Answer to the Question 01/03
BRICKS AND STICKSThe question was:
Two bricks are separated by three parallel wooden sticks of circular cross sections. The sticks are positioned as shown in the picture, separated from the ends of the brick and from each other by distances l1, l2, l3, l4. The weight of the top brick is P. What are the forces applied by each stick on the brick.
(6/03) The problem has been solved (21/1/2003) by Victor Ivanov, a teaching assistant at the Faculty of Physics of Sofia University (Bulgaria) (e-mail vgi@phys.uni-sofia.bg) (the solution below follows (to large extent) his derivation), (9/3/2003) by Gareth Pearce a Ph.D. student at U. Sydney (e-mail tilps@hotmail.com), (22/3/2003) by Regis Lachaume, a PhD student at Grenoble Observatory (UJF, Grenoble, France) lachaume@obs.ujf-grenoble.fr) (see his solution at the following postscript file, and (16/6/2003) by Ansgar Esztermann, from Heinrich-Heine-Universitaet Duesseldorf (Germany) (e-mail ansgar@thphy.uni-duesseldorf.de).
The answer:
For the sake of simplicity we introduce the coordinates Xi (i=1...3) of the points, where the rods touch the surface of the upper brick. They are defined as the horizontal positions of those points with respect to the center of the brick. The Y-axis is directed downwards from the brick's center. If we assume that the bricks and the rods are perfect rigid bodies then the conditions for equilibrium read:
(1) SUM(Fi) = P (force balance), and:
(2) SUM(Fi*Xi) = 0 (torques balance),
where Fi are the forces of normal reaction exerted from the rods on the upper brick. It is evident that if the number of rods is 3 or more, the two equations are insufficient to determine F's unequivocally (the problem is statically undefined).
To cope with the problem we must go beyond the rigid body approximation. Let us to assume that the rods acquire some small deformation when the upper brick is put on them. We will neglect the deformations of the surfaces of the bricks since as a rule the ceramic material of the bricks is rather stiff compared to the wood.
It can be shown that the deformation of the rod is a linear function of the force, despite the fact that it has only a "point contact" with the brick. (This is explained in "Comment 1" at the bottom of this page.) Thus we can treat the rods as elastic springs of equal stiffness K, and assume that the Hook's law is satisfied: Fi = K*Ui, where Ui is the deformation of the i-th rod. We assume that when the upper brick is laid onto the rods, due to their deformations, it's center is lowered by a small amount Y, and its surface tilts a little bit at an angle A with respect to the horizontal (the later effect takes place if the rods are placed asymmetrically with respect to the center of the brick). Then the deformation of the i-th rod is:
Ui = Y + A*Xi.
By substituting the forces Fi obtained from the Hook's law in Eqn. (1) and (2) we obtain the values of Y and A:
Y = (P/(3K))*D/(D - M2);
A = - (P/(NK))*M/(D - M2);
where D = SUM(Xi2)/3 and M = SUM(Xi)/3. Finally we obtain the forces of reaction:
(3) Fi = (P/3)*(D - Xi*M)/(D - M2).
It is evident from the last expression that if the rods are placed symmetrically with respect to the center of the brick, then the weight of the brick is spread uniformly among them with equal portions of P/3. A*Xi.
Comment 1: Contacts involving cylinders or spheres were treated in 19th century by H. Hertz. You can find this treatment in most elasticity textbooks (e.g., the book by Landau and Lifshitz). What is special in this case is that we have a point contact, which upon application of force broadens into a contact with finite area. For the case of a contact between two spheres, or a contact between a sphere and infinitely rigid plane, this leads to conclusion the the displacement (deformation of the sphere) is proportional to the force in the power 2/3. (The prefactor in this relation depends both on the elastic properties of the sphere and on its radius.)
The situation is quite different in the "two-dimensional" case of cylinder. It can be derived from the general formalism that treats contact of ellipsoids, or can be considered directly as was done by H. Poritsky in J. Appl. Mech. 17, p. 191 (1950). In this paper it has been shown that the displacement is simply proportional to the force applied to the cylinder. Thus the simple linear relation between the force and deformation used in the above derivation is justified. The only pathology caused by the presence of the point contact is related to the dependence of the proportionality constant on the properties of the cylinder: As in the case of a "regular" contact the force constant is proportional to the elastic constant (Young modulus) of the stick. However, for the point the force constant very weakly (logarithmically) depends on the diameter of the stick.
Comment 2: Regis Lachaume, a PhD student at Grenoble Observatory (UJF, Grenoble, France) also considered (23/1/2003) a very different case, in which the bricks and the sticks are rigid enough so that their deformation is small. However, the "brick" is so long that its transverse deformations become significant. This situation is not suited, for our "brick and sticks" case, but probably correctly describes a long board supported by sticks. See his solution in postscript file.
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