Irodov 1.79: Excess Tension Turns into Backward Friction

Problem Irodov 1.79

A cart $A$ carries body $1$ on its horizontal top surface and body $2$ pressed against its vertical right-hand face; a thread joins them over a pulley fixed at the top right corner of the cart. Bodies $1$ and $2$ have equal mass $m$ , and the coefficient of friction between each body and the cart is $k$ (the pulley itself is massless and frictionless). Find the minimum horizontal acceleration at which the cart $A$ must be driven so that neither body slides relative to it.

Body 1 sits on cart A's top surface and body 2 is pressed against its vertical right face. The thread leaves body 1 horizontally, passes over the pulley mounted at the cart's top right corner, and drops straight down onto body 2. — Body $1$ sits on cart $A$ ’s top surface and body $2$ is pressed against its vertical right face. The thread leaves body $1$ horizontally, passes over the pulley mounted at the cart’s top right corner, and drops straight down onto body $2$ .

Given

$m_1=m_2=m$
$k$

Find

the minimum acceleration

w_{\min}

of cart

A

Solution

Step 1Body 2 needs a normal force to exist at all

Since body $2$ shares the cart’s horizontal acceleration $w$ , and $N$ is the only horizontal force on it, its horizontal equation of motion fixes $N$ directly.

N = mw

(1)

Step 2Body 2’s vertical equation, friction at its limit

At the minimum acceleration, body $2$ is on the verge of slipping down, so friction acts upward at its full value $kN$ .

mg - T - kN = 0

Substituting $N=mw$ from (1) and solving for the tension,

T = mg - kmw

(2)

Step 3Body 1’s horizontal equation, friction at its limit

Body $1$ shares the cart’s acceleration $w$ too. It rides the cart’s horizontal top face, so it has no vertical acceleration and its vertical equation is a pure balance that fixes the normal force there,

\begin{align*}N_1 - mg &= 0\\ N_1 &= mg\end{align*}

The tension pulls body $1$ toward the pulley with more force than $mw$ alone requires, so at the critical point friction resists that excess, acting backward at its full value $kN_1 = kmg$ ,

mw = T - kmg

Substituting $T$ from (2),

\begin{align*}mw &= mg - kmw - kmg \\ w(1+k) &= g(1-k)\end{align*}

w_{\min} = \dfrac{g(1-k)}{1+k}

(3)

Check

At $k=0$ the result gives $w_{\min}=g$ : with no friction at all, body $1$ ’s horizontal equation reduces to $mw=T$ while body $2$ ’s vertical equation reduces to $T=mg$ , and combining the two forces $w=g$ regardless of either body’s mass. As $k\to1$ , $w_{\min}\to0$ : friction on body $1$ , approaching its own ceiling $kmg\to mg$ , is by itself almost enough to balance the thread’s pull, so almost no push is needed to keep body $2$ from sliding down; body $2$ ’s own reserve $kN=kmw$ is not the mechanism here, since $N=mw$ is itself small at small $w$ .

Remark

The minimum, not maximum, acceleration is the binding constraint at the low end because less push means less normal force $N=mw$ on body $2$ , hence less friction available there to help support its weight. But pushing harder does not leave unlimited friction in reserve on both bodies: body $1$ rests on the horizontal top, so its normal force is fixed at $mg$ independent of $w$ , and with it its friction ceiling $kmg$ ; the pull it must resist, however, only grows with $w$ . A maximum acceleration therefore exists too: past $w_{\max}=g(1+k)/(1-k)$ (for $k<1$ ), found by the same pair of equations with both frictions reversed at their limits, body $1$ can no longer keep pace with the cart and slips backward relative to it, dragging body $2$ up the face. The two thresholds bound a window $w_{\min}\le w\le w_{\max}$ within which both bodies stay put; the present problem asks only for its lower edge.