# Basic Static Equilibrium of a rigid body

I'm trying to understand static equilibrium in 3-dimensions, but I feel like I should understand it in 2-dimensions first

let's assume we have a rock with inhomogeneous density (black rectangle) with center of mass C

and are hanging it to the roof using two cables and two ball+socket supports, like in the following image: by definition, ball and sockets supports do not create any momentum-reaction so I would think to simplify the above diagram with the left one in the following image: but I am sure this is wrong, as it doesn't take into account the position of the off-centered center of mass C of the rock

I think the correct solution should be something like the right one

I can't completely understand how reaction moments Ma and Mb are generated, or how to calculate them

I'm trying to understand how to simplify this problem, I'm also watching youtube videos of static engineering to try to figure this out, but every lecture is about forces and reaction applied directly on the center of mass of a rigid body

what happens if reactions and forces are applied somewhere else on the body itself?

thank you all for your help!

## 1 Answer

This is a two dimensional, three force equilibrium problem which can be solved by using a triangle of forces.
As you have pointed out the lines of action of the three forces must meet at a point.

You can find the angles between the three forces by using the geometry of the arrangement which as well as distances $D_1$ and $D_2$ must include the vertical distance between the centre of mass and the top support (ceiling).
Without that distance you cannot solve such a problem.

Once you have done that you can draw a diagram to show the vector addition of the three forces which must be a triangle as the net force on the rock is zero.
With the angles of the triangles known you can solve such a problem.
This link may help you?