Kvant Physics Problem 77

Verified: yes
Verdicts: PASS + PASS
Solve time: 5m52s
Source on kvant.digital

Problem

F77. A rectangular block whose height is much greater than its length and width stands on a horizontal surface. How can one determine the coefficient of friction between the block and the surface using only a single measuring instrument—a ruler?

Setup and Assumptions

A rectangular block of height $H$ and base dimensions $a \times b$ stands on a horizontal rigid surface. The coefficient of static friction between the block and the surface is $\mu$. The only measuring instrument available is a ruler.

The block is homogeneous, so its center of mass lies at its geometric center. Gravitational acceleration is $g$. The surface is rigid and horizontal, and the contact is described by Coulomb friction with a uniform coefficient $\mu$.

The key feature is that the block is very tall compared to its base dimensions, so it can either slide or tip under a horizontal force applied at different heights.

Physical Principles

Static equilibrium holds as long as the resultant force and resultant torque on the block vanish, and the friction force at the base satisfies $F_f \le \mu N$.

Sliding begins when the applied horizontal force $F$ reaches the maximum static friction force

$F_{\text{slide}} = \mu mg.$

Tipping begins when the torque of the applied force about the edge of the base exceeds the restoring torque of gravity about that same edge. If the force is applied horizontally at height $h$, the tipping threshold satisfies

$F_{\text{tip}} \cdot h = mg \cdot \frac{a}{2},$

since the center of mass is horizontally displaced by $a/2$ from the pivot edge.

This comparison between two critical forces determines whether the block slides or tips first.

Derivation

A horizontal force is applied to the block using the ruler, with the contact point chosen at a controlled height $h$ above the base. The ruler provides no force measurement, but it allows controlled application of the force and accurate measurement of the height $h$.

For a fixed height $h$, two instability thresholds exist.

The sliding threshold is

$F_{\text{slide}} = \mu mg.$

The tipping threshold is obtained from torque balance about the lower edge of the base,

$F_{\text{tip}} h = mg \frac{a}{2},$

which gives

$F_{\text{tip}} = \frac{mg,a}{2h}.$

The actual failure mode corresponds to the smaller of these two forces. If $F_{\text{slide}} < F_{\text{tip}}$, the block slides first; if $F_{\text{tip}} < F_{\text{slide}}$, it tips first.

The transition between the two regimes occurs when both thresholds coincide,

$\mu mg = \frac{mg,a}{2h}.$

Canceling $mg$ yields

$\mu = \frac{a}{2h}.$

The experiment is carried out by varying the height $h$ at which the ruler applies the horizontal force. There exists a critical height $h^\ast$ separating two observable behaviors: for $h < h^\ast$ the block tips first, and for $h > h^\ast$ it slides first.

At this boundary,

$\mu = \frac{a}{2h^\ast}.$

The value $h^\ast$ is determined directly by measurement with the ruler.

Result

The coefficient of static friction is determined by measuring the critical application height $h^\ast$ at which the block changes from tipping-first to sliding-first under a horizontal push from the ruler. The result is

$\boxed{\mu = \frac{a}{2h^\ast}},$

where $a$ is the base dimension of the block in the direction of pushing and $h^\ast$ is the measured critical height.