Force Balance Calculations

At equilibrium conditions the disturbing force FH is balanced by the resisting force FR. When FH exceeds FR, 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

FR = Resisting Force to horizontal motion

FR = μN

μ = coefficient of friction

FH = 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.

1 comment
1. Kim said: This is cool!