Now consider a situation where we have a string pulled at both its
ends with two equal forces.
Unless it has equal forces on both ends, it will accelerate in one direction, so this is always true for a string at rest. There's no difference between this case and the case where one end is fixed to a wall.
Tension is normally dealt with as a scalar in a string, not a force with a specific direction. For a string at rest, the tension at any point in the string is equal to the forces at each end.
In my opinion, the requirement that the string be nonextensible creates conceptual issues.
On the one hand, it is stated that the string is nonextensible. On the other hand, it is stated on the diagram the "External force $F$ is applied such that the block remains at rest". The problem is if the string is nonextensible, and initially has no slack, then the block cannot move to the left, i.e., it will always be "at rest", regardless of the force $F$ to the left.
But more importantly, it makes it difficult to explain (1) why the static friction force counters the applied force before the tension in the string and (2) why when equilibrium is reached the friction force no longer exists.
To facilitate the answers to these questions I will replace the string with an ideal spring (see the figures below). An ideal spring, like the string, is massless. But unlike the string, it is extensible to the degree allowed by the spring constant. The magnitude of the tension in the spring is equals the magnitude of its restoring force, or $T=k\Delta x$ where $k$ is the spring constant and $\Delta x$ is its extension beyond the "relaxed" stated.
Now consider the following where the block is considered a rigid (nonextensible) body:
The spring, like the string, is initially relaxed so there is no tension. FIG 1 shows the block with no external force and the relaxed spring attached to the wall
In Fig 2 we gradually apply an increasing external force $F$ that is less than the maximum static friction force. Since the block cannot move, the spring cannot extend and thus the tension in the spring is still zero. This explains, for a physically real scenario, why the applied force $F$ is countered first by the static friction force.
In FIG 3 the applied force reaches the maximum static friction force and the friction force becomes kinetic friction, which is generally considered constant. Since kinetic friction is usually less than static friction, if the applied force $F$ is maintained at the value of the maximum static friction there will initially be a net force to the left causing the block to move to the left. (Note that actual value of the friction force during the transition from static to kinetic is undefined for the standard model of friction). At the same time, however, the spring extends creating an opposing tension force. So during this phase before the extension of the spring is a maximum, we have
$$F-f_{k}> T$$
$$\mu_{s}mg-\mu_{k}mg>k\Delta x$$
and the block is moving to the left.
- When the extension of the spring is such that the tension in the spring equals the applied force $F$, it's extension is maximized and we have
$$F=T=k\Delta x_{max}$$
Substituting into the first equation,
$$f_{k}=0$$
Meaning there is no net force for friction to oppose.
Note that in this example, the stiffer the spring (the greater $k$ is) the less the block needs to move before the tension equals the applied force, i.e., the quicker the tension rises. The nonextensible string is simply a spring with an infinite $k$.
Hope this helps.
Best Answer
Answer to the first question: This depends to some extent on the 'models' used for the forces of friction and tension. A typical model for string tension is as a restoring force that obeys Hooke's law: $$T = - kx$$ at least for a small positive extension $x$ in the length of the string, or equivalently, displacement of the block along the length of the string. You can go to the limit of the ideal string by making $k$ very large.
Similarly, friction is modeled as a force that opposes relative velocity between two surfaces in contact. You can consider it to instantaneously play a role when there is a "risk" of developing a small velocity.
Now, we must argue that, under a force $F$, the velocity developed on a free object is a stronger effect than displacement, if we were to conclude that "friction acts first". For this, let us look at what happens in a small time $dt$. The object develops a velocity $$dv = \frac{F}{m} dt$$ However, the displacement within the same time cannot be larger than $(dv)dt$, which goes as $(dt)^2 << dt$. Therefore, the displacement is really negligible in this time scale.
Thus, if we now turn 'on' both the tension and the friction, we expect the string tension to be negligible compared to the friction. The friction itself balances the applied force (if it is less than the limiting friction), and we effectively have a situation of zero tension.
Granted that this is a hand-wavy argument, but it may provide some intuition for why we take friction to act first in such problems, and why, as a real world example, you won't feel someone weakly pushing a heavy rock (on rough ground) that you may be leaning against from the other side.
For the other two questions, note that there is no limit on the tension that can build up in an ideal and in-extensible string, and this tension always acts along the length of the string.