Irodov 1.103: Waiting Out the Friction Threshold – Amit & Deepali Sinha Foundation

Problem Irodov 1.103

A body of mass $m$ rests on a horizontal plane, the friction coefficient between them being $k$ . Starting at the moment $t=0$ , a horizontal force $\vect F=\vect a t$ is applied to it, where $\vect a$ is a constant vector. Find the distance the body has covered during the first $t$ seconds after the force starts to act.

The block of mass m on the horizontal plane, with the one force the statement names: the horizontal F= a t, which grows in proportion to the time elapsed since it was switched on. — The block of mass $m$ on the horizontal plane, with the one force the statement names: the horizontal $\vect F=\vect a t$ , which grows in proportion to the time elapsed since it was switched on.

Given

$m$
$k$
$\vect F = \vect a t$

Find

The distance

s

covered during the first

t

seconds after the force begins

Solution

Step 1The body does not move until the force overcomes friction

Free body of the block, with all four forces drawn concurrent at the center of mass: the normal N, the weight m g, the applied F= a t and the friction f, which opposes the push. — Free body of the block, with all four forces drawn concurrent at the center of mass: the normal $\vect N$ , the weight $m\vect g$ , the applied $\vect F=\vect a t$ and the friction $\vect f$ , which opposes the push.

Perpendicular to the plane there is no acceleration, so the normal force balances gravity, $N=mg$ , and the maximum friction available is $kmg$ . The applied force $F=at$ grows from zero, so the body stays put until $F$ first reaches that maximum.

a t_0 = kmg

t_0 = \dfrac{kmg}{a}

(1)

Step 2Newton’s second law once the body is sliding

For $t\ge t_0$ the body slides, so kinetic friction has the constant magnitude $kmg$ and opposes the motion, i.e. it acts against $\vect F$ . Newton’s second law along the direction of $\vect F$ reads

m\frac{dv}{dt} = at - kmg.

Integrating from rest at $t_0$ (using (1), $kmg=at_0$ ) to a general $t\ge t_0$ ,

\begin{align*}v(t) &= \frac{1}{m}\int_{t_0}^{t}\bigl(at'-kmg\bigr)\,dt' \\ &= \frac{1}{m}\left.\left(\frac{at'^2}{2}-kmg\,t'\right)\right|_{t_0}^{t} \\ &= \frac{a}{m}\left(\frac{t^2}{2}-t_0 t+\frac{t_0^2}{2}\right)\end{align*}

v(t) = \dfrac{a}{2m}(t-t_0)^2

(2)

Step 3Distance covered since the force began

The body covers no distance before $t_0$ (Step 1), so integrate (2) from $t_0$ to $t$ to get the total distance since the force started acting:

s = \int_{t_0}^{t} \frac{a}{2m}(t'-t_0)^2\,dt' = \frac{a}{2m}\left.\frac{(t'-t_0)^3}{3}\right|_{t_0}^{t}.

s = \dfrac{a(t-t_0)^3}{6m}, \quad t_0=\dfrac{kmg}{a}

(3)

This holds for $t\ge t_0$ ; for $t<t_0$ the body has not moved yet, so $s=0$ .

Check

Dimensionally, $[a][t_0]^3/[m] = (\mathrm{N/s})\cdot\mathrm s^3/\mathrm{kg}=\mathrm{N\cdot s^2/kg}=\mathrm m$ , a length, as required. Also, differentiating (2) gives the acceleration $a(t-t_0)/m$ , which is exactly zero at $t=t_0$ : the body sets off with zero acceleration, consistent with $F(t_0)=kmg$ exactly canceling friction at that instant, and grows smoothly afterward as $F$ keeps increasing.

Remark

Because $v(t)$ in (2) is a perfect square in $(t-t_0)$ , the motion after $t_0$ looks locally like release from rest under zero initial acceleration rather than under a sudden jolt: a constant force switched on abruptly at $t_0$ would instead give an acceleration that jumps immediately to a nonzero value. The linear growth of $F=at$ is what smooths the acceleration to zero exactly at the threshold, even though the body’s speed and position are still discontinuous in their higher derivatives at that instant.