Solution to Question 3
Problem Analysis & Constraints
Goal: Find the region on a square plate of side $l$ where a large load $W$ can be placed such that all four suspension wires remain taut.
Geometric Constraint: For a rigid plate, corners A, B, C, D remain coplanar. This implies $z_A + z_C = z_B + z_D$. Since wires are elastic ($T \propto z$), the tensions satisfy:
The diagonals of a square bisect each other at the center. The vertical displacement of the center is the average of the displacements of the diagonal corners.
Since $z_{center} = \frac{z_A + z_C}{2} = \frac{z_B + z_D}{2}$, it follows that $z_A + z_C = z_B + z_D$.
By Hooke’s Law ($T \propto z$), the tensions follow the same relationship.
Force Equilibrium:
Combining these, the sum of diagonal pairs is constant:
$$ T_A + T_C = \frac{W}{2} \quad \text{and} \quad T_B + T_D = \frac{W}{2} $$
Torque Analysis
Let the load be placed at $P(x,y)$ relative to the center. We calculate torques about axes passing through $P$.
Axis parallel to x-axis ($y’ = y$):
The moment arm for line AB is $(l/2 + y)$ and for CD is $(l/2 – y)$.
Solving this yields:
$$ T_A + T_B = \frac{W}{l}\left(\frac{l}{2} – y\right) $$
Axis parallel to y-axis ($x’ = x$):
Similarly for sides AD and BC:
$$ T_A + T_D = \frac{W}{l}\left(\frac{l}{2} – x\right) $$
Deriving the Region
We solve for $T_A$. Using $2T_A + T_B + T_D = (T_A+T_B) + (T_A+T_D)$:
Simplifying for $T_A$:
$$ T_A = \frac{W}{4} – \frac{W(x+y)}{2l} $$
For the wire to remain taut, $T_A \ge 0$, which implies $x + y \le l/2$. Applying this to all four corners:
The region is a square rotated by $45^\circ$, bounded by the lines joining the midpoints of the sides of the plate ($|x| + |y| \le l/2$).