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

[Hak]

Lower limit for the normal force
To find the lower limit for $N$, we solve the torque equation for $ma$
$ma=\frac{3}{2}(F-\frac{1}{2}mg)\dfrac{x^2}{x^2+y^2}.$
Then we substitute in the force equation
$F+N=mg+\frac{3}{2}(F-\frac{1}{2}mg)\dfrac{x^2}{x^2+y^2}$
and solve for $N$, $$N=\frac{1}{2}F+mg\left[1-\frac{3x^2}{4(x^2+y^2)}\right].$$

Shouldn't it be $N = F \left[\frac{3}{2} \frac{x^2}{x^2+y^2} -1 \right]+mg\left[1-\frac{3x^2}{4(x^2+y^2)}\right]$? It seems to me that you have omitted to consider a term $\frac{x^2}{x^2+y^2}$ at $F$. I suggest you review the calculations; it may well be that I am wrong. In case I am right, how would your subsequent considerations change? They would no longer be the same, right? How would you modify them?

 

[kuruman]

Sorry to bother and reopen this thread. By what steps did you arrive at this equation? Thank you very much.

It's the torque equation about corner P.
$\tau_{net}=I_P~\alpha$
$F$ is applied at lever arm $2x$ and the weight is applied at lever arm $x$ in the opposite direction so the left hand side is
$LHS=F(2x)-mgx$.
On the right the moment of inertia about P is obtained using the parallel axis theorem

$\cancel{I_P=I_{cm}+Mr^2=\frac{1}{12}m(2x^2+2y^2)+m(x^2+y^2)=\frac{4}{3}m(x^2+y^2)}.$

$I_P=I_{cm}+Mr^2=\frac{1}{12}m\left[(2x)^2+(2y)^2\right]+m(x^2+y^2)=\frac{4}{3}m(x^2+y^2).$
The angular acceleration is related to the tangential acceleration of the center of mass by $\alpha = \dfrac{a}{x}.$ So the right hand side is
$RHS=\frac{4}{3}m(x^2+y^2)\dfrac{a}{x}.$

(Edited to fix typo in the equation for $I_P.$ See post #43.)

 

[Hak]

Shouldn't it be $N = F \left[\frac{3}{2} \frac{x^2}{x^2+y^2} -1 \right]+mg\left[1-\frac{3x^2}{4(x^2+y^2)}\right]$? It seems to me that you have omitted to consider a term $\frac{x^2}{x^2+y^2}$ at $F$. I suggest you review the calculations; it may well be that I am wrong. In case I am right, how would your subsequent considerations change? They would no longer be the same, right? How would you modify them?

Lower limit for the normal force
To find the lower limit for $N$, we solve the torque equation for $ma$
$ma=\frac{3}{2}(F-\frac{1}{2}mg)\dfrac{x^2}{x^2+y^2}.$
Then we substitute in the force equation
$F+N=mg+\frac{3}{2}(F-\frac{1}{2}mg)\dfrac{x^2}{x^2+y^2}$
and solve for $N$, $$N=\frac{1}{2}F+mg\left[1-\frac{3x^2}{4(x^2+y^2)}\right].$$ Now we see what's going on. The tipping condition says that $F$ must be greater than $\frac{1}{2} mg$ which sets a lower limit of $\frac{1}{4}mg$ to the first term on the RHS. The second term on the RHS has a lower limit also of $\frac{1}{4}mg$ in the thin-stick limit $x\rightarrow \infty.$ Therefore the lower limit of $N$ is $\frac{1}{2}mg~$ and the normal force cannot vanish.

I just noticed that in the stick limit $x \to \infty$, the first term on the RHS becomes $\frac{1}{2}F$, and since you imposed the condition $F > \frac{1}{2}mg$, then $\frac{1}{2}F > \frac{1}{4}mg$, the same result you arrived at. At first glance, it seems that in the expression of $N$ you have (I don't know whether voluntarily) computed the limit for $x \to \infty$ for the first term on the RHS, but did not do so for the second term on the RHS (only later). However, I believe that a correct approach should not involve an expression with a mixture of terms for which the limit has already been made and terms in which $x$ is still present. I do not know if I have explained myself well, if I am wrong, please do not hesitate to correct me.

