Question

An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of$\mathbf{2000}\mathbf{}\mathbf{kg}$. When that occupant drops a coin, its acceleration relative to the cab is$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ downward. What is the tension in the cable?

Short answer

The tension in the cable is $16.0\mathrm{kN}$.

Step-by-step solution

Given information

• The mass of the elevator cab with its occupant is, $\mathrm{M}=2000\hspace{0.33em}\mathrm{kg}$.
• The acceleration of the coin relative to the elevator cab is,${\mathrm{a}}_{\mathrm{ce}}=-8.00\mathrm{m}/{\mathrm{s}}^2.$

Understanding the concept of force and free body diagram

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

Draw the free-body diagram of a lamp according to given conditions. Apply Newton’s law for each FBD. Then by solving equations calculate the 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.

Calculate the tension in the cable

The acceleration of the coin with respect to the ground ( $a_{cg}$) is,

$a_{cg}=a_{ce}+a_{eg}$

Here,$a_{ce}$is the acceleration of the coin with respect to the elevator, its value is $-8.00\mathrm{m}/{\mathrm{s}}^2$.

$a_{eg}$is the acceleration of the elevator with respect to ground, its value is $-9.8\mathrm{m}/{\mathrm{s}}^2$as the coin is in the free fall with respect to ground and acted upon by the gravitational acceleration.

Therefore, we can write,

$$\begin{gathered} -9.8\mathrm{m}/{\mathrm{s}}^2=-8.0\mathrm{m}/{\mathrm{s}}^2+{\mathrm{a}}_{\mathrm{eg}} \\ {\mathrm{a}}_{\mathrm{eg}}=-1.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

FBD for elevator cab with its occupant:

From FBD we can write as,

$$\begin{gathered} \mathrm{T}-\mathrm{Mg}={\mathrm{Ma}}_{\mathrm{eg}} \\ \mathrm{T}=\mathrm{Mg}+{\mathrm{Ma}}_{\mathrm{eg}} \\ =\mathrm{M}\left(\mathrm{g}+{\mathrm{a}}_{\mathrm{eg}}\right) \end{gathered}$$

Substitute the values in the above equation to calculate the tension.

$$\begin{gathered} \mathrm{T}=2000\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2-1.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =16000\mathrm{N} \\ \approx 16\mathrm{KN} \end{gathered}$$

Therefore, the tension in the cable is $16\mathrm{KN}$.