Stability

Often times summing the torque of a system of objects around an axis to determine if the system is in equilibrium or not is often tedious. It turns out we also have another method to determine the stability of a system by looking at the total center of mass of the system and its location to the furthest most normal force.

Mass is distributed within a rigid body, however, we can define a "center of mass" (com) location for an object or system of objects. This is useful because if our object is in a uniform gravitational field (much like the approximate uniform gravitation field near the surface of the Earth), then the force of gravity that acts over the entire body of the object can be simplified and treated as if it acts through only the center of mass of the object. The center of mass will have a x, y, and z-component, but often times we can just consider the x and y-component because the z-component is of no importance to our problem. Mathematically we find the center of mass of a system of objects as follows:

$x^{tot}_{com} = \frac{x^{1}_{com}m_1 +x^{2}_{com} m_2 +x^{3}_{com} m_3 +...}{m_1 + m_2 + m_3 + ...} = \frac{\sum{x^{i}_{com} \, m_i}}{\sum{m_i}}$

$y^{tot}_{com} = \frac{y^{1}_{com}m_1 +y^{2}_{com} m_2 +y^{3}_{com} m_3 +...}{m_1 + m_2 + m_3 + ...} = \frac{\sum{y^{i}_{com} \, m_i}}{\sum{m_i}}$

As an example, the figure below shows a system of 4 objects with each of their individual center of masses identified by the small circle with the dotted block line going through it. The condition as shown below is the case where the system will not tip over, but if the center of mass shifted just slightly to the right, then it would tip.

Now, take a look at the pre-lecture reading and videos below.