Irodov 1.61: The Trailing Bar Pushes Back – Amit & Deepali Sinha Foundation

Problem Irodov 1.61

Two bars, labeled 1 and 2, sit touching each other on a plane inclined at angle $\alpha$ to the horizontal, bar 1 lower down the slope and bar 2 directly above it, so that bar 2 can push bar 1 further downhill through the contact but cannot pull it back. Their masses are $m_1$ and $m_2$ , and their respective coefficients of friction against the incline are $k_1$ and $k_2$ , with $k_1>k_2$ (bar 1, the lower one, is the more resistive of the two). Find (a) the contact force between the bars while they slide down together, and (b) the smallest angle $\alpha$ at which the pair, released from rest, begins to slide.

Bar 1 (friction k_1) sits lower on the slope, touching bar 2 (friction k_2) directly above it; the two move down the incline together. — Bar 1 (friction $k_1$ ) sits lower on the slope, touching bar 2 (friction $k_2$ ) directly above it; the two move down the incline together.

Given

$m_1$ , $m_2$
$k_1$ , $k_2$ ( $k_1>k_2$ )
$\alpha$

Find

(a) contact force

F

, (b)

\alpha_{\min}

for sliding

Solution

Step 1The combined system’s acceleration down the slope

Treating the touching pair as one system of mass $m_1+m_2$ , gravity supplies $(m_1+m_2)g\sin\alpha$ down the slope and kinetic friction removes $k_1 m_1 g\cos\alpha+k_2 m_2 g\cos\alpha$ , each bar’s own friction acting on its own weight.

(m_1+m_2)g\sin\alpha-(k_1 m_1+k_2 m_2)g\cos\alpha = (m_1+m_2)a

Dividing through by the combined mass $m_1+m_2$ leaves the shared acceleration, which both later steps use.

a = g\sin\alpha-\dfrac{(k_1 m_1+k_2 m_2)g\cos\alpha}{m_1+m_2}

(1)

Step 2Isolate bar 1 for the contact force

Bar 1 alone feels gravity, its own friction $k_1$ , the normal, and the push $F$ from bar 2 (down the slope). Applying Newton’s second law to bar 1 along the slope and substituting the shared acceleration (1) isolates $F$ .

\begin{align*}m_1 g\sin\alpha - k_1 m_1 g\cos\alpha + F &= m_1 a \\ F &= m_1\left(a - g\sin\alpha + k_1 g\cos\alpha\right) \\ &= m_1 g\cos\alpha\left(k_1-\dfrac{k_1 m_1+k_2 m_2}{m_1+m_2}\right) \\ &= m_1 g\cos\alpha\,\dfrac{k_1(m_1+m_2)-k_1 m_1-k_2 m_2}{m_1+m_2} \\ &= m_1 g\cos\alpha\,\dfrac{(k_1-k_2)m_2}{m_1+m_2}\end{align*}

F = \dfrac{(k_1-k_2)m_1 m_2 g\cos\alpha}{m_1+m_2}

(2)

Step 3The threshold for sliding

Motion begins when the acceleration (1) first turns positive, i.e. when the combined driving force equals the combined maximum friction.

(m_1+m_2)g\sin\alpha_{\min} = (k_1 m_1+k_2 m_2)g\cos\alpha_{\min}

\tan\alpha_{\min} = \dfrac{k_1 m_1+k_2 m_2}{m_1+m_2}

(3)

Check

$\tan\alpha_{\min}$ is a weighted average of $k_1$ and $k_2$ , so it lies strictly between them, $k_2<\tan\alpha_{\min}<k_1$ : steeper than what would let bar 2 alone slide, shallower than what bar 1 alone would need, exactly the compromise a contact force enforces. Also, setting $k_1=k_2=k$ in both boxed results gives $F=0$ and $\tan\alpha_{\min}=k$ , the ordinary single-bar result, as it must when there is no friction mismatch to push against.

Remark

The contact force $F$ found in Step 2 is exactly the surplus by which bar 1’s own friction, being the larger of the two coefficients, would hold bar 1 back more than the combined system’s actual acceleration allows, transmitted mechanically through the touching faces; it is the same bookkeeping that hands a locomotive’s excess pull to a lighter trailing wagon through the coupling between them, here compressive contact standing in for a coupling that can only push, never pull.