Domino Fall
Physics Root
The Forces and torques in the falling domino problem.
Lean one domino into its neighbor and ask if the first domino will push over the second.
The first domino has height H_{1}, width W_{1}, and thickness T_{1}.
The second domino H_{2},W_{2}, and T_{2}.
The forces on the first domino include the force
of gravity, F_{g1} at its center of mass.
The Normal force from the floor, F_{N1}.
The friction force with the floor, F_{f1}.
And, the contact force from the second domino, F_{C21}.
There may also be a friction force where the first domino touches the
second. This force is not shown.
If we choose the pivot point for torque calculations to be the point where the first domino touches the floor then we find that F_{C21} *H_{1} cosTh = F_{g1}*H_{1}/2 sin Th
Where Th is the angle between the dominos.
The distance between dominos is the width of the domino W_{1}.
So Sin Th = W_{1}/H_{1} and Cos Th = H_{C}/H_{1}
Substituting these for the sin and cos we get
F_{C21} *H_{1} *H_{C}/H_{1} = F_{g1}*H_{1}/2 W_{1}/H_{1} or
F_{C21} = (F_{g1}/2) W_{1}/H_{C}
If the first domino is at rest then the torque produced by the force of gravity and the normal force from the floor, must equal the opposite torque from the friction force from the floor and the contact force of the second domino.
The contact force of the first domino on the second, F_{C12}, is equal and opposite to the contact force of the second on the first F_{C21} by Newton's law that for every action there is an equal and opposite reaction.
This force F_{C12} provides the torque
which topples the second domino.
The second domino pivots about the corner above the caret ^ .
So the torque from the first domino is F_{C12} times the
height at which the first domino hits the second h_{C}. The
domino topples if this torque is greater than the restoring torque
from gravity F_{g2} times the distance of the center of mass
from the pivot point, half the thickness of the domino
T_{2}/2.
So the domino topples if F_{C12} * h_{C} > F_{g2} * T_{2}/2
plugging in for F_{C12} from above
(F_{g1}/2) W_{1}/H_{C} > F_{g2} * T_{2}/2
or H_{C} > F_{g1}W_{1}/F_{g2}T_{2}
Now:
F_{g2} = 3.375 F_{g1}
W_{1} = 6
T_{2} = 1.5
so
If H_{C} > 6/(3.375*1.5)
If H_{C} > 1.185
And H_{C} is about 4.5
Scientific Explorations with Paul Doherty 

17 April 2001 