Impact of decreasing weight from a part of body w.r.t tipping

Question

If I decrease the weight from top portion of the body keeping all dimensions same, the center of mass shifts downwards, but does it makes it more stable against tipping from perpendicular forces ?

Background: I have a nice sturdy easel style wooden stand which has iron frame for VESA mounting TVs. Till now I've mounted a 55 pound TV on it and it seems pretty stable (w.r.t a general impact from back or front).

I am about to change to a exact same size TV new model which is 40% lighter & is 33 pounds.

Per my rough calculation the center of mass is shifting about 2-3 inches downwards but is that factor sufficient ? As per generic family opinion, if I switch to a lighter TV, my stand will be 'more 'prone to tipping. Any advice?

P.S: Dont need any calculations here.

Answer 1

Even though You ask for no calculations a simple equation is helpful here.

A mass, m at a C.G. height, H, on a stand with base, B needs a force, F to topple.

$$F\geq \frac{m*B}{2H}$$ So if the quantity of $m/H$ decreases the stand is less stable if it increases it is more stable.

You can plug in your numbers and see what you get!

Answer 2

It will require less force to topple the lighter TV. But a lower center of gravity will mean that the whole stand can tip over more before it actually falls. If the center of mass was already high, then 2-3 inches won't be that different. If the center of mass is already relatively low then those 2-3 inches will make it significantly more stable. But with out knowing how wide the base is, how high the actual center of mass is, and how heavy the stand is, it's pretty impossible to give a good answer.

My best guess? If a 22 pound difference only lowers the center of mass by 2-3 inches, then that means the center of mass is already high or the stand weighs way more than the TV. In either case it'll probably be just about as stable (or unstable) with the new TV as it was with the old one.