An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at, 12 m/s is brought to rest with constant acceleration in a distance of 42 m.

• The mass of the cab and load is m = 1600 kg
• The initial velocity of the cab in the downward direction is${\mathrm{v}}_{\mathrm{o}}=-12\mathrm{m}/\mathrm{s}$
• The distance covered by the cab in the downward direction is d = - 42 m
• The final velocity of the cab is v = 0 m/s

The free-body diagram represents all the forces acting on the body. It gives an overall view of the directions and magnitude of the forces. The force on the body is equal to the product of mass and the acceleration of the body.

Draw the free-body diagram of the cab. Use the third kinematic equation and find out the acceleration of the cab. By using Newton’s second law, find tension in the cable of the cab.

Formulae:

$$\begin{gathered} {\overrightarrow{\mathbf{F}}}_{\mathbf{net}\mathbf{}}\mathbf{=}\mathbf{m}\overrightarrow{\mathbf{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ {\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{ad}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

The elevator cab is moving downward direction with velocity $v_o$and stops after a distance. Hence its final velocity will be v = 0 m/s. By using the third kinematical equation we can find the acceleration of the cab. Substitute the given values in equation (2)

role="math" localid="1657022683859" $$\begin{gathered} {\mathrm{v}}^2={\mathrm{v}}_0^2+2\mathrm{ad} \\ {(0\mathrm{m}/\mathrm{s})}^2={(-12\mathrm{m}/\mathrm{s})}^2-2\mathrm{a}\left(42\mathrm{m}\right) \\ \mathrm{a}=\frac{{(-12\mathrm{m}/\mathrm{s})}^2}{2\times 42\mathrm{m}} \\ \mathrm{a}=1.71\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

According to Newton’s second law,

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{net}}=\mathrm{m}\overrightarrow{\mathrm{a}} \\ \mathrm{T}-\mathrm{mg}=\mathrm{ma} \end{gathered}$$

Use sign convention according to the motion of the cab as shown in the free body diagram. Substitute the values in the above equation to find the tension.

$$\begin{gathered} \mathrm{T}=\mathrm{ma}+\mathrm{mg} \\ =\mathrm{m}(\mathrm{a}+\mathrm{g}) \\ =1600\mathrm{kg}\times (1.71\mathrm{m}/{\mathrm{s}}^2+9.8\mathrm{m}/{\mathrm{s}}^2) \\ =1.8\times {10}^4\mathrm{N} \end{gathered}$$

Hence, the tension in the supporting cable of an elevator cab is$1.8\times {10}^4\mathrm{N}$