91Ó°ÊÓ

The mass of both the cars $m=1500kg$

The coefficient of rolling friction ${\mathrm{μ}}_k=0.02$

The angle between horizontal and cable$\mathrm{θ}=35°$

Free body diagram of the lower car:

The net force on the lower car

$$\begin{gathered} T_2+f=mg\sin \left(\mathrm{θ}\right) \\ \text{where}f\text{isthefrictionforce} \\ f={\mathrm{μ}}_{r}mg\cos \left(\mathrm{θ}\right) \\ {T}_2+{\mathrm{μ}}_{r}mg\cos \left(\mathrm{θ}\right)=mg\sin \left(\mathrm{θ}\right) \\ {T}_2=mg\sin \left(\mathrm{θ}\right)-{\mathrm{μ}}_{r}mg\cos \left(\mathrm{θ}\right) \\ {T}_2=mg\left[\sin \left(\mathrm{θ}\right)-{\mathrm{μ}}_r\cos \left(\mathrm{θ}\right)\right] \\ {T}_2=1500kg\left(9.8m/s^2\right)\left[\sin \left(35°\right)-\left(0.02×\cos \left(35°\right)\right)\right] \\ {T}_2=8190.74N \end{gathered}$$

The tension in the connecting cable is $8190.74N$.

The mass of both the cars$1500kg$

The coefficient of rolling friction$35°$

The angle between horizontal and cablelocalid="1649168031616" $\mathrm{θ}=0.02$

The free body diagram of the upper car:

The net force on the upper car

$$\begin{gathered} T_1+T_2=mg\sin \left(\mathrm{θ}\right)+{\mathrm{μ}}_{r}mg\cos \left(\mathrm{θ}\right) \\ {T}_1=mg\sin \left(\mathrm{θ}\right)+{\mathrm{μ}}_{r}mg\cos \left(\mathrm{θ}\right)-T_2 \\ {T}_1=mg\left[\sin \left(\mathrm{θ}\right)+{\mathrm{μ}}_r\cos \left(\mathrm{θ}\right)\right]-T_2 \\ {T}_1=\left(1500kg\right)\left(9.8m/s^2\right)\left[\sin \left(35°\right)+\left(0.02×\cos \left(35°\right)\right)\right]-8190.74N \\ =481.66N \end{gathered}$$

The tension in the motor cable is$481.66N$