A mine elevator (cage) of mass m = 600 kg may move with…

A mine elevator (cage) of mass $m = 600\ \mathrm{kg}$ may move with accelerations up to $a = \pm 1.2\ \mathrm{m/s^2}$. Find (a) the greatest tension in the cable; (b) the tension if $a = g$; (c) the tension when the cage is at rest and when it moves at constant speed.

Explanation of Method

Cable tension of a mine elevator.

Step 1: Newton's second law for the cage

With $a$ positive upward: $T - mg = ma \;\Longrightarrow\; T = m\,(g + a)$.

Step 2: (a) Greatest tension

It occurs when the acceleration points up (speeding up on the way up, or braking on the way down): $T_{\max} = 600\,(9.81 + 1.2) = 6606\ \mathrm{N}$.

(The smallest, with $a = -1.2\ \mathrm{m/s^2}$, is $600 \cdot 8.61 = 5166\ \mathrm{N}$.)

Step 3: (b) $a = g$

An acceleration of magnitude $g$ downward is free fall: $T = m\,(g - g) = 0 \ \ (\text{the cable is slack})$.

If instead the cage accelerated upward at $g$, the tension would be $2mg = 11\,772\ \mathrm{N}$.

Step 4: (c) At rest or at constant speed

\[T = mg = 5886\ \mathrm{N}\]

$0$ $2000$ $4000$ $6000$ $T\ (\mathrm{N})$ $\text{up, a = +1.2}$ $6606$ $\text{rest / const v}$ $5886$ $\text{down, a = -1.2}$ $5166$ $\text{free fall, a = g ↓}$ $0$

Cable tension $T = m(g + a)$ for the 600 kg cage, with $a$ taken positive upward. In free fall the cable is slack.

Result

\[\boxed{\,T_{\max} = 600\,(9.81 + 1.2) = 6606\ \mathrm{N}\,}\]

\[\boxed{\,T = m\,(g - g) = 0 \ \ (\text{the cable is slack})\,}\]

\[\boxed{\,T = mg = 5886\ \mathrm{N}\,}\]

Remark

About the original solution. Parts (a) and (c) are correct. For (b) it assumed an upward acceleration of $g$ (11 772 N). In elevator problems, "$a = g$" normally means the cage falling freely, when the cable carries no load at all. Both values are given above.