Moment & Equilibrium

cos() > 1?
Make sure your right angle is in the right place? But yes, the values are 3/5 and 4/5 for sin and cos.
In ghostwalkers diagram, its the triangle with vertices
O, A and the point where the green and purple lines cross (midpoint of cone base).

Can you check my diagram please. I would be really grateful. I know I must have done a stupid mistake... sorry

The bottom one is fine 3 : 4 : 5.
The middle one is wrong as you're assuming the COM (hypotenuse length) is over the 5r point. This was sorted out previously?

What do you mean by being over the 5r point? I have drawn COM (C) directly about A. Is that wrong?

(above - bold) Yes. I thought you'd understood that. The COM will be to the left (x-coordinate) of A. It is not directly above A.

As in the other post, use cos() and either
* calculate the base using the COM (hypotenuse)
* calculate the hypotenuse using the 5r base

Both are equilvalent, the marking scheme does the first one. Both show that the COM is not directly above A.

Sorry, I thought that the centre of mass (C) will either be exactly above A, or to the left of A on the symmetrical line. Which would imply, that the x co-ordinate is less than or equal to 5r. Why cant COM be exactly above A? It would still be in equilibrium perhaps.

When the x coordinate of the COM is <= 5r, its in equilibrium. Agreed.
However, for this problem the COM is not directly above A. It is to the left of it. You seem to be assuming it is over the A in your working which seems to be causing the error.
Try writing it up a bit clearer, with explanations, and it will be easier to pinpoint your misunderstandings.

Ok i will write it in a clearer way and show it. But how do we get to know that COM is to the left of A. Is there some hint?

No you dont assume anything. You evaulate it using simple trig. You have a 3 : 4 : 5 triangle where the hypotenuse is 21r/4 and the base is the "4" side. How long is it? You simply need to show its <= 5r. That is all this question part requires. It shows the COM is to the left of A so in equilibrium.

Really sorry for bothering again, but I wanted to improve my understanding. Supposedly a question says that
" the object is in equilibrium ".

What I understand by equilibrium, is that all these 3 conditions will be True simultaneously

1) The net resultant force in any direction is zero.

2) If there is any force parallel to the ground, the friction would be limiting. (i.e. limiting equilibrium)

3) There will be an about to topple scenario where The Normal Reaction R is at the point about which the solid object will topple and the Centre Of Mass is directly above R.

Is it wrong?

If an object is in (static) equilibrium, the net forces acting on the object and the net moments must be zero. Basically, it can't move linearly or rotate.
1) Agree, this means no linear change in equilibrium.
2) Partially covered by 1), but friction does not have to be in limiting equilibrium. A bit like this question, friction is like a sponge and will soak up any excess force up to a maximum value - its limiting value. However, the force due to friction can be less than this limiting value, even zero if the net additional forces are already zero. Many questions do consider the limiting equilibrium scenario though.
3) In equilibrium, there must be a vertical reaction directly under the COM (assuming just these two forces act on the object). Otherwise, take moments about the reaction point and the net moment due to the COM will be non-zero and hence topple.

ok fair enough. But one thing I am still struggling to grasp, when I am attempting the question is, how do I know whether COM is to the left of A on the symmetrical line or exactly above A on the symmetrical line. That's what I am struggling with. sorry again....

You don't know it, you calculate it. This is precisely what the question part is asking.

The x coordinate of the COM is
21r/4 * 4/5 = 21r/5 = 4.2r < 5r

Or project the 5r "base" onto the hypotenuse which gives
5r * 5/4 = 25r/4 > 21r/4

So the COM is to the left of the 5r point on the base.

Thanks for keeping up with me. Now I am like getting what you are trying to state. Basically, it is equilibrium, but we don't know whether it is toppling equilibrium ( where it just about to topple and R is acting on the edge about which the solid will topple ) or just simple equilibrium ( not about to topple) . We got to find it ourself to see what is happening.

I don't think the question distinguishes between toppling/simple equilibrium. But simply whether the the COM is to the right of the A point and hence will definitely rotate. Or to the left of A (including A) and hence be in equilibrium. You have to work that out.

just one more thing, when you say:
" Or project the 5r "base" onto the hypotenuse which gives
5r * 5/4 = 25r/4 > 21r/4 "

What do you mean? Are you comparing the adjacent sides of both triangles?

All are similar 3 : 4 : 5 triangles. Projecting the 5r base onto the hypotenuse works out the point directly above A, hence the limiting COM position. It should be clear from the diagram? If you project the 21r/4 hypotenuse onto the base, you see its to the left of A. Both are equivalent ways of showing equilibrium.

ok got it. Is it ok what I have done? I think it looks ok.....