 

[kuruman]

Shouldn't it be $N = F \left[\frac{3}{2} \frac{x^2}{x^2+y^2} -1 \right]+mg\left[1-\frac{3x^2}{4(x^2+y^2)}\right]$? It seems to me that you have omitted to consider a term $\frac{x^2}{x^2+y^2}$ at $F$. I suggest you review the calculations; it may well be that I am wrong. In case I am right, how would your subsequent considerations change? They would no longer be the same, right? How would you modify them?

You are correct that I missed a factor multiplying $F$. Please note that in post #24 I redid this but using the CM as reference for the torques and the moment of inertia in order to bring my calculations in line with what the other contributors to this thread were doing. The expressions in post #24 are correct.

In case I am right, how would your subsequent considerations change?

Feel free to see for yourself. I am sure you can do it.

They would no longer be the same, right?

Right.

How would you modify them?

Sorry, I have no desire to modify them and spend more time on this. I have said all I have to say about this problem, which as @haruspex found out with his inquiries at AIP, was a hastily constructed problem.

 

[Hak]

You are correct that I missed a factor multiplying $F$. Please note that in post #24 I redid this but using the CM as reference for the torques and the moment of inertia in order to bring my calculations in line with what the other contributors to this thread were doing. The expressions in post #24 are correct.

Feel free to see for yourself. I am sure you can do it.

Right.

Sorry, I have no desire to modify them and spend more time on this. I have said all I have to say about this problem, which as @haruspex found out with his inquiries at AIP, was a hastily constructed problem.

I made my remarks on this in post #38. Could you please tell me if they are correct? I thank you.

 

[Hak]

Right.

According to my remarks in post #38, actually not. They should be the same considerations, so I would have been wrong to think otherwise. Why do you claim, then, that the subsequent considerations change? Thank you.

 

[Hak]

$I_P=I_{cm}+Mr^2=\frac{1}{12}m(2x^2+2y^2)+m(x^2+y^2)=\frac{4}{3}m(x^2+y^2).$

There is probably an error here. The equation fits only if in the quantity within parentheses the $2$ is replaced with $4$. After all, $2$ must also be squared, not just $x$ and $y$.

 

[haruspex]

There is probably an error here. The equation fits only if in the quantity within parentheses the $2$ is replaced with $4$. After all, $2$ must also be squared, not just $x$ and $y$.

Some missing parentheses around 2x, 2y, I suggest.

 

[kuruman]

Some missing parentheses around 2x, 2y, I suggest.

Yes, it should be $$I_P=I_{cm}+Mr^2=\frac{1}{12}m\left[(2x)^2+(2y)^2\right]+m(x^2+y^2)=\frac{4}{3}m(x^2+y^2).$$ Thanks.

 

[Hak]

Eliminating a instead:
$N(1+\frac 3{1+\frac{y^2}{x^2}})=mg-F(1-\frac 3{1+\frac{y^2}{x^2}})$.
Same problem. If $y>x\sqrt 2$ then, for sufficiently large F, N vanishes.

I'm so sorry, but I can't understand this passage. What chains of inequalities starting from $y>x\sqrt 2$ lead to the conclusion that $N$ vanishes? Thank you.

 

[haruspex]

I'm so sorry, but I can't understand this passage. What chains of inequalities starting from $y>x\sqrt 2$ lead to the conclusion that $N$ vanishes? Thank you.

$N(1+\frac 3{1+\frac{y^2}{x^2}})=mg-F(1-\frac 3{1+\frac{y^2}{x^2}})$.
If $y>x\sqrt 2$ then $1-\frac 3{1+\frac{y^2}{x^2}}>0$. So increasing $F$ reduces the RHS, without limit. For sufficiently large F, the RHS becomes zero.

 

[Hak]

$\dfrac{J_2}{J_1}=1-\dfrac{3x^2}{x^2+y^2}.$

Shouldn't it be $\dfrac{J_1}{J_2}=1-\dfrac{3x^2}{x^2+y^2}$? I can't get to the bottom of it. My correction, unfortunately, would unhinge everything.

 

[Hak]

Yes, you're right. I turned the threshold around in my head. I have been intrigued by this threshold for the past fast and I came up with a model for the initial motion of the thick stick a.k.a. $(2x)\times(2y)$ rectangle.

