Why are these two tension forces equal? 
I am new to this topic and my teacher told me that these two forces in the cable will be equal. I would like to know why this is case.
 A: Perhaps the easiest way to see this is to use the idea of work. Suppose that the lift moves down by distance $D$. Then
$$\text{Work done on rope} = T_{\text{right}} D$$
But the balancing weight will move up by the same distance, $D$, so
$$\text{Work done by rope} = T_{\text{left}} D.$$
But there is no gain or loss of energy by the rope or pulley, assuming no friction, and assuming  rope and pulley have negligible mass so KE isn't gained. Therefore there is no net work done by the rope, so
$$T_{\text{left}} D-T_{\text{right}} D=0\ \ \ \ \ \ \ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ \ \ \ T_{\text{left}}=T_{\text{right}}.$$
A: If the tension force along the cable wasn't equal everywhere it would mean the cable would separate into pieces. Imagine the cable consisted of a string of many blocks attached by glue of some sort (see left). You could also imagine them as many sections of rope but for me it's more clear that way. Each block exerts a force on the neighbouring blocks: the tension. If two blocks experience different forces it means that they will experience different acceleration and this will mean the blocks will fly apart or get closer together. So at every junction the forces must balance or otherwise the cable can't be a cable . This argument holds even if the cable is draped over the pulley (see right).

A: 
from the free body diagram
$$m_1\,a_1=-m_1\,g+T_1\\
m_2\,a_2=-m_2\,g+T_2\\
T_1-T_2=0$$
if the rope (length l) is inelastic then
$$z_1+z_2=l\quad \Rightarrow a_1+a_2=0$$
you obtain 4 equations for the 4 unknows $~a_1,a_2,T_1,T2~$ the solution
$$T_1=2\,g\frac{m_1\,m_2}{m_1+m_2}\\
T_2=2\,g\frac{m_1\,m_2}{m_1+m_2}$$
thus $$~T_1\overset{!}{=}T_2$$

with pulley (inertia I)

from the free body diagram
$$ I\,\alpha=F_c\,r\\
m_1\,a_1=-m_1\,g+T_1\\
m_2\,a_2=-m_2\,g+T_2\\
T_1-T_2=-F_c$$
and the rolling conditions
$$a_1=\alpha\,r\\
a_2=-\alpha\,r$$
6 equations for 6 unknows. the solution
$$T_1=g\frac{m_1(2\,m_2\,r^2+I)}{m_1\,r^2+m_2\,r^2+I}\\
T_2=g\frac{m_2(2\,m_1\,r^2+I)}{m_1\,r^2+m_2\,r^2+I}$$
thus for $~I=0~$
$$~T_1\overset{!}{=}T_2$$
