Irodov 1.80: Too Much Push Slides the Bar Uphill – Amit & Deepali Sinha Foundation

Problem Irodov 1.80

Prism $1$ , a right triangular wedge of base angle $\alpha$ with a vertical face on the side away from the slope, is driven with a horizontal acceleration of magnitude $w$ , directed toward the foot of the slope, away from the vertical face. Bar $2$ , of mass $m$ , rests on the prism’s sloped surface, held there only by friction, with coefficient $k$ , satisfying $k<\cot\alpha$ . Find the largest value of $w$ for which bar $2$ still does not slide relative to prism $1$ .

Prism 1, driven along the horizontal floor it stands on, accelerates to the left at w while bar 2 rides its slope of base angle α; the floor is hatched here because it is fixed ground, and its own friction plays no part. — Prism $1$ , driven along the horizontal floor it stands on, accelerates to the left at $w$ while bar $2$ rides its slope of base angle $\alpha$ ; the floor is hatched here because it is fixed ground, and its own friction plays no part.

Given

$m$
$\alpha$
$k<\cot\alpha$

Find

the maximum acceleration

w_{\max}

Solution

Step 1Equilibrium in the prism’s own frame

Free body of bar 2 in the prism's own frame: the normal N, the weight m g, the pseudo-force m w opposite the prism's acceleration, and the friction f, drawn down-slope. — Free body of bar $2$ in the prism’s own frame: the normal $\vect N$ , the weight $m\vect g$ , the pseudo-force $m\vect w$ opposite the prism’s acceleration, and the friction $\vect f$ , drawn down-slope.

Resolve along the slope’s outward normal and along the slope itself, taking up-slope as positive. The pseudo-force contributes $mw\cos\alpha$ up-slope and $mw\sin\alpha$ into the slope, while gravity contributes $mg\cos\alpha$ into the slope and $mg\sin\alpha$ down-slope, so the two equilibrium conditions read

\begin{align*}N - mg\cos\alpha - mw\sin\alpha &= 0 \\ f + mw\cos\alpha - mg\sin\alpha &= 0\end{align*}

The first of these fixes the normal reaction, which Step 2 needs.

N = m(g\cos\alpha+w\sin\alpha)

(1)

The second gives the friction the bar needs in order to stay put,

f = mg\sin\alpha - mw\cos\alpha

Step 2The critical friction condition

As $w$ grows, the pseudo-force’s up-slope component grows with it while gravity’s own down-slope pull stays fixed, so $f$ falls and eventually reverses: bar $2$ is on the verge of sliding up the slope, and friction, at its full value $kN$ , acts down-slope to hold it there.

mg\sin\alpha - mw\cos\alpha = -kN

Substituting (1) and dividing through by $m$ ,

\begin{align*}g\sin\alpha - w\cos\alpha &= -k(g\cos\alpha+w\sin\alpha) \\ g\sin\alpha + kg\cos\alpha &= w\cos\alpha - kw\sin\alpha \\ g(\sin\alpha+k\cos\alpha) &= w(\cos\alpha-k\sin\alpha)\end{align*}

Step 3Solve for

w_{\max}

Dividing numerator and denominator by $\cos\alpha$ ,

w = \dfrac{g(\tan\alpha+k)}{1-k\tan\alpha}

Multiplying top and bottom by $\cot\alpha$ puts it in the book’s own form.

w_{\max} = \dfrac{g(1+k\cot\alpha)}{\cot\alpha-k}

(2)

Check

At $k=0$ this reduces to $w_{\max}=g\tan\alpha$ , exactly the acceleration at which a frictionless block sits in equilibrium on a slope of angle $\alpha$ (the pseudo-force’s up-slope pull exactly balances gravity’s down-slope pull with no friction needed at all), which is the expected $k\to0$ limit. The stated condition $k<\cot\alpha$ is exactly what keeps the denominator $\cot\alpha-k$ positive, so $w_{\max}$ stays finite and positive.

Remark

Too much acceleration is not the only failure mode here, only the one this problem’s stated condition $k<\cot\alpha$ isolates. If in addition $k<\tan\alpha$ , the same two equations, with both frictions reversed at their limits, give a lower bound too: below $w_{\min}=g(\tan\alpha-k)/(1+k\tan\alpha)$ , gravity’s down-slope pull overwhelms the pseudo-force’s up-slope help and friction’s own reserve together, and bar $2$ slides down. Whether a given accelerating incline problem has a floor, a ceiling, or both on $w$ depends on which way the pseudo-force resolves along the slope relative to gravity and friction, not on any fixed rule of thumb.