Consider a block of weight W lying on the horizontal board.
As in the case when $\theta=0$ and we apply force parallel to board, friction force will increase until the impending of motion and equals to μN.
Does friction force fs increases also when inclined at some angle?
I think fs, the static friction force decrease because when the board is inclined Normal force becomes N=W $\cos\theta$, normal force decreases with increase in $\theta$.
Am I thinking right? If not then explain.
Or If I am right then it should be written in books that friction force decreases with inclination.
[Physics] Static friction force increases or decreases with increase in inclination until impending of motion
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Related Solutions
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.
As long as the object does not slide, the static frictional force is equal to $mg\sin \theta$. Once the frictional force also becomes equal to the normal force $mg\cos \theta$ times the coefficient of static friction, the object begins to slide. But the static friction law is an inequality, not an equality. $$F\leq \mu_s N$$ As long as the F is less than $\mu_sN$, the object won't slide. The static friction coefficient only tells you the value of the friction force at the point where the object is just on the verge of sliding.
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Best Answer
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.