Irodov 1.141: Elastic Cord Angle Signals Slipping Onset

Problem Irodov 1.141

A plank rests on a horizontal plane. A bar of mass $m=1.0\text{ kg}$ sits on top of the plank and is joined by a light elastic cord to a fixed point $O$ above the plane; the cord’s unstretched (natural) length is $l_0=40\text{ cm}$ , and initially it hangs vertically from $O$ to the bar. The coefficient of friction between the bar and the plank is $k=0.20$ . The plank is then pulled slowly (quasi-statically) to the right; friction between the bar and the plank drags the bar along with it, which stretches and tilts the cord away from vertical. The bar begins to slide relative to the plank exactly when the cord has tilted to an angle $\theta=30^\circ$ from the vertical. Find the work done by the friction force on the bar, in the frame fixed to the plane, from the start of the motion up to that instant.

The bar m sits on the plank, joined by the elastic cord of natural length l_0 to the fixed point O above; the plank is pulled slowly to the right, dragging the bar with it through friction at the plank's upper face. — The bar $m$ sits on the plank, joined by the elastic cord of natural length $l_0$ to the fixed point $O$ above; the plank is pulled slowly to the right, dragging the bar with it through friction at the plank’s upper face.

Given

$m=1.0\text{ kg}$
$l_0=40\text{ cm}$
$k=0.20$
$\theta=30^\circ$

Find

the work done by friction on the bar up to the onset of sliding

Solution

Step 1Quasi-static force balance on the bar

At a general angle $\theta$ the cord’s tension $T$ pulls the bar up and back toward $O$ , so its vertical component reduces the normal force and its horizontal component must be balanced by friction, as the free body in the next figure shows.

Free body of the bar at angle θ from vertical, forces concurrent at its center of mass: tension T along the cord, weight m g, the plank's normal reaction N, and the friction f that keeps the bar moving with the plank. — Free body of the bar at angle $\theta$ from vertical, forces concurrent at its center of mass: tension $\vect T$ along the cord, weight $m\vect g$ , the plank’s normal reaction $\vect N$ , and the friction $\vect f$ that keeps the bar moving with the plank.

\begin{aligned} N + T\cos\theta &= mg \\ f &= T\sin\theta \end{aligned} \tag{1}

Step 2The tension at the critical angle

Sliding begins exactly when the required friction reaches its maximum, $f=kN$ . Substituting (1) into this condition,

\begin{align*}T\sin\theta &= k(mg-T\cos\theta) \\ T(\sin\theta+k\cos\theta) &= kmg\end{align*}

T = \dfrac{kmg}{\sin\theta+k\cos\theta}

(2)

Step 3The cord’s elongation at that angle

With the vertical distance from $O$ to the bar fixed at $l_0$ , the cord’s stretched length at angle $\theta$ is $l_0/\cos\theta$ , so its elongation beyond the natural length $l_0$ is

\begin{aligned} \Delta l(\theta) &= \frac{l_0}{\cos\theta}-l_0 \\ &= l_0\,\frac{1-\cos\theta}{\cos\theta}. \end{aligned} \tag{3}

Step 4Work done by friction, from the work-energy theorem

The bar’s speed is negligible throughout this quasi-static process, so the net work done on it by every force, from the start to the critical angle, is zero. Both $N$ and $mg$ are vertical while the bar’s motion is purely horizontal, so neither does any work; the elastic tension is conservative, doing work equal to minus the elastic energy it stores; friction supplies the rest.

W_N + W_{mg} + W_T + W_f = 0 \;\Longrightarrow\; 0+0-U_{\mathrm{el}}(\theta)+W_f = 0.

W_f = U_{\mathrm{el}}(\theta)

(4)

Step 5The stored elastic energy at the critical angle

For a linear elastic cord, the energy stored at elongation $\Delta l$ under tension $T$ is $T\Delta l/2$ . Substituting the tension (2) and the elongation (3),

\begin{align*}U_{\mathrm{el}}(\theta) &= \frac{T\,\Delta l(\theta)}{2} \\ &= \frac{1}{2}\cdot\frac{kmg}{\sin\theta+k\cos\theta}\cdot l_0\,\frac{1-\cos\theta}{\cos\theta}\end{align*}

Using (4), this is the work done by friction, evaluated numerically with $g=9.8\text{ m/s}^2$ , $\cos30^\circ=\sqrt3/2$ and $\sin30^\circ=1/2$ .

A = \dfrac{kmgl_0}{2}\cdot\dfrac{1-\cos\theta}{(\sin\theta+k\cos\theta)\cos\theta} = 0.09\text{ J}

(5)

Check

Dimensionally $kmgl_0$ is a force times a length, an energy, matching the answer. As $k\to0$ (a frictionless plank), the boxed formula’s numerator $k\to0$ while the bar could never be dragged along in the first place, consistent with there being no friction to do any work; and as $\theta\to0$ , $1-\cos\theta\to0$ so $A\to0$ , matching a critical angle that never develops because sliding would begin immediately.

Remark

The friction force itself changes throughout the process, from zero at $\theta=0$ (no horizontal pull to resist yet) up to its maximum $kN$ only at the very end; Step 4 sidesteps integrating that varying force over the bar’s actual path by using energy conservation instead, converting the unknown work $W_f$ into the elastic energy $U_{\mathrm{el}}$ , which Step 5 then evaluates from $T$ and $\Delta l$ alone. The cord’s unstated spring constant cancels because $U_{\mathrm{el}}=T\Delta l/2$ needs only the tension and elongation AT the critical angle, both already fixed by Steps 2 and 3, never the spring constant or the path between the start and that angle.