The tangential force acting on the pulley is the friction F between the pulley and the string.How is that the torque applied by friction is equal to the torque applied by the difference in tensions ?
**The previous answer I had posted was completely wrong
Consider an elemental length of string wrapped around the pulley. We know that the string is massless(light - its an approximation). There is a tension acting on the string $T+dT$ from one side and $T$ from the other side.(There is a small change in tension because we have considered a small part of the string only). There is a small amount of friction $df$ acting on this string.
The force equation of this string is as follows.
$$T+dT-T-df=dm\cdot a$$
We know that $dm=0$ so,
$$dT=df$$
Integrate both sides and you get
$$f_{\mathrm{total}}=\Delta T=T_2-T_1$$
On the pulley on the left, there are 4 forces applied, $T_1'$, $T_2'$, the gravitational acceleration on the pulley (its weight) $m' g$ (directed downwards), and the tension of the rope at the center of the pulley $T$, which is the one that you draw, but directed upwards. Now, the tension $T$ balances the weight $mg$ and the other two tensions $T_1'$ and $T_2'$, and the pulley don't move.
However, the toques of the tensions $T_1'$ and $T_2'$ may not balance, and may result in a rotation of the pulley. In fact, if $L=I\omega$ is the angular momentum of the pulley, $I$ the momentum of inertia, and $\omega$ the angular velocity, one has
$$
\frac{d L}{dt}
=I\frac{d \omega}{dt}=r T_1'-r T_2'
$$
where $r$ is the radius of the pulley and the terms at the right side of the equations are the torques of the tension forces applied to the pulley.
If your problem is just to determine the static equilibrium of the system, and not its dynamics, you may want to assume $\frac{d L}{dt}=0$ and therefore balance the two torques $r T_1'=r T'_2$, that is, $T_1'=T_2'$.
Trying to understand tensions in a 2 pulley rope system, where M1>M2. Is it true that T1>T2>T3?
Best Answer
If pulleys are not weightless and maybe rope is not weightless. Weight $Q_1=m_1 g$ accelerates whole setup. Tension $T_1$ is responsible for acclereating two pulleys, rope and mass $m_2$, $T_2$ accelerates one pulley less to the same acceleration, so that it is lower. By the same reason $T_3$ is lower than $T_2$.