Why will the maximum value of static friction act? [closed]

Question

I viewed a similar question but that didn't solve my query.In the shown problem, while solving the equations my teacher just assumes that the static friction will have its maximum value and solves the question,(NOTE: there is no friction on the inclined surface, it is only on the flat one)

What I can't understand is why he assumes that the static friction will attain its maximum value when we don't even know the value of tension in the rope ?

Is it possible to solve for the tension in rope and the acceleration of the 2 blocks without knowing the friction ?

Answer 1

This is probably an approximation for a physical system, which doesn't exist in real life. He probably maid that clarification only to be even able to solve the problem. In general we have to distinguish static $F_s$ and dynamic $F_d$ friction. They are both approximation, which quite well explains experiments. They are both calculated with same equation $F=\mu F_\perp$, where $F_\perp$ is force perpendicular to surface. Usually, the coefficient $\mu$ is bigger for static friction. So what normally happens is that we start pushing object with some force, and when the pushing force $F_{p}$ is greather than dynamic friction, the object starts to move, and instead of static friction we than have dynamic. The object than starts moving with $a=\frac{F_p-F_d}{m}$. The transition happen almost imidiately, and it is discontinuous. Good example of Force vs time for such a experiment is here!.

In order to be able to solve the problem, we normally assume, that the object is already moving with negligible velocity ($v \rightarrow 0$), so we don't have to bother with static friction. Other way is to just assume, that the maximal force of static friction is also the force of dynamic friction, which your teacher did.

Answer 2

Is it possible to solve for the tension in rope and the acceleration of the 2 blocks without knowing the friction ?

No.
Without knowing the value of friction, it is impossible to calculate the acceleration of the $1\ kg$ block, because the acceleration itself depends on the friction.
Let us call the $1\ kg$ block as block 1 and $2\ kg$ block as block 2. There is a systematic way to solve your problem, and it is by considering every part of the system one by one.

Consider the following points-
1. Either the $1\ kg$ block will move towards right, or it won't move at all.
2. Let us calculate the net force on the block of mass $2\ kg$. Let the system's acceleration be $a$. $$2g\ \text{sin}\ 53^{\circ}+16-T_1=2a\ ...(1)$$ which gives us- $$T_1=32 - 2a\ ...(2)$$ considering $g=10\ m/s^2$. From this equation, we see that if $a=0$ i.e. for the system at rest, the tension $T_1$ has a maximum value which is $32\ N$.

3. The pulley either rotates clockwise, or does not rotate at all. ( Again, because of point 1). Therefore, the torque towards the clockwise direction is always greater than or equal to the torque towards anticlockwise direction. This gives us- $T_1 \geq T_2.$

Now, let us suppose that block 2 is at rest, i.e. $a=0$ in $(2)$. This gives us $T_1=32\ N$. Now, because block 2 is at rest, block 1 is also at rest, due to the string's length being conserved. This tells us that the pulley is not rotating, which means- $$T_2=T_1=32\ N$$.
But the limiting value of friction on block 1 is $2\ N$. Hence, it is not under equilibrium.
We assumed that block 2 does not move, and it turned out that in that case, block 1 will move. We thus have a contradiction. But where does this contradiction stem from?
The assumption that block 2 can stay at rest.

We thus conclude that block 2 can not stay at rest, which means block 1 will move too, giving rise to limiting value of friction i.e. $2\ N$.

Answer 3

In this problem, T1 is well above the weight of the 1 kg mass. The system will accelerate. You write a force equation for each mass, and a torque equation for the pulley, and then solve for the two tensions and the linear (and angular) acceleration. The given coefficient is for kinetic (not static) friction.