Kvant Physics Problem 555

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Verdicts: SKIP + SKIP
Solve time: 2m45s
Source on kvant.digital

Problem

F555. (a) A light rigid rod of length $l$ with a load of mass $m$ at its upper end is hinged at its lower end and held in a vertical position by two horizontal springs of stiffness $k$ attached to its upper end (Fig. 1). The rod cannot move in the direction perpendicular to the springs. For what value of the load mass will the vertical position of the rod cease to be stable?

(b) A steel wire of length $l$ is fixed into the floor so that it stands vertically. If a small horizontal force $F$ is applied to its upper end, the upper end of the wire is displaced by $a \ll l$. Estimate the mass of the load that the wire can support at its upper end without bending. Neglect the difference between the shapes assumed by the wire under the action of horizontal and vertical forces.

Setup and Assumptions

Part (a) concerns a light rigid rod of length $l$ hinged at its lower end. A load of mass $m$ is attached to the upper end. Two identical horizontal springs, each with stiffness $k$, are attached to the upper end and hold the rod in the vertical position. The rod can rotate in the plane of the springs but cannot move in the perpendicular direction. The rod itself is massless. The springs obey Hooke's law and are initially unstretched when the rod is vertical.

The unknown is the value of the mass $m$ at which the vertical position loses stability.

Part (b) concerns a vertical steel wire of length $l$. A small horizontal force $F$ applied at the upper end produces a horizontal displacement $a$, where $a \ll l$. The wire is treated as an elastic element. The unknown is the maximum load mass that can be placed at the upper end without causing buckling. The difference between the wire shapes under horizontal and vertical loading is neglected.

Gravity acts with acceleration $g$.

Physical Principles

The stability of an equilibrium position is determined by the potential energy. An equilibrium is stable if the potential energy has a minimum. For a small angular displacement $\theta$ from equilibrium, the coefficient of $\theta^2$ in the expansion of the total potential energy must be positive.

The elastic potential energy of a spring is

$$U_{\text{spring}}=\frac{kx^2}{2},$$

where $x$ is the extension or compression.

The gravitational potential energy of a mass $m$ at height $h$ is

$$U_g=mgh.$$

For a compressed vertical elastic member, loss of stability occurs at the Euler critical load

$$P_{\text{cr}}=\frac{\pi^2 EI}{4l^2},$$

for a rod fixed at one end and free at the other. Here $E$ is Young's modulus and $I$ is the second moment of area.

For a small horizontal force applied to the free end of such a rod, the displacement is

$$a=\frac{Fl^3}{3EI}.$$

This relation follows from elementary beam theory.

Derivation

For part (a), let the rod be deflected through a small angle $\theta$ from the vertical.

The upper end then moves horizontally by

$$x=l\sin\theta.$$

Each spring changes its length by $x$. The total elastic energy of the two springs is

$$U_s=2\cdot\frac{kx^2}{2}=kx^2 =kl^2\sin^2\theta.$$

The height of the mass above the hinge is

$$h=l\cos\theta.$$

Hence the gravitational potential energy is

$$U_g=mgl\cos\theta.$$

The total potential energy is

$$U(\theta)=kl^2\sin^2\theta+mgl\cos\theta.$$

For small angles,

$$\sin\theta \approx \theta, \qquad \cos\theta \approx 1-\frac{\theta^2}{2}.$$

Substituting,

$$U(\theta) \approx kl^2\theta^2 + mgl\left(1-\frac{\theta^2}{2}\right).$$

Collecting terms,

$$U(\theta) \approx mgl + \left( kl^2-\frac{mgl}{2} \right)\theta^2.$$

The vertical position is stable when the coefficient of $\theta^2$ is positive:

$$kl^2-\frac{mgl}{2}>0.$$

The limiting case corresponds to vanishing coefficient,

$$kl^2-\frac{mgl}{2}=0.$$

Solving for $m$,

$$m=\frac{2kl}{g}.$$

This is the critical mass.

For part (b), let the critical load be $P_{\text{cr}}=mg$.

The beam-deflection formula gives

$$a=\frac{Fl^3}{3EI},$$

hence

$$EI=\frac{Fl^3}{3a}.$$

Substituting this into Euler's formula,

$$P_{\text{cr}} = \frac{\pi^2}{4l^2} \cdot \frac{Fl^3}{3a}.$$

After simplification,

$$P_{\text{cr}} = \frac{\pi^2Fl}{12a}.$$

Since $P_{\text{cr}}=mg$,

$$m = \frac{\pi^2Fl}{12ag}.$$

This is the required estimate for the maximum supported mass.

Result

For part (a), the vertical position ceases to be stable when

$$\boxed{m_{\text{cr}}=\frac{2kl}{g}}.$$

For part (b), the maximum load mass that can be supported without buckling is

$$\boxed{ m_{\max} = \frac{\pi^2Fl}{12ag} }.$$

No numerical values are provided in the statement, so the answers remain in symbolic form.

Sanity Checks

For part (a), the expression

$$m_{\text{cr}}=\frac{2kl}{g}$$

has dimensions

$$\frac{(N/m),m}{m/s^2} = \frac{N}{m/s^2} = kg,$$

which is the correct dimension for mass.

If $k\to 0$, then

$$m_{\text{cr}}\to 0.$$

Without springs, the inverted rod is unstable for any positive load. If $k$ increases, the springs provide stronger restoring action and the critical mass increases proportionally.

For part (b),

$$\frac{Fl}{ag} = \frac{N\cdot m}{m\cdot m/s^2} = kg,$$

so the dimensions are correct.

A stiffer wire corresponds to a smaller displacement $a$ under the same force $F$. The formula predicts a larger admissible mass, which agrees with physical intuition. Increasing the wire length decreases stability because the critical load varies as $1/l^2$ at fixed stiffness $EI$; after eliminating $EI$ through the measured deflection, the resulting estimate scales as $l/a$, consistent with the beam-deflection relation.

The most common source of error in part (a) is the sign of the gravitational contribution. The mass is above the hinge, so a small tilt lowers its height, producing the term

$$-\frac{mgl}{2}\theta^2,$$

which destabilizes the vertical position. Reversing this sign would give an incorrect stability condition. In part (b), the factor $1/4$ in Euler's formula for a member fixed at one end and free at the other is the numerical factor most easily omitted.