I am not an expert in such fields, but I'll give you an overview of how I've learnt it.
The main point to realize is that, on a microscopic scale, the surfaces we initially thought of as "smooth" contain actually a great many irregular protuberances.
Coming back to the surface area between the two objects, one must carefully distinguish between the microscopic area of contact and the macroscopic upon which the friction force is independent, meaning they can be lying on top of each other with their larger cross sections or their smaller parts, it will not matter. Of course this seems surprising at first because friction results from adhesion, so one might expect the friction force to be greater when objects slide on their larger sides, because the contact area is larger. However, what determines the amount of adhesion is not the macroscopic contact area, but the microscopic contact area, and the latter is pretty much independent of whether the objects lie on a large face or on a small face.
Key idea is that the normal force puts pressure on the protuberances of one surface against those of the other which causes the protuberances/junctions to undergo a certain flattening (elastic deformation e.g.), and this increases the effective area of contact between the "rough" parts (before, you can imagine that only the tip points where actually bonding), as illustrated in these two pictures:
![enter image description here](https://i.stack.imgur.com/MvZNY.png)
Second picture: larger effective area of contact or in other words higher number of contact points between the protuberances, also as pointed out by Jim.
To conclude, we now can tell that for large macroscopic contact areas, the number of protuberances in contact is larger but since the normal force is distributed over all of them, their deformations are less important (smaller effective microscopic area), whereas the opposite will hold for smaller macroscopic surfaces, where the deformations are very strong and maximize the contact between the junctions, but their numbers is comparatively lower. All of which explains why macroscopic areas don't matter.
As for larger normal forces, it will increase the deformation of junctions and make the coupling between the surfaces stronger.
![Different normal forces](https://i.stack.imgur.com/1qlTw.jpg)
Its true that the truck's acceleration does depend on the engine specifications.
However, the maximum acceleration the truck can have is the same as the maximum friction of the wheels from the force of static friction. We don't want the truck wheels to slip. If it accelerates above the static friction force then the wheels will begin to slip. Without this force of friction the truck would not move at all.
$$a=g[\mu_s\cos \theta − \sin \theta]$$ is correct; plugging the values in will give you the maximum acceleration of the truck.
As an additional afterthought to the problem. The fact that the maximum acceleration does not depend on mass implies that it can be said that the maximum acceleration of the truck does not depend on the number of wheels that it has. Assume a truck with 8 wheels (an APC weighing m1=4 tons), versus a truck with 4 wheels (a SUV weighing m2=1 ton), versus a bike with 2 wheels (m3=20 lbs); the number of wheels reduces the mass parameter used in Newton's second law by a fraction of the total. The m parameter in your equations could be written $$m = m1/8$$, or $$m = m2/4$$, or $$m = m3/2$$ This mass parameter would still cancel out in the end, the APC, the SUV, and the Bike all would have the same maximum acceleration.
Best Answer
If you push on each side of the book with force $F_N$, then what is holding the book? The static friction of course.
Formula for static friction: $F_{s}\leq\mu_s F_N$
Since it works on both sides of the book, two times this friction will hold up the book's weight: $2 F_s=31\:\mathrm{N}$
So far so good.
But now, think about what force is actually needed to hold that book. If you reduce your pressing force $F_N$ on the book, then you also reduce the maximum force the static friction $F_s$ can reach. The actual friction is constant, though, because of $2 F_s=31\:\mathrm{N}$, since the weight to be held doesn't change. But the maximum that the friction can reach now gets closer to the value it must take as you ease on your pressing force.
At some point you have reduced $F_N$ so much that the maximum limit for $F_s$ is equal to the needed value to hold the book. Now the friction is at its maximum, and this corresponds to the minimum pressure you must provide to avoid the book from falling.