The normal force on a thin stick being lifted on a surface

[haruspex]

This means that N goes to 0 when F_0 goes to 0

Yes, and $F_0$ is just shorthand for $\frac 12mg$, so, for $\beta\geq 1$, N=0 implies m=0.

 

[haruspex]

I obtain: $\delta F = \frac{1+\beta}{1-\beta} F_0$, or $F = \frac{2}{1-\beta} F_0$. What does this tell me? Is there any particular information?

It says that , for $\beta\geq 1$, assuming N can be zero gives a silly answer, so it can't go to zero.

 

[Hak]

Yes, and $F_0$ is just shorthand for $\frac 12mg$, so, for $\beta\geq 1$, N=0 implies m=0.

Why for $\beta \ge 1$ and not for only $\beta = 1$? Thanks.

 

[haruspex]

Why for $\beta \ge 1$ and not for only $\beta = 1$? Thanks.

Because for $\beta>1$ the equation you quoted has the form $N=F_0+(positive factor)\delta F$.

 

[Hak]

It makes sense because otherwise you have made two contradictory assumptions: $N>0, \beta=1$. The only way both can be true is mg=0.

Reviewing post #53, I cannot understand the above statement. Why the only way to make $N > 0$ and $\beta = 1$ is $mg = 0$? Perhaps you meant to say $N = 0$? I cannot understand it.

 

[haruspex]

Perhaps you meant to say N=0?

Yes, good catch. Corrected.