Irodov 1.65: Friction's Ceiling Freezes the Plank – Amit & Deepali Sinha Foundation

Problem Irodov 1.65

A bar of mass $m_2$ rests on top of a plank of mass $m_1$ , and the plank itself rests on a smooth (frictionless) horizontal floor. A horizontal force that grows with elapsed time as $F = at$ , for a constant $a$ , is applied to the bar, and the coefficient of friction between the plank and the bar is $k$ . Find how the plank’s acceleration $w_1$ and the bar’s acceleration $w_2$ each depend on $t$ , and sketch the approximate shape of both curves.

The bar (mass m_2) rides on the plank (mass m_1), which stands on a smooth floor, so only the bar-plank interface carries friction. The applied force F grows with time and acts on the bar. — The bar (mass $m_2$ ) rides on the plank (mass $m_1$ ), which stands on a smooth floor, so only the bar-plank interface carries friction. The applied force $\vect F$ grows with time and acts on the bar.

Given

$m_1$
$m_2$
$F=at$
$k$

Find

w_1(t)

w_2(t)

Solution

Step 1While the bodies move as one

As long as friction can hold the bar to the plank, the two move with a common acceleration $w$ . The floor is smooth, so the only horizontal force on the whole system is $F$ itself, and Newton’s second law for the combined mass gives

(m_1+m_2)w = at.

w = \dfrac{at}{m_1+m_2}

(1)

Step 2The friction the plank actually needs

The friction pair at the bar-plank interface, which this step introduces: the bar drags the plank forward with f and the plank pulls back on the bar with - f, equal and opposite. The floor under the plank is smooth and exerts nothing horizontal. — The friction pair at the bar-plank interface, which this step introduces: the bar drags the plank forward with $\vect f$ and the plank pulls back on the bar with $-\vect f$ , equal and opposite. The floor under the plank is smooth and exerts nothing horizontal.

The floor exerts no horizontal force on the plank, so the plank’s own acceleration can only come from the friction the bar exerts on it. Newton’s second law for the plank alone reads

m_1 w = f.

Substituting the common acceleration (1) turns that into the friction as an explicit function of time.

f = \frac{m_1 a t}{m_1+m_2}.

Step 3Where static friction runs out

The bar presses on the plank with its own weight, so the interface carries a normal force $N=m_2 g$ and the friction between them cannot exceed $kN=km_2g$ . The locked motion of Step 1 holds only while the demanded friction of Step 2 stays under that ceiling,

\frac{m_1 a t}{m_1+m_2} \le km_2g.

Setting the two sides equal gives the time $t_0$ at which slipping begins.

t_0 = \dfrac{km_2g(m_1+m_2)}{am_1}

(2)

Step 4Two decoupled equations once slipping starts

For $t\ge t_0$ (Eq. (2)) the bar slides over the plank, so the friction between them is kinetic and pinned at its limiting value $km_2g$ : forward on the plank, backward on the bar. The two bodies now obey separate equations of motion.

\begin{align*} m_1 w_1 = km_2g, \qquad m_2 w_2 = at-km_2g \tag{3}\end{align*}

Dividing each by its own body’s mass gives the two accelerations in the slipping regime.

\begin{aligned} w_1 &= \dfrac{km_2g}{m_1}, & w_2 &= \dfrac{at-km_2g}{m_2}. \end{aligned}

Step 5Assembling the plots

Both accelerations start at the origin along the same line of slope $a/(m_1+m_2)$ , split apart at $t_0$ , and past it, by (3), $w_2$ climbs away with the steeper slope $a/m_2$ while $w_1$ levels off.

The two accelerations against time: w_1 (gold) rises with the pair, then holds constant past t_0; w_2 (royal) rises with the pair, then climbs more steeply, the two curves meeting at t_0 and diverging afterward. — The two accelerations against time: $w_1$ (gold) rises with the pair, then holds constant past $t_0$ ; $w_2$ (royal) rises with the pair, then climbs more steeply, the two curves meeting at $t_0$ and diverging afterward.

\begin{aligned} w_1 &= \begin{cases} \dfrac{at}{m_1+m_2}, & t\le t_0\\[6pt] \dfrac{km_2g}{m_1}, & t\ge t_0\end{cases} \\[8pt] w_2 &= \begin{cases} \dfrac{at}{m_1+m_2}, & t\le t_0\\[6pt] \dfrac{at-km_2g}{m_2}, & t\ge t_0\end{cases} \end{aligned}

(4)

For $t\le t_0$ the bar and plank move together: friction is strong enough to drag the plank along, and both share the acceleration $at/(m_1+m_2)$ .
For $t\ge t_0$ the bar slips: friction is pinned at its kinetic value $km_2g$ , so the plank coasts at the constant rate $km_2g/m_1$ while the bar’s own acceleration keeps rising as $(at-km_2g)/m_2$ .

Check

In the frictionless limit $k\to0$ , Eq. (2) gives $t_0\to0$ : slipping starts at once, $w_1\to0$ for every $t$ (nothing can push the plank at all) and $w_2\to at/m_2$ (the bar alone under $F$ ), exactly the free bar on a frictionless plank.

Remark

The two regimes are two different one-body problems stitched together at $t_0$ : before it, the pair moves as a single rigid body of mass $m_1+m_2$ under $F$ ; after it, friction is pinned at its ceiling and the bar effectively decouples, so $w_2$ becomes the acceleration of the bar ALONE under $F=at$ , reduced by the constant drag $km_2g$ that the plank keeps pulling back on it. The plank, meanwhile, stops caring about $F$ altogether and just coasts at whatever constant acceleration that same friction ceiling can still give it.