An elevator cab that weighs $\mathbf{}\mathbf{27}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{KN}$moves upward. (a) What is the tension in the cable if the cab’s speed is increasing at a rate of $\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$and (b) What is the tension in the cable if the cab’s speed is decreasing at a rate of $\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

• The acceleration of the cab is,$\mathrm{a}=2.4\mathrm{m}/{\mathrm{s}}^2$.
• The weight of the elevator cab is$\mathrm{W}=27.8\mathrm{kN}\mathrm{or}\hspace{0.33em}27800\mathrm{N}$

Newton’s second law states that the net force on the body is the vector sum of all the forces acting on the body. The force is equal to the product of mass and the acceleration of the body. Using the free-body diagram, we can determine the forces and their direction acting on the body.

Draw the free-body diagram of a elevator cab according to given conditions. Apply Newton’s law for each FBD. Then by solving equations calculate unknown mass and tension.

Formulae:

$\sum \mathrm{F}=\mathrm{ma}$

Here, F is force acting on the body, m is mass and a is the acceleration of the body.

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{cab}=\frac{\mathrm{W}}{\mathrm{g}} \\ =\frac{27800\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =2837\mathrm{kg} \end{gathered}$$

When the elevator cab is descending the net force on it is in the downward direction. The forces on the cab are tension in the cord in the upward direction and weight of the cab in the downward direction. Since, the elevator is speeding upward, the net force is in the upward direction. Using the free body diagram I, and applying Newton’s law to the situation we get,

$$\begin{gathered} \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\ \mathrm{T}=\mathrm{m}\left(\mathrm{g}+\mathrm{a}\right) \end{gathered}$$

Substitute the given values in the equation and calculate the tension.

$$\begin{gathered} \mathrm{T}=2837\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2+1.22\mathrm{m}/{\mathrm{s}}^2\right) \\ =3.13\times {10}^4\mathrm{N} \end{gathered}$$

Hence, the tension in the elevator cable is $3.13\times {10}^4\mathrm{N}$.

The cab is moving upward, but the speed is decreasing. It means the cab is decelerating and the net force is acting in the downward direction. Using free body diagram II and applying Newton’s second law to this situation, we get

$$\begin{gathered} \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\ \mathrm{T}=\mathrm{m}\left(\mathrm{g}+\mathrm{a}\right) \end{gathered}$$

Now, substitute the given values in the above equation and calculate the tension.

$$\begin{gathered} \mathrm{T}=2837\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2-1.22\mathrm{m}/{\mathrm{s}}^2\right) \\ =2.43\times {10}^4\mathrm{N} \end{gathered}$$

Hence, the tension in the elevator cable is$2.43\times {10}^4\mathrm{N}$ .