USAPhO 1997, Четвъртфинал — Задача 1

• (a) $T = F_o(1 - n/N)$

• (b) $T = F_o(1 - n/N)$

  The same as without friction.

(a, 8)

The force $F_o$ must accelerate $N$ blocks each of mass $m$:
$F_o = Nma$, so that $a = F_o/Nm$. (4)

Let $T$ be the tension dragging $N$-$n$ blocks to the right. $F = ma$ gives
$T = (N-n)ma$ (2)

Using $a$,
$T = (N-n)\,F_o/N = F_o\,(1 - n/N)$. (2)

(b, 12)

Each block has exerted on it the friction force $\mu mg$. (2)

$F = ma$ applied the the entire $N$-block system gives
$F_o - N\mu mg = Nma$ (2)

Solving for acceleration,
$a = (F_o/Nm) - \mu g$ (2)

Let $T$ be the tension dragging $N$-$n$ blocks to the right. $F=ma$ gives
$T - (N-n)\mu mg = (N-n)ma$ (2)

so that upon substituting for $a$,
$T = (N-n)m\,[\mu g + (F_o/Nm) - \mu g\,]$ (2)

which gives the final answer,
$T = F_o(1 - n/N)$, the same as without friction. (2)