must be something about setting up this problem that I don't understand so I would appreciate an hint about how to reason about it.

 

[haruspex]

Where does the 30° come from?

normal force of the surface and gravitational force) will act on the center of mass which is in the geometric center of the figure

The normal force from the ground acts on the flat surface, but at what point of it?

 

[Steve4Physics]

There must be something about setting up this problem that I don't understand so I would appreciate an hint about how to reason about it.

If the centre of gravity is over the base, the cylinder is stable (why?).

If the centre of gravity is to the right of the base, the cylinder will topple over (why?)

So where would the centre of gravity be when the cylinder is on the verge of toppling over?

 

[hutchphd]

Homework Statement:: A slanted cylinder of negligible mass has radius $R=0.3m$ and an internal angle of $\alpha=70°$ lies on a flat surface and is filled up with a liquid up to a height $H$. Find the maximum value of $H$ such that the cylinder remains in equilibrium
Relevant Equations:: $\vec{\tau}=\vec{r}\times\vec{F}, \sum\vec{\tau}=\vec{0}, \sum{\vec{F}}=\vec{0}$

The cylinder will cease to be in equilibrium when the sum of the torques on the cylinder calculated with respect to the rightmost point of contact of the cylinder with the plane will be unbalanced.

Yes. So draw a picture of that situation (in your head or on paper) and notice the symmetry about a vertical line from the pivot which must then be true. The rest is just a little bit of geometry. Do you see it?

 

[lorenz0]

Yes. So draw a picture of that situation (in your head or on paper) and notice the symmetry about a vertical line from the pivot which must then be true. The rest is just a little bit of geometry. Do you see it?

from the drawing I would say that the maximum $H$ is the one such that $mg$ and the normal force $N$ acting on the pivot are collinear, otherwise the torque due to $mg$ wrt the pivot point will cause the cylinder to gain a clockwise angular acceleration that will make it fall. EDIT: it seems to me now that it would be sufficient to impose that the x-coordinate of the center of mass be at most $R$ to the right of the center of the cylinder so it should be sufficient to impose that $H\tan(20°)=R$ i.e. $H=\frac{R}{\tan(20°)}$. Is this correct?

 

[hutchphd]

I would write it as $$\frac H R=\tan \alpha$$ but that is the same thing. Good.

 

[TSny]

EDIT: it seems to me now that it would be sufficient to impose that the x-coordinate of the center of mass be at most $R$ to the right of the center of the cylinder so it should be sufficient to impose that $H\tan(20°)=R$

Does $H$ here represent the height of the top surface of the water, or the height of the center of mass of the water?

 

[lorenz0]

Does $H$ here represent the height of the top surface of the water, or the height of the center of mass of the water?

The height of the top surface of the water

 

[haruspex]

The height of the top surface of the water

Then "double" check your answer.

 

[hutchphd]

Or realize that I am correct about "half" the time. For reasons that escape me I assumed that R was the diameter of the container. Physics correct, geometry wrong. Thanks for check.

 

[lorenz0]

Then "double" check your answer.

Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose $H=\frac{T}{\tan(20°)}$.

 

[TSny]

Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose $H=\frac{T}{\tan(20°)}$.

What are the lengths of the base and height of the red triangle? In particular, what is the height of the center of mass in terms of $H$?

 

[haruspex]

Could you please explain why? it seems to me that to have the x-coordinate of the center of mass be to the right of the pivot point it should be sufficient to impose $H=\frac{T}{\tan(20°)}$.

I assume you mean $H=\frac{R}{\tan(20°)}$, which is what you posted before.
H is the full height, not the height to the mass centre. R is a radius, not a diameter.