Irodov 1.102: The Friction That Catches Up – Amit & Deepali Sinha Foundation

Problem Irodov 1.102

A small block is released from rest on an incline that makes angle $\alpha$ with the horizontal. Unlike ordinary friction, its coefficient here is not constant but grows in proportion to the distance $x$ already covered, $k = ax$ for some constant $a$ . Find the distance the block slides before the growing friction brings it to rest, and find the greatest speed it reaches at any point along that slide.

The setup: a small bar on a plane inclined at angle α to the horizontal, released from rest at the top of its slide and sliding down the slope. — The setup: a small bar on a plane inclined at angle $\alpha$ to the horizontal, released from rest at the top of its slide and sliding down the slope.

Given

$\alpha$
$k = ax$

Find

Find

s

, the distance the bar slides before stopping, and

v_{\mathrm{max}}

, the greatest speed it reaches over that distance

Solution

Step 1Newton’s second law along the slope

Free body of the bar at a general point x along the slide: the normal N, the weight m g, and the friction f, which opposes the sliding and so points up the slope. — Free body of the bar at a general point $x$ along the slide: the normal $\vect N$ , the weight $m\vect g$ , and the friction $\vect f$ , which opposes the sliding and so points up the slope.

Let $x$ be the distance already slid, measured down the slope from the start, and let the bar have mass $m$ . Perpendicular to the slope there is no acceleration, so the normal force balances the perpendicular component of gravity:

N = mg\cos\alpha.

The friction coefficient at that point is $k = ax$ , so the friction force has magnitude $f = kN = axmg\cos\alpha$ and, since the bar slides down, it points up the slope (see the free body figure above). Newton’s second law along the slope, with the down-slope direction taken as positive $x$ , then reads

m\frac{dv}{dt} = mg\sin\alpha - axmg\cos\alpha,

and dividing through by $m$ ,

\frac{dv}{dt} = g\sin\alpha - agx\cos\alpha \tag{1}

Step 2Change the variable from

t

x

Since $v=dx/dt$ , the chain rule rewrites the acceleration in terms of $x$ instead of $t$ .

\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\,\frac{dv}{dx}.

Substituting into the equation of motion (1),

v\frac{dv}{dx} = g\sin\alpha - agx\cos\alpha,

which separates into

v\,dv = \left(g\sin\alpha - agx\cos\alpha\right)dx.

Step 3Integrate from the start

The bar starts from rest at $x=0$ , so integrate the left side from $0$ to $v$ and the right side from $0$ to $x$ :

\int_0^v v'\,dv' = \int_0^x \left(g\sin\alpha - agx'\cos\alpha\right)dx',

\frac{v^2}{2} = gx\sin\alpha - \frac{1}{2}agx^2\cos\alpha.

Multiplying through by 2 gives $v$ as a function of position,

v^2 = 2gx\sin\alpha - agx^2\cos\alpha

(2)

Speed v against distance slid x (illustrative curve for α=30°, a=1 m^-1, g=9.8 m/s^2). v rises from zero, peaks at x^*, and returns to zero at the stopping distance s. — Speed $v$ against distance slid $x$ (illustrative curve for $\alpha=30^\circ$ , $a=1\,\mathrm{m^{-1}}$ , $g=9.8\,\mathrm{m/s^2}$ ). $v$ rises from zero, peaks at $x^*$ , and returns to zero at the stopping distance $s$ .

By (2), $v^2$ is a downward parabola in $x$ with one root at $x=0$ : the other root is the stopping distance, and the parabola’s vertex, midway between the two roots, is where the speed is greatest, exactly the shape the $v(x)$ plot above shows.

Step 4Stopping distance

The bar stops when $v=0$ again. Setting the right side of (2) to zero at $x=s$ ,

0 = 2gs\sin\alpha - ags^2\cos\alpha = gs\left(2\sin\alpha - as\cos\alpha\right).

Since $s \neq 0$ , divide through by $gs$ :

2\sin\alpha - as\cos\alpha = 0 \quad\Longrightarrow\quad as\cos\alpha = 2\sin\alpha.

Solving for $s$ ,

s = \frac{2}{a}\tan\alpha

(3)

Step 5Where the speed is greatest

Differentiate the right side of (2) with respect to $x$ and set the result to zero:

\frac{d(v^2)}{dx} = 2g\sin\alpha - 2agx\cos\alpha = 0.

Solving for the position $x^*$ of maximum speed,

x^* = \frac{\sin\alpha}{a\cos\alpha} = \frac{\tan\alpha}{a},

which is exactly half of the stopping distance (3), as the symmetry of the parabola in the $v(x)$ plot already shows.

Step 6Maximum speed

Substitute $x^*$ back into (2):

\begin{aligned} v_{\mathrm{max}}^2 &= 2g\sin\alpha\cdot\frac{\tan\alpha}{a} - ag\cos\alpha\cdot\frac{\tan^2\alpha}{a^2} \\ &= \frac{2g\sin\alpha\tan\alpha}{a} - \frac{g\cos\alpha\tan^2\alpha}{a} \\ &= \frac{g\tan\alpha}{a}\left(2\sin\alpha - \cos\alpha\tan\alpha\right) \\ &= \frac{g\tan\alpha}{a}\left(2\sin\alpha - \sin\alpha\right) \\ &= \frac{g}{a}\sin\alpha\tan\alpha. \end{aligned}

Taking the square root,

v_{\mathrm{max}} = \sqrt{\dfrac{g}{a}\sin\alpha\tan\alpha}

(4)

Check

$a$ carries units of inverse length, so that $k=ax$ is dimensionless; then $s=(2/a)\tan\alpha$ comes out in units of length and $v_{\mathrm{max}}^2=(g/a)\sin\alpha\tan\alpha$ in units of $(\mathrm{length/time})^2$ , both as they must. In the limit $a\to0$ (friction never develops) both $s$ and $v_{\mathrm{max}}$ diverge, consistent with the bar sliding forever under the constant acceleration $g\sin\alpha$ with nothing left to bring it to rest.

Remark

The coefficient $k=ax$ grows without bound as the bar slides, and by the stopping point itself it has reached $k(s)=as=2\tan\alpha$ , which already exceeds $1$ once $\alpha>26.57^\circ$ . Nothing in the mechanics breaks (the derivation never assumed $k\le1$ ), but it is a reminder that ”friction coefficient growing linearly with distance” is a mathematical idealization rather than a coefficient bounded the way $\mu<1$ usually is for ordinary material pairs.