91Ó°ÊÓ

It is given that,

The mass of each car with occupants is$M=2800\mathrm{kg}$

The acceleration of each car is$a=0.81\mathrm{m}/{\mathrm{s}}^2$

The angle made by cable with horizontal is$\mathrm{θ}=35°$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram the difference in the tension between adjacent sections of the pull cable can be found.

Formula:

The Newton’s second law is,

$F_{net}=\underset{}{∑Ma}$

where,$F_{net}$is the net force, Mis mass and a is an acceleration.

FBD for the system is,

According to the Newton’s second law,

$F_{net}=\underset{}{∑}Ma$

From FBD, write for the net force acting in x-direction as,

$$\begin{gathered} F_{net,x}=T_2-T_1-Mg\sin \mathrm{θ} \\ =Ma \\ {T}_2-T_1=Ma+Mg\sin \mathrm{θ} \\ {T}_2-T_1=2800\left(0.81+9.8\sin \left(35°\right)\right) \\ {T}_2-T_1=1.8×{10}^4 \end{gathered}$$

Hence, the difference in the tension between adjacent sections of pull cable is$1.8×{10}^4\mathrm{N}$.