What net force must act on an elevator of mass m = 1000 kg…

What net force must act on an elevator of mass $m = 1000\ \mathrm{kg}$ to give it an acceleration $a = 2\ \mathrm{m/s^2}$? What cable forces are needed to accelerate it upward and downward with this acceleration?

Explanation of Method

Cable force on an accelerating elevator.

Step 1: Net force

\[F_{\text{net}} = ma = 2000\ \mathrm{N}\]

Step 2: Cable force, acceleration upward

\[ F_1 - mg = ma \;\Longrightarrow\; F_1 = m\,(g + a) = 1000 \cdot 11.81. \]

\[F_1 = 11\,810\ \mathrm{N} \approx 11.8\ \mathrm{kN}\]

Step 3: Cable force, acceleration downward

\[ mg - F_2 = ma \;\Longrightarrow\; F_2 = m\,(g - a) = 1000 \cdot 7.81. \]

\[F_2 = 7810\ \mathrm{N} \approx 7.81\ \mathrm{kN}\]

$F$ $mg$ $a$ $a \uparrow:\ F = 11.8\ \mathrm{kN}$ $F$ $mg$ $a$ $a \downarrow:\ F = 7.81\ \mathrm{kN}$

Cable force (red) and weight (blue), to scale. To accelerate upward the cable must pull more than the weight; to accelerate downward, less.

Result

\[\boxed{\,F_{\text{net}} = ma = 2000\ \mathrm{N}\,}\]

\[\boxed{\,F_1 = 11\,810\ \mathrm{N} \approx 11.8\ \mathrm{kN}\,}\]

\[\boxed{\,F_2 = 7810\ \mathrm{N} \approx 7.81\ \mathrm{kN}\,}\]

Remark

Check. The original results are correct. It is the direction of the acceleration, not of the motion, that matters: $F_1$ also applies when the elevator is moving down and braking.