Imagine the rectangle lying flat on a frictionless surface at rest. At t = 0 two impulses $J_1$ and $J_2$ are delivered respectively at the CM and the upper right corner (see figure below). Find the initial velocity of the lower left corner. The rationale is that insight might be gained if one looks at the circumstances under which the velocity of that point is zero. Specifically, if the vertical component of the velocity is zero, that means no vertical impulse is delivered at its point of support, i.e. the normal force will not change from its static value of $\frac{1}{2}mg.$
View attachment 332433
From linear and angular momentum conservation
$\vec{V}_{cm}=(\dfrac{J_2}{m}-\dfrac{J_1}{m})\mathbf{\hat y}$

$\vec{\omega}=\dfrac{3J_2~ x}{m(x^2+y^2)}\mathbf{\hat z}.$

The velocity of point P is
$\vec V_P=\vec{V}_{cm}+\vec{\omega}\times (-x~\mathbf{\hat x}-y~\mathbf{\hat y})=\dfrac{3J_2 ~xy}{m(x^2+y^2)}\mathbf{\hat x}+\left[\dfrac{J_2}{m}\left(1-\dfrac{3x^2}{x^2+y^2}\right)-\dfrac{J_1}{m} \right]\mathbf{\hat y}$
We see that the horizontal component of point P is zero only in the thin stick limit $y=0$. The vertical component is zero when
$\left[\dfrac{J_2}{m}\left(1-\dfrac{3x^2}{x^2+y^2}\right)-\dfrac{J_1}{m} \right]=0\implies \dfrac{J_2}{J_1}=1-\dfrac{3x^2}{x^2+y^2}.$

In the problem posed, we are looking for the normal force at t = 0 when the forces are "turned on". We identify $J_2=F~\Delta t$ and $J_1=mg~\Delta t$ to get $$\frac{F}{mg}=1-\dfrac{3x^2}{x^2+y^2}=1-\beta$$ as the condition for the initial vertical velocity to be zero.

What does this mean? Here's what I think.

1. When the RHS turns negative, the vertical component of corner P cannot be zero. We know that because in the thin stick approximation ($y=0$) that's the case and RHS = - 2.
2. When the RHS is positive, it gives the ratio of the applied force to the weight that will result in the corner accelerating horizontally but not vertically.

We can replace $F=mg(1-\beta)$ in $~N(\beta+1)=mg+F(\beta-1)~$ to get the normal force when the velocity of P has no vertical component $~N_0=\dfrac{\beta(2-\beta)}{1+\beta}.$

It is interesting to note that at the threshold $\beta = 1$ ($y^2=2x^2$) $\dfrac{F}{mg}=0$ and $N_0=\frac{1}{2}mg$. Thus, at threshold point P will have zero vertical velocity only if the stick is left alone at equilibrium.

If it is not left alone at threshold, point P will have vertical velocity $V_{P,y}=-\dfrac{J_1}{m}=-\dfrac{mg\Delta t}{m}=-{g\Delta t}$ which is (perhaps not surprisingly) the free fall result for the CM. Curiouser and curiouser.

With my correction, at the threshold, we would have $mg = 0$ and $N_0= 0$. A nonsense physical situation. Maybe we have to switch $J_1$ and $J_2$ within brackets, it would make much more sense.

 

[haruspex]

With my correction, at the threshold, we would have $mg = 0$ and $N_0= 0$. A nonsense physical situation. Maybe we have to switch $J_1$ and $J_2$ within brackets, it would make much more sense.

Which correction? I agree post #30 switches $J_1, J_2$. It should arrive at $mg=F(1-\beta)$. But how do you get mg=0?

 

[Hak]

Which correction? I agree post #30 switches $J_1, J_2$. It should arrive at $mg=F(1-\beta)$. But how do you get mg=0?

Pay attention to this proposition in @kuruman's post.

It is interesting to note that at the threshold $\beta = 1$ ($y^2=2x^2$) $\dfrac{F}{mg}=0$ and $N_0=\frac{1}{2}mg$. Thus, at threshold point P will have zero vertical velocity only if the stick is left alone at equilibrium.

At the threshold, $\beta =1$. If you substitute in the expression you correct, you will have $mg = F(1-1) = 0$. I am not convinced.

 

[haruspex]

Pay attention to this proposition in @kuruman's post.At the threshold, $\beta =1$. If you substitute in the expression you correct, you will have $mg = F(1-1) = 0$. I am not convinced.

I do find post #30 confusing. In the equations, there is no normal force at P, so the model has replaced mg with some other force of the same magnitude but the whole is free floating.
If P has no vertical movement in that model, what does that correspond to in the original? Seems to me it is N=0, not $N=\frac 12mg$. But at the threshold value, $N=\frac 12mg$. If you insist on N=0 but $\beta=1$ then, yes, you will deduce mg=0.

