TITLE: Geometric equations derived from the block on a plane problem
QUESTION [1 upvotes]: From the above diagram, you get
$$N=Mg\cos\theta \tag{1}$$
But drawing the components of $N$ gives you
$$N\cos\theta=mg \tag{2}$$
$$N=\frac{mg}{\cos\beta}$$
Obviously, both of them can't be correct, I would like to know what is wrong in 2nd equation. Sorry for such an elementary question but it seems I don't really understand components of a vector

REPLY [1 votes]: The second equation is wrong due to assumptions.  
Writing $N \cos \theta = mg$ is not a true statement in most conditions.
The equation $N = mg \cos \theta$ comes from assuming that the mass is supported by the incline, so if those two are not equal, there would be acceleration into or away from the slope, which is not something you would see in real life.

REPLY [0 votes]: You cant write $ N cos\theta = mg $ as block will have some acceleration along plane of inclination which will have some component in downward direction. If block has acceleration a downwards along plane, then we must write $$ mg - N cos\theta = masin\theta $$  and this equation indeed is satisfied.

REPLY [0 votes]: The mass is accelerating down the slope  so there is a component of the acceleration along the slope $a$ vertically downwards.
Let this component of acceleration  be $a_{\rm vertical}=a\sin\theta$.
Using Newton's second law $F=ma$ gives $mg-N \cos \theta = m a_{\rm vertical}$ not zero as you have assumed.
{"set_name": "stack_exchange", "score": 1, "question_id": 314481}