91Ó°ÊÓ

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

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:

$∑\mathrm{F}=\mathrm{ma}$

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

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.

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

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