Irodov 1.123: The Turning Point Sets the Minimum – Amit & Deepali Sinha Foundation

Problem Irodov 1.123

Two bars of masses $m_1$ and $m_2$ lie on a horizontal surface, joined by a light spring that is initially at its natural, unstretched length. Between each bar and the surface the coefficient of friction is $k$ . A constant force is applied to the bar of mass $m_1$ , horizontally along the line joining the bars. Find the minimum value of that force which will eventually set the bar of mass $m_2$ into motion.

The two bars joined by the initially unstretched spring, with the constant force F applied to m_1 so as to stretch it; the coefficient of friction k is the same under both bars. — The two bars joined by the initially unstretched spring, with the constant force $\vect F$ applied to $m_1$ so as to stretch it; the coefficient of friction $k$ is the same under both bars.

Given

$m_1$
$m_2$
$k$

Find

F_{\min}

Solution

Step 1The net force stretching the spring

While $m_2$ is still at rest, $m_1$ moves under three forces along the spring’s line: the applied force $F$ , the spring’s own restoring pull, and kinetic friction opposing its motion. Writing $x$ for the spring’s stretch beyond its natural length and $\kappa$ for its stiffness gives the net force on $m_1$ as a function of $x$ .

F_{\mathrm{net}}(x) = F-\kappa x-km_1g \tag{1}

Step 2Kinetic energy of

m_1

as it stretches the spring

The work-energy theorem, applied from release, where $m_1$ is at rest, to a general stretch $x$ , integrates the net force (1) over the displacement.

\begin{align*}\frac12m_1v^2-0 &= \int_0^xF_{\mathrm{net}}(x')\,dx' \\ &= \int_0^x\bigl[F-\kappa x'-km_1g\bigr]\,dx'\end{align*}

\frac12m_1v^2 = (F-km_1g)\,x-\frac{\kappa x^2}{2}

(2)

Step 3The stretch needed to drag the second bar

The second bar begins to slide the instant the spring’s pull on it reaches its own limiting friction. This fixes the critical stretch $x_0$ , independent of $F$ .

\kappa x_0 = km_2g

(3)

Step 4The minimum force

Left alone, $m_1$ ’s stretch grows until its speed, (2), returns to zero at the greatest extension the applied force can ever produce; call it $x_{\max}$ . The least force that succeeds is the one for which $x_{\max}$ lands exactly on the threshold stretch $x_0$ of (3), so setting $v=0$ at $x=x_0$ in (2) and eliminating $\kappa x_0$ with that same relation gives $F_{\min}$ .

\begin{align*}0 &= (F-km_1g)\,x_0-\frac{\kappa x_0^2}{2} \\ F-km_1g &= \frac{\kappa x_0}{2} \\ &= \frac{km_2g}{2}\end{align*}

F_{\min} = kg\left(m_1+\dfrac{m_2}{2}\right)

(4)

Check

Dimensionally $[k][m_1][g]=(\text{dimensionless})(\mathrm{kg})(\mathrm{m/s^2})=\mathrm N$ , a force, as required. As $m_2\to0$ , $F_{\min}\to km_1g$ , exactly the force needed to overcome $m_1$ ’s own limiting friction, since a vanishingly light second bar needs no stretch at all to be dragged.

Remark

The spring’s stiffness $\kappa$ never entered the final result: it appears only as the product $\kappa x_0$ , which (3) always eliminates in favor of $km_2g$ . The same threshold-extension argument therefore gives the same $F_{\min}$ for any linear spring joining the two bars, whatever its stiffness (a nonlinear, cubic restoring force would not cancel the same way, and gives $m_2/4$ in place of $m_2/2$ ); a stiffer spring only reaches the threshold pull $km_2g$ after a smaller stretch $x_0=km_2g/\kappa$ , not after a different force.