Answer
a) $8.19\;\rm kN$
b) $482\;\rm N$
Work Step by Step
First, we need to draw the force diagram of the two cars.
And, as we did in the previous pulley problems, we assume that the pulley is frictionless and the masses of the two cables are negligible.
This means that the tension force along the pulley's rope is constant, as shown in the figures below.
a) We need to find the tension in the connecting cable $T_P$ which is the cable that is rolling on the pulley.
Using the force diagram of the descending car; where we know that the net force exerted on it is zero since it is moving down the incline at a constant speed.
$$\sum F_{xD}=mg\sin\theta-T_P-F_{rD}=ma_x=m(0)=0$$
where $D$ is for the descending car.
Thus,
$$ T_P=-F_{rD}+mg\sin\theta$$
we know that the rolling friction is given by $F_r=\mu_r F_n$
$$ T_P=mg\sin\theta-\mu_r F_{nD} \tag 1$$
$$\sum F_y=F_{nD}-mg\cos\theta=ma_y=m(0)=0$$
Thus, $$F_{nD}=mg\cos\theta$$
Plugging into (1);
$$ T_P=mg\sin\theta-\mu_r mg\cos\theta=mg\left[\sin\theta-\mu_r \cos\theta\right] $$
Plugging the known;
$$ T_P= 1500\times9.8\times \left[ \sin35^\circ-0.02\cos35^\circ \right] $$
$$T_P=\color{red}{\bf 8.19\times10^3}\;\rm N$$
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b) Now we need to find the tension in the second wire $T_M$ which is the one that is connected to the motor.
Using the force diagram of the ascending car; where we know that the net force exerted on it is zero since it is moving up the incline at a constant speed.
$$\sum F_y=F_{nA}-mg\cos\theta=ma_y=m(0)=0$$
where $A$ is for the ascending car.
Thus,
$$F_{nA}=mg\cos\theta\tag 2$$
$$\sum F_{xA}=T_M+T_P-mg\sin\theta -F_{rA}=ma_x=m(0)=0$$
Thus,
$$T_M=mg\sin\theta +F_{rA}-T_P$$
And from the rolling friction law;
$$T_M=mg\sin\theta +\mu_{r}F_{nA}-T_P$$
Plugging from (2);
$$T_M=mg\sin\theta +\mu_{r}mg\cos\theta-T_P$$
$$T_M=mg\left[\sin\theta +\mu_{r} \cos\theta\right]-T_P$$
Plugging the known;
$$T_M= 1500\times9.8\times\left[\sin35^\circ+0.02 \cos35^\circ\right]-(8.19\times10^3) $$
$$T_P=\color{red}{\bf 482 }\;\rm N$$
