Irodov 1.64: Which Way Once It Finally Moves – Amit & Deepali Sinha Foundation

Problem Irodov 1.64

A body $m_1$ rests on an incline of angle $\alpha=30^\circ$ with friction coefficient $k=0.10$ against the slope, joined by a thread over a frictionless pulley at the top of the incline to a body $m_2$ hanging freely off the incline’s vertical back face, with mass ratio $m_2/m_1=\eta=2/3$ . Both bodies start from rest. Find the magnitude and direction of the acceleration of $m_2$ once the system begins to move.

m_1 on the incline at α=30°, tied by a thread that runs up the slope, over the apex pulley and down to the hanging m_2. — $m_1$ on the incline at $\alpha=30^\circ$ , tied by a thread that runs up the slope, over the apex pulley and down to the hanging $m_2$ .

Given

$\alpha=30^\circ$
$\eta=m_2/m_1=2/3$
$k=0.10$

Find

magnitude and direction of

m_2

’s acceleration

Solution

Step 1Which way the system moves

With $\alpha=30^\circ$ , $\sin\alpha=0.500$ and $\cos\alpha=0.866$ , the descending threshold from 1.63(a) is

\sin\alpha+k\cos\alpha = 0.500+0.10\times0.866 = 0.587.

The given ratio compares against it as

\eta = \frac{2}{3} = 0.667 > 0.587,

so $m_2$ is heavy enough to descend, pulling $m_1$ up the slope; friction on $m_1$ then opposes that upward slide and acts down the slope.

Step 2Newton’s second law for each body

The forces, in the sense Step 1 established: N, the weight m_1 g and friction f pointing down the slope on m_1, which is being dragged up it, with the thread tension T holding back the descending m_2 against its own weight. — The forces, in the sense Step 1 established: $\vect N$ , the weight $m_1\vect g$ and friction $\vect f$ pointing *down* the slope on $m_1$ , which is being dragged up it, with the thread tension $\vect T$ holding back the descending $m_2$ against its own weight.

Let $T$ be the thread tension and $w$ the common acceleration magnitude, positive in the sense $m_2$ descends and $m_1$ climbs.

\begin{aligned} m_2 g - T &= m_2 w \\ T - m_1 g\sin\alpha - k m_1 g\cos\alpha &= m_1 w \end{aligned} \tag{1}

Step 3Eliminate

T

and solve for

w

Adding the two equations of (1) cancels $T$ , leaving a single equation in $w$ .

\begin{align*}m_2 g - m_1 g\sin\alpha - k m_1 g\cos\alpha &= (m_1+m_2) w \\ w &= g\,\frac{m_2/m_1-\sin\alpha-k\cos\alpha}{1+m_2/m_1} \\ &= g\,\frac{\eta-\sin\alpha-k\cos\alpha}{1+\eta}\end{align*}

Substituting $\eta=2/3$ , $\sin\alpha=0.500$ , $k\cos\alpha=0.0866$ ,

w = g\,\frac{0.667-0.500-0.0866}{1.667} = 0.0482\,g.

w = g\,\dfrac{\eta-\sin\alpha-k\cos\alpha}{1+\eta} \approx 0.048g \approx 0.05g

(2)

directed vertically downward, since Step 1 found $m_2$ to be the heavier side: $m_2$ descends while $m_1$ is drawn up the slope.

Check

$0.048g$ is small and positive, consistent with Step 1’s finding that $\eta=0.667$ only just clears the descending threshold $0.587$ : the system is barely past the point of moving at all, so a small acceleration is exactly what is expected. As a genuine limiting case, setting $k=0$ in (2) gives $w=g(\eta-\sin\alpha)/(1+\eta)=g(0.667-0.500)/1.667=0.100g$ , a clean frictionless check on the same expression. Dimensionally $w$ is $g$ times a dimensionless ratio, as required.

Remark

The rate $dw/dk=-g\cos\alpha/(1+\eta)$ is fixed once $\alpha$ and $\eta$ are fixed, so it is no more sensitive to $k$ near this threshold than anywhere else; what changes near the threshold is only that $w$ itself is small, so the same absolute shift in $k$ is a large fractional change in it. Raising $k$ to $0.20$ drops the numerator to $0.667-0.500-0.173=-0.006$ , but $\eta=0.667$ then sits inside the rest band $(0.327,0.673)$ from 1.63(c), not past its other threshold, so the system simply stops rather than reversing direction. A problem’s numbers sitting near a case boundary is a signal to check the boundary first, exactly as Step 1 did, rather than to plug into a single formula on faith.