Block at rest on an inclined surface

In summary, In the problem (ii),the friction force is ##Mgcosθμ##,the component of weight in the inclined surface is ##Mgsinθ## and the component of the applied force ##F## along the inclined surface is ##Mgcosθ##.Here the component of ##F## is acting opposite to the direction of the component of weight.Then what will be the direction of friction force.Please clear me this without any question.If it would be a plane surface,then there would work only ##F## and the friction force and to remain the block in rest for that case,the friction force had to be greater or equal to ##F##.But being the surface inclined,
  • #36
Sorry for that,I have edited my post.Now see it and clearly answer the questions.
 
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  • #37
Akash47 said:
Sorry for that,I have edited my post.Now see it and clearly answer the questions.
Without having worked the problem your way, those formulas look plausible. But they go singular when ##\mu## = 1.

Hint: With ##\mu## = 1, the inclined plane could be vertical, the tangent would be infinite and the object would not fall. [Think of pressing a book against a wall to prevent its fall].

With ##\mu## > 1, is there still a limiting angle?

There is a different approach to the problem that removes the singularity. What if you started by combining W+F into one combined force and rotated your coordinate system by 45 degrees? [Don't forget that 45 degree offset when reporting your computed result]
 

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