**Hint** : Friction opposes tendency to move. Tension is produced if the string is stretched $very$ slightly. So, increase friction to maximum and then tension will act if necessary.

Ironically, you are thinking absolutely right. Give yourself a cookie.

From part $a$, we know that the blocks will be at rest at all angles below that.

You are also right as at very small angles there is no need of tension and we can ignore it to solve for, again, an angle condition. You have done excellent work. Congrats.

Now we come to the middle angles. Oh... they drive you insane, don't they?

Let's start. We can start our analysis from 2 blocks, 1 will give a contradiction and other will give a result, but I will start with the one giving contradiction. This will help you.

*All angles are in degrees* :

$\theta=35 $

Lets start by analysing Block A (No racism intended)

Gravity is trying to pull it down : $5*10*\sin(35)N=28.67N$

Friction comes to the rescue(up) : $50*\cos(35)N=16.38N$ // read my hint to know why friction is put max here

As it is at rest, $T=12.29N$

Now Block B is also at rest,

Weight = $114.71N$

max f= $81.92N$

$16.38+114.71=12.29+f$

$f=118.8N$

OOPS, it exceed max value. So, Lets start by analysing Block B. (I love alliteration)

Gravity trying : $114.71N$

Friction comes to the rescue(up) : $81.92N$

You can take from here I guess. calculate tension. Note that you have to revise your calculation for tension again as reaction friction force will be provided by A. Better assume it $f$ from starting FBD of B.

This will yield the correct answer. Friction will be less than max value for upper block. In most cases, You should start analysing with heavier block(my experience). Hope your doubts are cleared.

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, $T_1'=T_2'$.

## Best Answer

In a pulley we can assume that the friction can be neglected, so that the tension in the rope can be assumed to be constant. Thus, consider the following diagram, showing the forces acting on the rope due to a turning angle $\mathrm{d}\theta$:

From the symmetry of the problem (any part of the rope that is in contact with the pulley will feel the same pressure, since the curvature and the tension are constant), you conclude that the force per unit length in the normal direction is given by:

$p = \frac{\mathrm{d}F_N}{\mathrm{d}\theta R}$

since the length that supports the force $\mathrm{d}F_N$ is $\mathrm{d}\theta R$.

But since, from the diagram, $\mathrm{d}F_N = 2 T \sin{(\mathrm{d}\theta/2)} \sim T \mathrm{d}\theta$ when the angle tends to zero, the above equation becomes:

$p = \frac{T}{R}$

Therefore, the pressure does not depend on the total angle bent, but rather on the curvature at which you are bending it (or one over the radius of the pulley). Obviously, the total force on the pulley

willdepend on the total angle, which can be calculated easily by integrating the projection of $p$ over the angle bisector along the total arc along which there is contact with the rope.