Block at rest on an inclined surface

[haruspex]

Yes, that is all correct, except I do not understand this line:

there is still a case when Fcosθ=Wsinθ

In what sense is this a special case? Isn't it just encompassed by the range of values for which the system is static?

 

[Akash47]

Oh! It's not a special case.When $Fcosθ=Wsinθ$,then there is no friction force working.How will you explain it?

 

[haruspex]

Oh! It's not a special case.When $Fcosθ=Wsinθ$,then there is no friction force working.How will you explain it?

What's to explain? If the net of the other forces tending to make the surfaces slide against each other is zero then there is no frictional force.

 

[Akash47]

I was saying that there is a case $θ=45^°$ when also the block will not move and there will be no friction.And the range of the values of θ is$:tan^{-1}\frac{1-μ}{1+μ} ≤θ≤ tan^{-1}\frac{1+μ}{1-μ}$.And to get the value θ=45,you have to put μ=0 in either the lower bound or the upper bound.But can a static friction co-efficient of a surface ever
be zero(at least not in this problem where μ is clearly mentioned)?Then what is your answer to this?
Though the question asks for the range of values in terms of μ (which don't really need the value θ=45),I think the full solution also need the value $θ=45^°$.

 

[jbriggs444]

I was saying that there is a case θ=45 when also the block will not move and there will be no friction.And the range of the values of θ is$: \frac{1-μ}{1+μ} ≤θ≤ \frac{1+μ}{1-μ}$.And to get the value θ=45,you have to put μ=0 in either the lower bound or the upper bound.But can a static friction co-efficient of a surface ever
be zero(at least not in this problem where μ is clearly mentioned)?Then what is your answer to this?
Though the question asks for the range of values in terms of μ (which don't really need the value θ=45),I think the full solution also need the value θ=45.

It is clear that $\theta$ = 45 degrees falls within the solution set regardless of the value of $\mu$

You are going to need to write down some equations so that we can see where you are going wrong. It is pretty clear that those formulas for $\theta$ are not correct. For one thing, they give a result that is unitless, but you are using degrees, so $\theta$ should not be unitless.

 

[Akash47]

Sorry for that,I have edited my post.Now see it and clearly answer the questions.

 

[jbriggs444]

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]