Irodov 1.68: When the Normal Force Vanishes – Amit & Deepali Sinha Foundation

Problem Irodov 1.68

A small body of mass $m$ rests on a smooth (frictionless) horizontal plane. Starting at $t = 0$ , a force of steadily growing magnitude $F = at$ ( $a$ a constant) is applied to the body, its direction fixed at a constant angle $\alpha$ above the horizontal throughout. Find (a) the body’s speed at the instant it leaves the plane, and (b) the distance it has traveled along the plane up to that instant.

The body resting on the smooth horizontal plane, with the growing force F held at a fixed angle α above the horizontal. — The body resting on the smooth horizontal plane, with the growing force $\vect F$ held at a fixed angle $\alpha$ above the horizontal.

Given

$F = at$
$\alpha$
$m$
no friction

Find

(a) the speed

v

at breakaway, (b) the distance

s

traveled up to breakaway

Solution

Step 1The instant the plane lets go

The free body while contact lasts: besides F the plane supplies a normal reaction N, and gravity supplies the weight m g. Breakoff is the instant N dies. — The free body while contact lasts: besides $\vect F$ the plane supplies a normal reaction $\vect N$ , and gravity supplies the weight $m\vect g$ . Breakoff is the instant $\vect N$ dies.

While the body still touches the plane it has no vertical acceleration, so the vertical forces balance: $N$ , the weight $mg$ , and the growing vertical component of $\vect F$ . The body breaks off exactly when $N$ first reaches zero.

N = mg - F\sin\alpha = mg - at\sin\alpha

Setting $N=0$ and solving for the time $t_1$ ,

t_1 = \dfrac{mg}{a\sin\alpha}

(1)

Step 2Horizontal speed built up while sliding

The plane is frictionless and $N$ never acts horizontally, so the only horizontal force is $F\cos\alpha$ . Newton’s second law along the plane, with $F=at$ substituted and the variables separated,

\begin{align*}m\frac{dv}{dt} &= F\cos\alpha\\ m\frac{dv}{dt} &= at\cos\alpha\\ dv &= \frac{a\cos\alpha}{m}\,t\,dt\end{align*}

Integrating from rest at $t=0$ up to the speed $v$ at time $t$ ,

\begin{align*}\int_0^v dv' &= \frac{a\cos\alpha}{m}\int_0^t t'\,dt'\end{align*}

so that, evaluating both sides,

v(t) = \dfrac{a\cos\alpha}{2m}\,t^2

(2)

Step 3Speed at breakoff

Substituting the breakoff time (1) into (2),

\begin{align*}v &= \frac{a\cos\alpha}{2m}\,t_1^2 \\ &= \frac{a\cos\alpha}{2m}\left(\frac{mg}{a\sin\alpha}\right)^2\end{align*}

v = \dfrac{mg^2\cos\alpha}{2a\sin^2\alpha}

(3)

Step 4Distance traveled before breakoff

Integrating (2) from rest gives the distance traveled at any time,

\begin{align*}s(t) &= \int_0^t v(t')\,dt'\\ &= \frac{a\cos\alpha}{2m}\int_0^t t'^2\,dt'\\ &= \frac{a\cos\alpha}{6m}\,t^3\end{align*}

Substituting the breakoff time (1),

\begin{align*}s &= \frac{a\cos\alpha}{6m}\,t_1^3 \\ &= \frac{a\cos\alpha}{6m}\left(\frac{mg}{a\sin\alpha}\right)^3\end{align*}

s = \dfrac{m^2 g^3\cos\alpha}{6a^2\sin^3\alpha}

(4)

Check

As $\alpha\to90^\circ$ the force is purely vertical: $\cos\alpha\to0$ makes both boxed results vanish, correct, since a purely vertical pull that just balances $mg$ builds no horizontal speed at all. As $\alpha\to0$ , $\sin\alpha\to0$ and $t_1\to\infty$ : a force with no vertical component can never lift the body off a horizontal plane, so it never breaks away, consistent with both answers diverging.

Remark

Because $F=at$ makes the horizontal acceleration itself grow linearly in $t$ , $v(t)$ and $s(t)$ are locked together at every instant, not only at breakoff: Step 2 gives $v(t)=(a\cos\alpha/2m)t^2$ and Step 4 gives $s(t)=(a\cos\alpha/6m)t^3=(t/3)v(t)$ . So $s=(t_1/3)v$ at breakoff too, a signature of any motion started from rest under a force proportional to $t$ : it is the cubic-in- $t$ analogue of the familiar $s=(t/2)v$ for constant acceleration.