 

[Hak]

I do find post #30 confusing. In the equations, there is no normal force at P, so the model has replaced mg with some other force of the same magnitude but the whole is free floating.
If P has no vertical movement in that model, what does that correspond to in the original? Seems to me it is N=0, not $N=\frac 12mg$. But at the threshold value, $N=\frac 12mg$. If you insist on N=0 but $\beta=1$ then, yes, you will deduce mg=0.

And according to you it makes sense to say that $mg=0$? I don't understand...

 

[haruspex]

And according to you it makes sense to say that $mg=0$? I don't understand...

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.
Correction: two contradictory assumptions: $N=0, \beta=1$.

 

[Hak]

It makes sense because otherwise you have made two contradictory assumptions: $N>0, \beta=1$.

Why are they contradictory?

 

[Hak]

I don't understand why

at the threshold value, $N=\frac 12mg$.

According to the following equation initially pointed out:
$N(\beta+1)=mg+F(\beta-1)$
If we replace $mg=F(1-\beta)$ in it, we will have:

$$N(\beta+1) = mg + mg \frac{(\beta-1)}{(1-\beta)} = mg - mg \frac{\cancel{(\beta-1)}}{\cancel{(\beta -1)}} = 0$$.

However, the expression is not defined at $\beta = 1$, since the denominator would go to $0$. The expression would result $N = \frac{mg}{2}$, as you say, if and only if $mg \frac{(\beta-1)}{(1-\beta)} = 0$ in the case where $\beta = 1$: in fact, we would have $2N = mg$. But we can't do that, as far as I know: as already mentioned, the denominator of this expression is $0$, so it is inherently contradictory...

If we leave the expression as is, i.e., $N(\beta+1)=mg+F(\beta-1)$, substituting the threshold value $\beta = 1$, we will have:

$2N = mg$, that is, $N = \frac{mg}{2}$, regardless of $F$.

My conclusion is that it is not possible to apply the expression $\left[\dfrac{J_2}{m}\left(1-\dfrac{3x^2}{x^2+y^2}\right)-\dfrac{J_1}{m} \right]=0\implies \dfrac{J_1}{J_2}=1-\dfrac{3x^2}{x^2+y^2} \implies \frac{mg}{F}=1-\dfrac{3x^2}{x^2+y^2}=1-\beta$ in the case where $\beta = 1$, that is, at the threshold value. That is why this is a special case that is difficult to analyze: it is the limiting case.

Correct me where you think I am wrong. Thank you very much.

 

[haruspex]

I don't understand why

See my post #11.

According to the following equation initially pointed out:
$N(\beta+1)=mg+F(\beta-1)$
If we replace $mg=F(1-\beta)$ in it, we will have:

$$N(\beta+1) = mg + mg \frac{(\beta-1)}{(1-\beta)} $$.

No, we will have $N(\beta+1)=0$.

 

[Hak]

See my post #11.

No, we will have $N(\beta+1)=0$.

Why is my expression wrong? I substituted only and I obtained the same equation as you did...

Is my following reasoning correct?

 

[haruspex]

Why is my expression wrong? I substituted only and I obtained the same equation as you did...

Not quite. You first turned $mg=F(1-\beta)$ into $F=\frac{mg}{(1-\beta)}$, which is invalid if $\beta=1$. If you use $mg=F(1-\beta)$ to replace mg instead you get $N(1+\beta)=0$.

 

[Hak]

Not quite. You first turned $mg=F(1-\beta)$ into $F=\frac{mg}{(1-\beta)}$, which is invalid if $\beta=1$. If you use $mg=F(1-\beta)$ to replace mg instead you get $N(1+\beta)=0$.

Of course, you are absolutely right. But how do you judge my subsequent proceedings? It seems to me that the case where $\beta = 1$ is not covered by the previous equation, so that is why it is peculiar and surprising. It means that it cannot in any way be guessed by a normal analysis.

 

[kuruman]

It seems that I created a mess by switching $J_1$ and $J_2$ in post #30. It behooves me to offer my apologies and then clean up my mess. Here are my revised arguments.

Assume a model in which the "stick" lies on its side on a frictionless surface and that impulses $J_1=mg\Delta t$ and $J_2=F\Delta t$ are delivered simultaneously as shown in the diagram, post #30. Consider the y-component of the velocity of point P. As @haruspex pointed out in post #51, there is no normal force at that point in the model but there is in the actual problem. Below I explain why I think the model is applicable.

