Reading through the first ten posts, the original poster is asking for a rigid-body mechanics answer. A rigid, horizontal, square plate, having a vertical point load at its center, is suspended by four inextensional, vertical cables. One cable is removed. How is the load redistributed to the remaining three cables? The original poster also indicates he wants to consider the load to be slightly eccentric from the center of the plate.
Let a = length of plate side, h = half of diagonal length = a*cos(45 deg), P = applied load at center of plate, m = plate mass, and g = gravitational acceleration. Let F1, F2, F3, F4 = force in cables 1, 2, 3, and 4, respectively.
Before cable number 4 is removed, moment equilibrium (Newton's first law), and symmetry, dictates that cable number 2 must carry one-fourth of the load, as follows. Summing moments about connection 2 gives, summation(M2) = 0 = h*F1 + h*F3 + (2*h)*F4 - h*(P + m*g). But, by symmetry, F1 = F2 = F3 = F4, reducing four unknowns to one unknown. Therefore, substituting and solving for F2 gives, F2 = (P + m*g)/4.
As soon as cable 4 is removed, moment equilibrium, and symmetry, dictates that the force in cable 2 must now change to zero, as follows.
Summing moments about connection 1 gives, summation(M1) = 0 = 0*(P + m*g) - 0*F1 - 0*F3 + h*F2. Therefore, solving for F2 gives, F2 = 0.
Now let's say you instead want to solve this problem like a "bolt pattern," after cable 4 is removed. Let A = cross-sectional area of each cable, and yb = distance of "bolt pattern" centroid from line 1-3. Therefore, yb = summation(A*y)/summation(A) = (0 + 0 + A*h)/(3*A) = h/3. Now translate the applied loads to the centroid. The equivalent applied loads at the centroid are, Pc = (P + m*g) and Mc = (h/3)(P + m*g) = Pc*h/3. Summing moments about connection 1 for applied tensile load Pc gives, summation(M1t) = 0 = h*F2t - 0*(F1t + F3t) - (h/3)*Pc. Solving for F2t gives, F2t = Pc/3. Summation of vertical forces gives, summation(Ft) = 0 = F1t + F2t + F3t - Pc. By symmetry, F1t = F3t. Therefore, substituting and solving for F1t gives, F1t = F3t = (Pc - Pc/3)/2 = Pc/3. To distribute the applied centroidal moment Mc, sigma = M*y/I; therefore, F = M*y*A/I (Eq. 1). But I = integral[(y^2)(dA)] = A*summation(y^2); therefore, F = M*y/summation(y^2). But summation(y^2) = 2(yb^2) + (h-yb)^2 = 2*(h/3)^2 + (h - h/3)^2 = 2(h^2)/3. Therefore, substituting, Eq. 1 becomes, F = (3/2)(Mc)
/(h^2) = (3/2)(Pc*h/3)
/(h^2) = (Pc*h)
/(2*h^2). Therefore, the force in cable 1 due to applied moment Mc is, F1m = (Pc*h)(h/3)/(2*h^2) = Pc/6. And, likewise, the force in cable 2 due to applied moment Mc is, F2m = (Pc*h)(h/3 - h)/(2*h^2) = -Pc/3. Thus, by superposition, the total force in cable 1 is, F1 = F1t + F1m = Pc/3 + Pc/6 = Pc/2 = (P + m*g)/2. And the total force in cable 2 is, F2 = F2t + F2m = Pc/3 - Pc/3 = 0.
Both approaches give exactly the same answer; F2 = 0. Notice in the second approach, I distribute the applied tensile load and moment to the "bolt pattern" separately, using superposition. I.e., I distribute the applied centroidal moment using the familiar linear, elastic beam flexure formula, sigma = M*y/I. (This formula assumes a rigid, planar lamina and assumes a linear stress distribution.) For the applied tensile load, I can't use the flexure formula because the tensile load generates zero moment at the centroid. Therefore I use basic moment equilibrium to distribute the applied tensile load about one axis -- the same method you would use for calculating the reactions on a simply supported beam. This is how you distribute an applied tensile load to, say, a bolt pattern or structure in rigid-body mechanics -- sum moments about the orthogonal, principal axes, and solve for the unknowns. In your problem, we didn't need to sum moments about the other principal axis, because a symmetry condition reduced the number of unknowns. The problem was statically determinate with only two equations and two unknowns. But when the number of unknowns exceeds the number of statics equations, then Hooke's law (component stiffnesses) and compatibility equations must also be utilized to add additional relations, to, e.g., write the plate rotation angle in terms of component elongations.
The assumption of rigid-body mechanics can sometimes be reasonable, depending on the problem, depending on the stiffnesses (and deflections) of the components relative to the magnitude of the applied load. And it's realistic to assume there will be imperfect mass distribution in the plate, imperfect location of the cable attachment points, and/or slight eccentricity in the applied load. Therefore, it's reasonable to assume there will be a finite, minute eccentricity in this problem that generates either a positive or negative minute moment about a line from cable 1 to 3. If this slight eccentricity generates a minute positive moment about line 1-3, then when cable 4 is removed, the plate will rotate almost 90 degrees before coming to rest. And the force in cable 2 will be zero. If, on the other hand, this minute moment about line 1-3 (due to eccentricities) is negative, then when cable 4 is removed, the plate will not move at all (in rigid-body mechanics). And the force in cable 2 will be only a minute tensile force that balances the minute eccentricity moment. I.e., the force in cable 2 will be virtually zero.
The force in this problem is carried by only two of the three cables because the weight of the plate is almost perfectly balanced by gravity about a line from cable 1 to 3, and load P is applied at the midpoint of line 1-3. Therefore, moment equilibrium dictates there can be no force applied to cable 2 (except for a minute tensile force to balance the minute eccentricity moment). And notice that the force in cable 2, F2 = 0, doesn't change if you change the gravitational field, whereas the force in cable 1 does change, because F1 = (P + m*g)/2.
Lcubed, if you now want an answer different from the rigid-body mechanics answer you requested, let us know, because it would give a slightly different answer (though perhaps not significantly different). Sometimes rigid-body mechanics is applicable and is a reasonable approximation, depending on the problem; other times it's not. In your given problem (posts 8 and 10), it's a very reasonable, close approximation (even if the cables are elastic), provided the plate doesn't bend much. If the cables
are elastic (and the plate is still relatively rigid), then when you remove cable 4, as ivymike said, you'll actually see corner 2 rise slightly as the force in cable 2 changes from (P + m*g)/4 to zero; and you'll see corners 1 and 3 move downward slightly as the force in cables 1 and 3
doubles. In other words, the plate will rotate slightly and now be "crooked." But notice, even then, the force in cable 2 is still zero. The plate is "crooked" merely because cable 2 is now too short to let the plate hang perfectly level.
But also notice, once you remove cable 4, you are now in a state of unstable equilibrium; so the plate could rotate about line 1-3 to almost anywhere between, say, zero and 90 degrees and still satisfy equilibrium.
