I did it but a little more understanding needed

[neelakash]

Homework Statement

It is a sort of looking back...that I am doing these old problems.However,I have attached the figures in the pdf file...where the diagram of the two problems are given.I included two problems as my question is a single one that is common to these two problems.

In the 1st one,mass of the block=m.The normal force N acts upward.External force F inclined at an angle θ to the vertical.Friction co-eff μ.Force F needed to get the block just started is (This is the first part of the problem and I have calculated this)
F=μmg/[sinθ-μ cosθ]

In the 2nd one,I found out the maximum value of M/m so that the system is at rest.M is the mass on the inclined plane.Friction co-eff is μ.
the maximum value is M/m= μ/[sinθ-μ cosθ]


My question here is what will be the situation when tanθ<μ

Homework Equations

The Attempt at a Solution

The condition means (f/N)<μ or,f<μN or,f<f(max)

In the 1st problem it means that friction has not yet attained the maximum value and we must have the block in equilibrium...That is friction automatically adjusts itself to be equal and opposite to F sinθ all the time.I mean we increase the value of F and friction being equal and opposite,no motion results.And with tanθ<μ,we cannot increase F to that value as obtained...

Now I am not sure that my explanation conforms that given in the book.It is a worked out example,in fact.
The condition also means that F<0 as I calculated.They say that "for angles less than θ,one cannot push the block ahead,however large the force may be".

In the 2nd problem,I will say that friction< maximum possible value of friction and that means that it will adjust itself such that no motion would occur.That it friction will be equal and opposite to the resultant of (mg sinθ-T).

Here the book says "since at the condition M/m ratio<0,hence,no motion would occur"...


My point in these two problems is that F<0 or, M/m<0 might imply that tanθ<μ leads to such bizarre result...It may be that the two explanations are equivalent...please help me in better understanding...

 

[FedEx]

Homework Statement

It is a sort of looking back...that I am doing these old problems.However,I have attached the figures in the pdf file...where the diagram of the two problems are given.I included two problems as my question is a single one that is common to these two problems.

In the 1st one,mass of the block=m.The normal force N acts upward.External force F inclined at an angle θ to the vertical.Friction co-eff μ.Force F needed to get the block just started is (This is the first part of the problem and I have calculated this)
F=μmg/[sinθ-μ cosθ]

In the 2nd one,I found out the maximum value of M/m so that the system is at rest.M is the mass on the inclined plane.Friction co-eff is μ.
the maximum value is M/m= μ/[sinθ-μ cosθ]


My question here is what will be the situation when tanθ<μ

Homework Equations

The Attempt at a Solution

The condition means (f/N)<μ or,f<μN or,f<f(max)

In the 1st problem it means that friction has not yet attained the maximum value and we must have the block in equilibrium...That is friction automatically adjusts itself to be equal and opposite to F sinθ all the time.I mean we increase the value of F and friction being equal and opposite,no motion results.And with tanθ<μ,we cannot increase F to that value as obtained...

Now I am not sure that my explanation conforms that given in the book.It is a worked out example,in fact.
The condition also means that F<0 as I calculated.They say that "for angles less than θ,one cannot push the block ahead,however large the force may be".

In the 2nd problem,I will say that friction< maximum possible value of friction and that means that it will adjust itself such that no motion would occur.That it friction will be equal and opposite to the resultant of (mg sinθ-T).

Here the book says "since at the condition M/m ratio<0,hence,no motion would occur"...


My point in these two problems is that F<0 or, M/m<0 might imply that tanθ<μ leads to such bizarre result...It may be that the two explanations are equivalent...please help me in better understanding...

I am not able to see the figure

 

[neelakash]

I am sorry,but I can see the file named Doc1.pdf.

However,I attached it once more.
Hope now you will be able to see it.

<original file approved--Doc Al>