My book says that 'Between surface of contact between two bodies there are contact forces applied between each other (they are equal in magnitude but opposite in direction,obeying Newton's 3rd law) due to electromagnetic interaction. The perpendicular component of the contact force is the Normal force while the parallel component is the frictional force. Friction is caused due to the coldwelding ie. formation of bond between the micro contact areas. When they distort due to external force,they create vibration waves which become damped releasing heat.' But it didn't tell about the mechanism of static friction, how it happens in the microscopic level. As the external force on the body increases, static frictional force also increases its value (upto a level). How does the force of static friction increase ? What is the cause of it at the microscopic level? Does the bond become more strong? How does this increase in static friction happen?
[Physics] How does static friction increase with increase in the applied force
forcesfriction
Related Solutions
You have four forces acting on the body.
A horizontal force, a frictional force, a normal reaction and the weight of the body.
Your free body digram does not show all four forces.
You need to apply Newton's second law parallel to the plane and perpendicular to the plane and the perpendicular application should answer your question about the normal force and hence the frictional force.
Update
Here is my free body diagram and yours with the normal reaction and the frictional force added.
Note the $F \sin 30^\circ$ component of $F$ which is pushing down on the object and thus making $N$ greater than $mg \cos 30^\circ$ by that amount and so increasing the value of the frictional force along the slope.
Using Newton's second law you can two equations with two unknowns, $F$ and $N$, and hence solve for $F$.
Remember that the maximum static frictional force is $\mu\, N$.
If the board is horizontal no frictional force is necessary to keep the block stationary.
Inclining the board will require there to be a static frictional force acting on the block equal in magnitude but opposite in direction to the component of the weight down the slope $mg\sin \theta$.
As $\theta$ increase so the static frictional force increases.
This continues until the angle of the board is such that the maximum static frictional force $\mu mg\cos \theta$ is equal to the component of the weight of the block down the slope $mg\sin \theta$.
So $\mu = \tan \theta$ and this the basis of a method for finding the coefficient of static friction.
You are correct in saying that the maximum possible static frictional force $\mu mg \cos \theta$ decreases with increasing board angle.
Best Answer
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:
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.