Irodov 1.256: Pulling Down Raises the Friction Ceiling

Problem Irodov 1.256

A uniform solid cylinder of mass $m$ lies with its axis horizontal across two fixed horizontal planks, which act as rails and carry it from below along two lines of its curved surface. A thread is wound on the cylinder, and its free end hangs vertically down in the gap between the planks, where it is pulled straight down with a constant force $F$ . The coefficient of friction between the cylinder and the planks is $k$ . Find the largest magnitude $F_{\mathrm{max}}$ of that force for which the cylinder still turns on the planks without sliding on them, and the acceleration $w_{\mathrm{max}}$ of the cylinder’s axis when the force has that value.

The cylinder on its planks, seen along the axis. The two planks lie one behind the other and project onto the hatched support line, with the thread hanging down in the gap between them. The gray dimension marks the radius R. — The cylinder on its planks, seen along the axis. The two planks lie one behind the other and project onto the hatched support line, with the thread hanging down in the gap between them. The gray dimension marks the radius $R$ .

Given

$m$
$k$
$F$ , constant and vertically down

Find

F_{\mathrm{max}}

that still gives no sliding ·

w_{\mathrm{max}}

of the axis

Solution

Step 1Newton’s law for the cylinder, in two projections

Four forces act: the pull $F$ transmitted by the thread, the weight $mg$ at the axis, and, at the contacts with the planks, the total normal force $N$ and the total friction force $f$ . The axis rides along the planks and so has no vertical acceleration, which fixes $N$ at once.

N = mg + F. \tag{1}

Horizontally the thread and the weight contribute nothing, so the friction from the planks is the only force that can accelerate the axis, with acceleration $w$ .

f = m w. \tag{2}

The free body. The normal force and friction act at the plank contact, the weight at the axis, and the pull F where the thread leaves the rim. The curved arrow is the angular acceleration β, and v the direction of travel. — The free body. The normal force and friction act at the plank contact, the weight at the axis, and the pull $\vect F$ where the thread leaves the rim. The curved arrow is the angular acceleration $\beta$ , and $\vect v$ the direction of travel.

Step 2The torque equation about the axis

Both $F$ and $f$ have the same moment arm $R$ about the axis, since the thread leaves the rim tangentially and the contact lies on the rim as well, and they turn the cylinder in opposite senses. The weight and the normal force have no moment about the axis. With $I = mR^2/2$ for a uniform solid cylinder and $\beta$ its angular acceleration,

\begin{aligned} I\beta &= FR - fR \\ \frac{mR^2}{2}\,\beta &= (F - f)R. \end{aligned} \tag{3}

Step 3Rolling without sliding closes the system

Rolling on the fixed planks means the contact line is instantaneously at rest, so the axis speed and the spin are locked together and their rates are too. Substituting $w = \beta R$ into the torque equation (3), and then (2) for $mw$ , eliminates both $R$ and the kinematics.

\begin{align*}\frac{mR^2}{2}\cdot\frac{w}{R} &= (F - f)R \\ \frac{mw}{2} &= F - f \\ \frac{f}{2} &= F - f \\ \frac{3f}{2} &= F\end{align*}

f = \dfrac{2F}{3}

(4)

so exactly two thirds of the pull is passed to the planks as friction, whatever the cylinder’s radius. The acceleration of the axis follows from (2).

w = \dfrac{2F}{3m}

(5)

Step 4The largest pull the planks can hold without sliding

The planks can supply only as much friction as their roughness allows, and the ceiling is $kN$ with $N$ from (1). Rolling therefore survives exactly while the demanded friction (4) stays under that ceiling.

\begin{align*}f &\le kN \\ \frac{2F}{3} &\le k\,(mg + F) \\ 2F &\le 3kmg + 3kF \\ F\,(2 - 3k) &\le 3kmg\end{align*}

F_{\mathrm{max}} = \dfrac{3kmg}{2 - 3k}

(6)

The division assumes $3k < 2$ ; the case $k \ge 2/3$ is taken up in the Remark.

Step 5The acceleration of the axis at that pull

The largest acceleration reached without sliding is what (5) returns at $F = F_{\mathrm{max}}$ .

\begin{align*}w_{\mathrm{max}} &= \frac{2F_{\mathrm{max}}}{3m} \\ &= \frac{2}{3m}\cdot\frac{3kmg}{2 - 3k}\end{align*}

w_{\mathrm{max}} = \dfrac{2kg}{2 - 3k}

(7)

Check

At $F = F_{\mathrm{max}}$ the two sides of the sliding condition must meet exactly. The demanded friction is $2F_{\mathrm{max}}/3 = 2kmg/(2-3k)$ , while the available friction is $k(mg + F_{\mathrm{max}}) = kmg\,[(2-3k) + 3k]/(2-3k) = 2kmg/(2-3k)$ : the same number, as the boundary case requires. Both answers are also dimensionally sound, $F_{\mathrm{max}}$ being a mass times $g$ and $w_{\mathrm{max}}$ an acceleration, and both vanish with $k$ , since planks that cannot rub cannot make the cylinder roll at all.

Remark

Both the friction demanded and the friction available grow with $F$ , which is what makes the threshold a close-run thing. Pulling harder loads the planks harder, so the ceiling $k(mg+F)$ climbs with slope $k$ , while the demand $2F/3$ climbs with slope $2/3$ : the pull feeds its own grip, and the question is only which of the two lines climbs faster. The demand overtakes the ceiling at a finite $F$ only when $k < 2/3$ , and for $k \ge 2/3$ the ceiling rises at least as fast as the demand, so the cylinder rolls without sliding under any pull whatever, however large, which is exactly what the divergence of $3kmg/(2-3k)$ as $k \to 2/3$ is announcing.