**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.

## Best Answer

First of all, where are you getting the term $\cos(30^{\circ})$? What two things here have an angle of $30^{\circ}$?

By having a term $mg\sin(\theta)$, I think you are assuming that the normal force from the wall is completely described by the weight of the ball. In many simple cases, this is true. But because the ball also has the force from the string, it's not true here.

You have 3 forces on the ball, not 2: The tension in the string $T$, the weight of the ball $W$, and the normal force $N$. In your first equation, you are assuming facts about N that you do not yet have.

In your second equation you have $N$, but without a $\cos$ or $\sin$ term. The normal force does not act vertically, so this is incorrect.