The normal force is what is necessary to provide the observed acceleration. Since point P in the model starts from rest, the observed acceleration will be in the direction of the initial velocity. If in the model the initial velocity is in the negative y-direction, in the actual problem the normal force will increase above the static value of $\frac{1}{2}mg.$ Of more interest is the condition(s) under which the normal force will decrease and may even become to zero. This possibility could be realized if in the model the initial velocity of point P is in the positive y-direction. Then in the actual problem the normal force could be diminished below the static value of $\frac{1}{2}mg$, perhaps all the way down to zero.

The condition for positive vertical velocity at P is $${J_2}(1-\beta)-{J_1} >0~~~~\left(\beta \equiv \frac{3x^2}{x^2+y^2}\right).$$Let us consider a rod of fixed length $2x$ and variable height $2y$. The range of values for $\beta$ is $0<\beta<3.$ We substitute $J_1=mg\Delta t$ and $J_2=F\Delta t$ to get $$F(1-\beta) >mg~~~~(0<\beta<3).\tag{1}$$Inequality (1) is a condition that must be satisfied for the normal force to drop below the static value. It can be satisfied only in the range $0<\beta<1$ because $mg$ cannot be less than a negative number.

Specifically, note that $\beta =0$ corresponds to $y\rightarrow \infty$ at constant $x$ which is essentially describing a vertical stick standing on its tip. Then $F=mg$ is the threshold for the condition to be satisfied. Well, we know that the pulling force must be greater than the weight of the stick in order to lift it off the surface ($N=0$) so there is nothing new here. The CM is already above the "corner" P, there is no rotation of P about the CM, therefore P can only go up when the $F$ is greater than the weight.

What about sticks such that $0<\beta<1$? From post #24 we have the equation $$N(\beta+1)=mg+F(\beta-1)\implies N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right].\tag{2}$$We want to know what $N$ looks like when condition (1) is satisfied. If we replace $F(1-\beta)$ in the RHS of equation (2) with the smaller term $mg$ from inequality (1), then LHS > RHS: $$N>\frac{1}{1+\beta}(mg-mg)\implies N>0.$$It looks like the normal force will be positive and hence non-zero for all values of $\beta$. All this agrees with one's intuition. As the stick is lifted off the surface, point P rotates about the CM and pushes down on the surface thus maintaining a non-zero normal force. The normal force will not go to zero unless the CM is above P in which case the stick is lifted straight off the surface ($N=0$). The actual problem asks for the normal force "shortly after the right end of the stick leaves the surface". This happens before the CM and point P are lined up along the vertical unless, as we have seen, the stick is thin and already along the vertical.

 

[Hak]

It seems that I created a mess by switching $J_1$ and $J_2$ in post #30. It behooves me to offer my apologies and then clean up my mess. Here are my revised arguments.

Assume a model in which the "stick" lies on its side on a frictionless surface and that impulses $J_1=mg\Delta t$ and $J_2=F\Delta t$ are delivered simultaneously as shown in the diagram, post #30. Consider the y-component of the velocity of point P. As @haruspex pointed out in post #51, there is no normal force at that point in the model but there is in the actual problem. Below I explain why I think the model is applicable.

The normal force is what is necessary to provide the observed acceleration. Since point P in the model starts from rest, the observed acceleration will be in the direction of the initial velocity. If in the model the initial velocity is in the negative y-direction, in the actual problem the normal force will increase above the static value of $\frac{1}{2}mg.$ Of more interest is the condition(s) under which the normal force will decrease and may even become to zero. This possibility could be realized if in the model the initial velocity of point P is in the positive y-direction. Then in the actual problem the normal force could be diminished below the static value of $\frac{1}{2}mg$, perhaps all the way down to zero.

The condition for positive vertical velocity at P is $${J_2}(1-\beta)-{J_1} >0~~~~\left(\beta \equiv \frac{3x^2}{x^2+y^2}\right).$$Let us consider a rod of fixed length $2x$ and variable height $2y$. The range of values for $\beta$ is $0<\beta<3.$ We substitute $J_1=mg\Delta t$ and $J_2=F\Delta t$ to get $$F(1-\beta) >mg~~~~(0<\beta<3).\tag{1}$$Inequality (1) is a condition that must be satisfied for the normal force to drop below the static value. It can be satisfied only in the range $0<\beta<1$ because $mg$ cannot be less than a negative number.

