Irodov 1.63: Three Ways the Ratio Can Go – Amit & Deepali Sinha Foundation

Problem Irodov 1.63

A body of mass $m_1$ rests on a plane inclined at angle $\alpha$ to the horizontal, with coefficient of friction $k$ between $m_1$ and the incline. A thread runs from $m_1$ up the slope to a pulley mounted at the top of the incline, then straight down the far (vertical) side to a second body of mass $m_2$ hanging freely in the air. The pulley and both threads are massless, and the pulley is frictionless. Both bodies start at rest. Find the range of the mass ratio $\eta=m_2/m_1$ for which (a) $m_2$ begins to descend (pulling $m_1$ up the slope), (b) $m_2$ begins to rise (pulled up by $m_1$ sliding down the slope), and (c) the system remains at rest.

m_1 rests on the incline of angle α, tied by a thread that runs up the slope, over the apex pulley and down to m_2, which hangs freely off the vertical back face. — $m_1$ rests on the incline of angle $\alpha$ , tied by a thread that runs up the slope, over the apex pulley and down to $m_2$ , which hangs freely off the vertical back face.

Given

$\alpha$
$k$

Find

the ranges of

\eta=m_2/m_1

for (a)

m_2

descending, (b)

m_2

rising, (c) rest

Solution

Step 1

m_2

descends:

m_1

is on the verge of sliding up

Case (a): N, the weight m_1 g and friction f acting down the slope on m_1, together with the thread tension T on the hanging m_2 against its own weight. In case (b) the friction arrow on m_1 alone reverses. — Case (a): $\vect N$ , the weight $m_1\vect g$ and friction $\vect f$ acting *down* the slope on $m_1$ , together with the thread tension $\vect T$ on the hanging $m_2$ against its own weight. In case (b) the friction arrow on $m_1$ alone reverses.

At the threshold, $m_2 g$ just balances $m_1$ ’s weight component down the slope plus the maximum friction resisting the upward slide.

m_2 g = m_1 g\sin\alpha + k m_1 g\cos\alpha

Dividing through by $m_1 g$ turns the threshold into a statement about the mass ratio alone.

\frac{m_2}{m_1} = \sin\alpha + k\cos\alpha

Above that ratio the hanging weight exceeds everything the slope and friction can muster against it, so the case (a) condition is the strict inequality.

\dfrac{m_2}{m_1} > \sin\alpha+k\cos\alpha

(1)

beyond which $m_2$ genuinely accelerates downward.

Step 2

m_2

rises:

m_1

is on the verge of sliding down

At this threshold, $m_1$ ’s driving component along the slope just balances the pull needed to hold $m_2$ up plus the maximum friction resisting the downward slide.

m_1 g\sin\alpha = m_2 g + k m_1 g\cos\alpha

Isolating $m_2 g$ and dividing through by $m_1 g$ again gives the second threshold ratio.

\frac{m_2}{m_1} = \sin\alpha - k\cos\alpha

Only below that ratio is the hanging weight light enough for $m_1$ to win, so case (b) is again the strict inequality.

\dfrac{m_2}{m_1} < \sin\alpha-k\cos\alpha

(2)

beyond which (i.e. for smaller $\eta$ still) $m_1$ genuinely slides down and $m_2$ rises.

Step 3Neither threshold reached: the system stays at rest

Combining the complements of (1) and (2) (i.e. neither inequality holds) gives the resting range directly.

\sin\alpha-k\cos\alpha < \dfrac{m_2}{m_1} < \sin\alpha+k\cos\alpha

(3)

Check

The three open ranges from Steps 1 to 3 cover the whole positive $\eta$ axis except for the two boundary points $\eta=\sin\alpha\pm k\cos\alpha$ themselves, exactly the exhaustive case split the problem asks for: at either boundary, static friction sits at its limiting value and the system is on the verge of motion rather than genuinely inside case (a), (b) or (c). Setting $k=0$ merges the two boundaries into the single point $\eta=\sin\alpha$ and makes range (c) the empty interval, leaving (a) $\eta>\sin\alpha$ , (b) $\eta<\sin\alpha$ , the frictionless result where any imbalance moves the system.

Remark

The width of the resting range, $2k\cos\alpha$ , is exactly twice $k\cos\alpha$ , the friction term that enters each threshold’s own force balance ( $m_2 g$ weighed against $m_1 g\sin\alpha\pm k m_1 g\cos\alpha$ ): here friction has to fight gravity’s pull in either direction depending on which body is heavier, so its stabilizing effect shows up on both sides of the frictionless balance point $\eta=\sin\alpha$ rather than on just one.