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For a rigid object in contact with a fixed environment and acted upon by gravity in the vertical direction, its support polygon is a horizontal region over which the center of mass must lie to achieve static stability. For example, for an object resting on a horizontal surface (e.g. a table), the support polygon is the convex hull of its "footprint" on the table. The support polygon succinctly represents the conditions necessary for an object to be at equilibrium under gravity. That is, if the object's center of mass lies over the support polygon, then there exist a set of forces over the region of contact that exactly counteracts the forces of gravity. Note that this is a ''necessary'' condition for stability, but ''not a sufficient'' one.


Derivation

Let the object be in contact at a finite number of points C_1,\ldots,C_N. At each point C_k, let FC_k be the set of forces that can be applied on the object at that point. Here, FC_k is known as the ''friction cone'', and for the Coulomb model of
friction Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding (motion), sliding against each other. There are several types of friction: *Dry friction is a force that opposes the relative la ...
, is actually a cone with apex at the origin, extending to infinity in the normal direction of the contact. Let f_1,\ldots,f_N be the (unspecified) forces at the contact points. To balance the object in static equilibrium, the following Newton-Euler equations must be met on f_1,\ldots,f_N: * \sum_^N f_k + G = 0 * \sum_^N f_k \times C_k + G \times CM = 0 * f_k \in FC_k for all k where G is the force of gravity on the object, and CM is its center of mass. The first two equations are the Newton-Euler equations, and the third requires all forces to be valid. If there is no set of forces f_1,\ldots,f_N that meet all these conditions, the object will not be in equilibrium. The second equation has no dependence on the vertical component of the center of mass, and thus if a solution exists for one CM, the same solution works for all CM+\alpha G. Therefore, the set of all CM that have solutions to the above conditions is a set that extends infinitely in the up and down directions. The support polygon is simply the projection of this set on the horizontal plane. These results can easily be extended to different friction models and an infinite number of contact points (i.e. a region of contact).


Properties

Even though the word "polygon" is used to describe this region, in general it can be any convex shape with curved edges. The support polygon is invariant under translations and rotations about the gravity vector (that is, if the contact points and friction cones were translated and rotated about the gravity vector, the support polygon is simply translated and rotated). If the friction cones are
convex cone In linear algebra, a ''cone''—sometimes called a linear cone for distinguishing it from other sorts of cones—is a subset of a vector space that is closed under scalar multiplication; that is, is a cone if x\in C implies sx\in C for every . ...
s (as they typically are), the support polygon is always a convex region. It is also invariant to the mass of the object (provided it is nonzero). If all contacts lie on a (not necessarily horizontal) plane, and the friction cones at all contacts contain the negative gravity vector -G, then the support polygon is the convex hull of the contact points projected onto the horizontal plane.


References

{{DEFAULTSORT:Support Polygon Classical mechanics