An elevator, hanging from a single cable, moves upward at constant speed. Friction and air resistance are negligible. Is the tension in the cable greater than, less than, or equal to the gravitational force on the elevator? Explain. Include a free-body diagram as part of your explanation.

(1) Tension force $\left(\overrightarrow{T}\right)$in the upwards-pointing orientation of the cable.
(2) Gravitational force $\left({\overrightarrow{F}}_G\right)$a force acting on a vertically downward-pointing elevator.

The diagram is represented as:

We can write the expressions as :

$\begin{array}{r}{\overrightarrow{F}}_{nt}=\overrightarrow{T}-{\overrightarrow{F}}_G\\ m\overrightarrow{a}=\overrightarrow{T}-{\overrightarrow{F}}_G\end{array}$

When the elevator is travelling at a constant pace in the upward direction, it is said to be accelerating $\overrightarrow{a}=0$
When an elevator is travelling higher, the forces operating on the elevator are as follows:
(1) The cable has a tension force pointing up.
(2) Gravitational force $\left({\overrightarrow{F}}_G\right)$the cable has a tension force pointing up.

$\begin{array}{r}{\overrightarrow{F}}_{\mathrm{nt}}=\overrightarrow{T}-{\overrightarrow{F}}_G\\ m\overrightarrow{a}=\overrightarrow{T}-{\overrightarrow{F}}_G\\ \overrightarrow{T}-{\overrightarrow{F}}_G=0[\overrightarrow{a}=0]\\ \overrightarrow{T}={\overrightarrow{F}}_C\end{array}$

As a result, an elevator suspended on a single cable climbs upward at a constant pace, and the tension force in the cable equals the elevator's gravitational pull.