Kvant Physics Problem 620

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Problem

If an object with mass greater than $m_0$ is placed at the center of a square table, the table legs will break. Find the set of points on the table where an object with mass $\dfrac{m_0}2$ can be placed without risking breaking the table.

O. Batischev

Setup and Assumptions

A rigid square table of side $2a$ is supported by four identical legs located at its corners, modeled as point supports. A point object of mass $m$ is placed at a point $(x,y)$ on the tabletop, where the origin is at the center of the square and the coordinate axes are aligned with the sides. Each leg can withstand a maximum vertical reaction force corresponding to a load $m_0 g$ when the object is placed at the center of the table.

The gravitational acceleration is $g$. The tabletop is assumed rigid, so the load is distributed among the legs purely by static equilibrium. Deformations of the tabletop and legs are neglected, and frictional or horizontal forces are ignored. The system is treated as a statically determinate support problem in which the reaction forces are uniquely determined by equilibrium conditions.

Physical Principles

Static equilibrium requires that the sum of all vertical forces equals zero in the inertial frame of the table, giving $\sum_i R_i = mg$, where $R_i$ are the reactions of the four legs.

For a rigid square plate supported at its corners, the reaction forces under a point load vary linearly with position and satisfy the bilinear interpolation relations obtained from equilibrium of moments about the coordinate axes. If the corners are at $(\pm a, \pm a)$ and the load is applied at $(x,y)$, the reactions are given by

$$R_{\pm,\pm} = \frac{mg}{4}\left(1 \pm \frac{x}{a}\right)\left(1 \pm \frac{y}{a}\right),$$

where the signs correspond to the corner coordinates.

A leg fails when its reaction exceeds the critical value $m_0 g /4$, since at the threshold case of a central load $m_0$, symmetry implies equal sharing among four legs.

Derivation

For a mass $m_0$ placed at the center $(x,y)=(0,0)$, each leg carries a force $m_0 g /4$, which defines the maximum allowable reaction per leg. For a general mass $m=m_0/2$, the reactions become

$$R_{\pm,\pm} = \frac{m_0 g}{8}\left(1 \pm \frac{x}{a}\right)\left(1 \pm \frac{y}{a}\right).$$

The safety condition requires that for all four legs,

$$\frac{m_0 g}{8}\left(1 \pm \frac{x}{a}\right)\left(1 \pm \frac{y}{a}\right) \le \frac{m_0 g}{4}.$$

Dividing by $m_0 g /8$ gives

$$\left(1 \pm \frac{x}{a}\right)\left(1 \pm \frac{y}{a}\right) \le 2.$$

The most heavily loaded leg corresponds to the choice of signs matching the signs of $x$ and $y$, since this maximizes both linear factors simultaneously. The maximal reaction is therefore

$$R_{\max} = \frac{m_0 g}{8}\left(1 + \frac{|x|}{a}\right)\left(1 + \frac{|y|}{a}\right).$$

The safety condition reduces to

$$\left(1 + \frac{|x|}{a}\right)\left(1 + \frac{|y|}{a}\right) \le 2.$$

Expanding this inequality yields

$$1 + \frac{|x|}{a} + \frac{|y|}{a} + \frac{|x||y|}{a^2} \le 2,$$

which simplifies to

$$\frac{|x|}{a} + \frac{|y|}{a} + \frac{|x||y|}{a^2} \le 1.$$

This inequality defines the admissible region on the tabletop.

Result

The set of safe positions satisfies

$$\left(1 + \frac{|x|}{a}\right)\left(1 + \frac{|y|}{a}\right) \le 2.$$

Equivalently,

$$\frac{|x|}{a} + \frac{|y|}{a} + \frac{|x||y|}{a^2} \le 1.$$

The admissible region is

$$\boxed{\left{(x,y): |x|,|y|\le a,\ \left(1 + \frac{|x|}{a}\right)\left(1 + \frac{|y|}{a}\right) \le 2 \right}}.$$

Sanity Checks

At the center, $x=0$ and $y=0$ gives $(1)(1)=1\le 2$, which corresponds to a load $m_0/2$ safely below the failure threshold.

At the edge midpoint, $x=a$ and $y=0$ yields $(2)(1)=2$, which saturates the inequality and corresponds to a marginal case where one leg reaches the critical load $m_0 g /4$.

At a corner, $x=a$ and $y=a$ gives $(2)(2)=4>2$, which violates the condition and predicts failure, consistent with the strong concentration of load onto a single leg.

Dimensional consistency holds since all terms in the inequalities are dimensionless ratios of coordinates to the table half-side $a$. The reaction formula preserves units of force $mg$, ensuring correct scaling of the failure condition. The most sensitive point for error is the identification of the maximal-loaded leg, since an incorrect choice of sign combination would underestimate the peak reaction and enlarge the admissible region incorrectly.