**Force Balance Calculations**

At equilibrium conditions the disturbing force F_{H} is balanced by the resisting force F_{R}. When F_{H }exceeds F_{R}, the Rolling Stone will begin to move. A diagram and Force Balance calculation is shown below.

where a = horizontal acceleration from earthquake in g’s

W = Weight of the Stone = mg

m = mass of stone

g = acceleration due to gravity

N = Normal Reaction Force

F_{R} = Resisting Force to horizontal motion

F_{R} = μN

μ = coefficient of friction

F_{H} = Horizontal Force due to earthquake

Therefore, the Force required to roll the Rolling Stone can be expressed in units of g’s (gravitation force) and is equal (in g’s) to the coefficient of friction operating at the area of contact between the rolling stone and the floor. One can approximate this coefficient by taking the value of the coefficient of rolling friction of limestone on limestone which is approximately equal to 0.25. Thus, ~0.25 g of force is required to initiate rolling of the Rolling Stone

To visualize this, a force of 0.25 g is equivalent to feeling a jolt equal to about 25% of your body weight. If you weighed 100 pounds ( 45 kg.), you would feel a jolt of ~25 lb ( 11 kg.) ) during the aftershock.

This is cool!