Kvant Physics Problem 375

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Problem

On a horizontal surface there are two blocks with masses $m_1$ and $m_2$, connected by an undeformed spring. What is the minimum horizontal force $F$ that must be applied to one of the blocks so that the second block also starts to move? The coefficient of friction between the blocks and the surface is $k$.

G. Kotkin

Setup and Assumptions

Two blocks of masses $m_1$ and $m_2$ rest on a horizontal rough surface with coefficient of friction $k$, which acts as the coefficient of static friction limiting the onset of motion. The blocks are connected by a massless undeformed spring aligned horizontally along the line of motion. A horizontal force $F$ is applied to one of the blocks, taken without loss of generality to be $m_1$.

The spring is initially at its natural length, so the initial spring force is zero. The motion is analyzed in an inertial reference frame fixed to the ground. Gravitational acceleration is $g$. The spring is assumed ideal, with no mass and linear response, but its stiffness is irrelevant in the limiting condition since only the threshold of motion of the second block is required.

The unknown is the minimum force $F_{\min}$ such that the second block $m_2$ begins to move.

Physical Principles

The maximum static friction force on each block is

$$f_{s,i}^{\max} = k m_i g.$$

Newton’s second law for each block along the horizontal direction is

$$\sum F_x = m_i a.$$

At the threshold of motion, the static friction force attains its limiting value while acceleration remains vanishingly small in the limiting configuration used to determine the onset condition.

The spring force is uniform in magnitude for both blocks and transmits tension between them, acting in opposite directions on the two masses.

Derivation

The second block $m_2$ begins to move when the spring force $T$ reaches the maximum static friction force acting on it. At the threshold,

$$T = k m_2 g.$$

Before this point, the second block remains at rest, so the spring extension adjusts to transmit exactly this force. The first block is pulled by the external force $F$, opposed by both its friction and the spring tension.

At the instant that the second block is about to start moving, the system is still at the verge of static equilibrium, so the net force on the first block is zero in this limiting configuration. The forces acting on $m_1$ are the applied force $F$ in the direction of motion, the static friction force $k m_1 g$ opposing motion, and the spring tension $T$ also opposing motion of $m_1$.

Force balance on $m_1$ at the threshold gives

$$F = k m_1 g + T.$$

Substituting the threshold condition for the spring tension,

$$F = k m_1 g + k m_2 g.$$

Factoring common terms yields

$$F = k g (m_1 + m_2).$$

This value corresponds to the minimum applied force required to bring the system to the point where the second block loses static equilibrium and begins to move.

Result

The minimum force required is

$$F_{\min} = k g (m_1 + m_2).$$

Substitution of units gives $k$ dimensionless, $g$ in $\mathrm{m/s^2}$, and masses in $\mathrm{kg}$, so the force is in newtons:

$$F_{\min} = k (m_1 + m_2) g \ \mathrm{N}.$$

Sanity Checks

The expression has correct dimensional consistency since $kg \cdot m/s^2$ yields newtons.

In the limit $m_2 \to 0$, the result reduces to $F_{\min} = k m_1 g$, which corresponds to the force needed just to start sliding a single block, matching the expected single-body friction threshold.

In the limit $m_1 \to 0$, the result reduces to $F_{\min} = k m_2 g$, consistent with the force needed to initiate motion of the second block through the spring.

The critical step where an error would most easily arise is the force balance on the first block at the threshold. Neglecting the spring force or assuming kinetic instead of static friction would incorrectly change the threshold condition and produce an underestimate of the required force.