Moment & Equilibrium

Hi,
Can anyone help me with this question? I have successfully done the part (a) where it asks to find the centre of mass of the object from the end of cylinder.

Part(b) is where, i get stuck.

(b) Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.

My understanding tells me that when a body is in equilibrium, the net force in any direction is zero. Also by equilibrium, I tend to take the about to topple scenario and about to slide scenario. I have drawn a diagram but clueless about how to go further. My intuition tells me that for the object to be in equilibrium, clockwise moments must be equal to anti-clockwise moments. Just don't know how to progress. Any help would be appreciated.

In your diagram, the COM must be directly above the region of the cone directly in contact with the floor.
If it doesn,t there will be a clockwise moment.

I have drawn the diagram according to what you are saying, now what should I do next?

That will be the limit of equilibrium (obviously), where the COM is directly above the 5r point.
I couldn't easily read the question, but Id guess theyre looking tor a bit of trig (triangles).
The COM will also lie on the line connecting the vertex to the base centre.

I have posted the question again in better quality for you to read.

Thanks, going through it now - realize you'd posted a fair bit originally.

I have drawn the diagram according to what you are saying, now what should I do next?

You just need to forn a right angled triangle (as you've pretty much done) where the hypotenuse is
the COM distance from the cone vertex
and the base is the floor and origin (cone vertex) is one point of the triangle. Then show the base is <= 5r when in equilibrium. You have the side ratio to get the angle(s) in the triangle.

Or equivalently the COM (x,y) coordinates has an x coordinate <= 5r when the cone vertex is the origin.

For info, here's a diagram that uses the labelling of the markscheme.

I have done so far like this.... The base 0A in my diagram is already 5r, why are you saying that prove that the base is less than or equal to 5r. I don't get it, sorry for the hassle...

As long as the COM has an x coordiante <= 5r, its in equilibrium.
5r is the limiting value.

So to be exact you are saying that if the COM x co-ordinate is less than or equal to 5r, it's in equilibrium. Well one thing is confusing me, suppose that the x co-ordinate is less than 5r ( i.e x<5r ), then wouldn't this lead to the anti-clockwise moments about O (The Vertex Of The Cone) be greater than clockwise moments about O, which doesn't lead to equilibrium as anti-clockwise moments are not equal to clockwise moments.

No as there is solid support between the floor coordinates 0 and 5r. No moment would be generated. Just imagine it physically - your COM is over your feet shadown - you don't fall over.

umm, Can we say that if x<5r (x being the COM x co-ordinate), our COM will shift towards the left, will this lead to "R" (The Normal Reaction) being shifted as well towards the left, exactly below COM?

Yes. Id edited the previous reply as you were typing.

Not read through it fully, but you use the original cone triangle to get sin(theta) and cos(theta) as the cone triangle is a similar right angled triangle. Theyre the obvious simple side ratios.
Then either assume the base is 5r and calculate the hypotenuse length and show that the COM is less than this value.
Or assume the hypotenuse is the COM vertex distance and show that the base is less than 5r.

The mark scheme does the latter.

I used the bigger triangle with vertices at O,A and C (Centre Of Mass). It leads to cos(theta) = 5r which is weird, but if i use the cone triangle I get cos(theta) = 5/4. Getting confused as which triangle I should use, I think we should use the bigger triangle as we are interested in the centre of mass's x co-ordinate which I want to evaluate.