Kvant Physics Problem 183

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Problem

A dynamometer that slides on a smooth horizontal table is pulled with a constant force $F=4~\text{Н}$. What does the dynamometer read if the mass of its spring is equal to the mass of its casing, and the dynamometer was calibrated in the horizontal position?

G. L. Kotkin

Setup and Assumptions

A dynamometer consists of a casing of mass $m_c$ and a spring of mass $m_s$ inside it. The device is pulled along a smooth horizontal table by a constant external force $F=4~\text{N}$. The table is frictionless, so horizontal external forces act only through the pull.

The system accelerates as a rigid body, but internal stresses in the massive spring are nonuniform. The dynamometer is calibrated in a horizontal static configuration, meaning that when $F$ is applied quasistatically and the spring mass is irrelevant, the reading equals the spring tension.

The casing is assumed rigid, the spring is uniform with constant linear mass density, and the force is applied at one end of the spring. The reading of the dynamometer is identified with the tension transmitted to the casing end of the spring.

The given condition is $m_s=m_c$.

Physical Principles

Newton’s second law for the motion of the whole device gives

$F = (m_c + m_s)a.$

For an extended massive spring, the tension is not constant along its length during acceleration. In the non-inertial frame moving with acceleration $a$, each small element of mass experiences an effective body force $-\mathrm{d}m,a$, which produces a spatial gradient of tension.

For a uniform spring of mass $m_s$, the difference between tensions at its ends equals the total inertial force on the spring,

$T_{\text{pull}} - T_{\text{case}} = m_s a.$

The casing is acted on by the spring force, so its equation of motion is

$T_{\text{case}} = m_c a.$

Derivation

The casing is connected to one end of the spring, and the external force $F$ is applied to the opposite end. The spring end at the pull side transmits the external force, so

$T_{\text{pull}} = F.$

Across the spring, the tension drop equals the total inertial force of the spring mass,

$T_{\text{pull}} - T_{\text{case}} = m_s a.$

Substituting $T_{\text{pull}}=F$ gives

$F - T_{\text{case}} = m_s a.$

The casing is accelerated by the spring force, hence

$T_{\text{case}} = m_c a.$

Substituting this into the previous relation yields

$F - m_c a = m_s a,$

so

$F = (m_c + m_s)a,$

and therefore

$a = \frac{F}{m_c + m_s}.$

The dynamometer reading equals the force transmitted to the casing, hence

$T_{\text{case}} = m_c a = m_c \frac{F}{m_c + m_s}.$

Using $m_s=m_c$,

$T_{\text{case}} = m_c \frac{F}{2m_c} = \frac{F}{2}.$

Result

The dynamometer reading is

$T = \frac{F}{2}.$

Substituting $F=4~\text{N}$ gives

$T = \frac{4~\text{N}}{2} = 2~\text{N}.$

The final answer is

$\boxed{2~\text{N}}.$

Sanity Checks

The expression $T = m_c F/(m_c+m_s)$ has dimensions of force since it is a product of a mass ratio and force, ensuring dimensional consistency.

In the limit $m_s \to 0$, the result reduces to $T \to F$, matching the behavior of an ideal massless spring dynamometer. In the opposite limit $m_s \gg m_c$, the reading tends to zero, consistent with most of the force accelerating the spring mass rather than being transmitted to the casing.

The acceleration $a=F/(m_c+m_s)$ is consistent with the total inertia of the system, and any error would most likely arise from incorrectly assuming uniform tension in the massive spring, which would incorrectly give $T=F$ instead of $T=F/2$ for equal masses.