Specifically, note that $\beta =0$ corresponds to $y\rightarrow \infty$ at constant $x$ which is essentially describing a vertical stick standing on its tip. Then $F=mg$ is the threshold for the condition to be satisfied. Well, we know that the pulling force must be greater than the weight of the stick in order to lift it off the surface ($N=0$) so there is nothing new here. The CM is already above the "corner" P, there is no rotation of P about the CM, therefore P can only go up when the $F$ is greater than the weight.

What about sticks such that $0<\beta<1$? From post #24 we have the equation $$N(\beta+1)=mg+F(\beta-1)\implies N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right].\tag{2}$$We want to know what $N$ looks like when condition (1) is satisfied. If we replace $F(1-\beta)$ in the RHS of equation (2) with the smaller term $mg$ from inequality (1), then LHS > RHS: $$N>\frac{1}{1+\beta}(mg-mg)\implies N>0.$$It looks like the normal force will be positive and hence non-zero for all values of $\beta$. All this agrees with one's intuition. As the stick is lifted off the surface, point P rotates about the CM and pushes down on the surface thus maintaining a non-zero normal force. The normal force will not go to zero unless the CM is above P in which case the stick is lifted straight off the surface ($N=0$). The actual problem asks for the normal force "shortly after the right end of the stick leaves the surface". This happens before the CM and point P are lined up along the vertical unless, as we have seen, the stick is thin and already along the vertical.

Music to my ears! Thank you.

 

[haruspex]

$$F(1-\beta) >mg~~~~(0<\beta<3).\tag{1}$$

What about sticks such that $0<\beta<1$? From post #24 we have the equation $$N(\beta+1)=mg+F(\beta-1)\implies N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right].\tag{2}$$We want to know what $N$ looks like when condition (1) is satisfied. If we replace $F(1-\beta)$ in the RHS of equation (2) with the smaller term $mg$ from inequality (1), then LHS > RHS: $$N>\frac{1}{1+\beta}(mg-mg)\implies N>0.$$

When I use (1) in (2) I get N<0.

 

[Hak]

When I use (1) in (2) I get N<0.

Me too. Indeed:

$$F(1-\beta) > mg \Rightarrow \ - F(1-\beta) < - mg \Rightarrow \ mg - F(1-\beta) < mg - mg \Rightarrow \ mg+ F(\beta - 1) < 0 \Rightarrow \ N(\beta +1) < 0 \Rightarrow \ N<0 \ \ \text{if} \ \ \ 1+\beta > 0$$.

What do we do now? What happens?

 

[haruspex]

Me too. Indeed:

$$F(1-\beta) > mg \Rightarrow \ - F(1-\beta) < - mg \Rightarrow \ mg - F(1-\beta) < mg - mg \Rightarrow \ mg+ F(\beta - 1) < 0 \Rightarrow \ N(\beta +1) < 0 \Rightarrow \ N<0 \ \ \text{if} \ \ \ 1+\beta > 0$$.

@kuruman wrote:

Inequality (1) is a condition that must be satisfied for the normal force to drop below the static value.

I disagree. As I showed in post #11, it is the condition for N to go negative. If you plug in such a condition, finding that N is negative is not surprising.
I'll see if I can identify a flaw in @kuruman's argument; there has to be one.

 

[Hak]

@kuruman wrote:

I disagree. As I showed in post #11, it is the condition for N to go negative. If you plug in such a condition, finding that N is negative is not surprising.
I'll see if I can identify a flaw in @kuruman's argument; there has to be one.

Does this mean that @kuruman's model cannot be applied (as he said) to the problem, or that there is an error within the argument which, when fixed, allows the model to be saved and made valid?

 

[haruspex]

Does this mean that @kuruman's model cannot be applied (as he said) to the problem, or that there is an error within the argument which, when fixed, allows the model to be saved and made valid?

Too soon to say.

 

[Hak]

Too soon to say.

OK, now I'm curious. Let me know, @kuruman and @haruspex, what you can find out and conclude. Thanks.

 

[haruspex]

OK, now I'm curious.

I had the impression it was a permanent state.

 

[Hak]

I had the impression it was a permanent state.

Do you consider this a negative aspect? Haha

 

[haruspex]

Do you consider this a negative aspect? Haha

Not at all.... just a little time-consuming.