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

[Hak]

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

Yes, I agree. In this case, however, you have to admit that it is worth it, at least a little. I am sure it will bring good benefits.

 

[haruspex]

there is no normal force at that point in the model

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 12mg.

No, in the actual problem the normal force will be >0.
Consider F=0. In your model, that only leaves $J_1$, a downward impulse, so P will have an initial downward movement, yet in the actual problem the normal force remains at $mg/2$.

 

[kuruman]

No, in the actual problem the normal force will be >0.
Consider F=0. In your model, that only leaves $J_1$, a downward impulse, so P will have an initial downward movement, yet in the actual problem the normal force remains at $mg/2$.

Yes, I have been queasy about this point. This thread has become the bane of my existence with all the mistakes I have made. There is another issue that bothers me. We have derived the expression $$N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right]$$ from the force equation and the torque equation relative to the CM. For $\beta =2$ this expression gives $N=\frac{1}{3}(mg+F).$

Now let's find the expression for $N$ with the torque equation relative to the point contact P as I started doing earlier. $$\begin{align} & F+N=mg+ma \nonumber \\ & F(2x)-mgx=\frac{4}{3}m(x^2+y^2)\frac{a}{x}\implies 2F-mg=\frac{4}{\beta}ma\implies ma=\frac{\beta}{4}(2F-mg).\nonumber \end{align}$$Replace $ma$ in the top equation to get $$N=mg-F+\frac{\beta}{4}(2F-mg)=mg\left(1-\frac{\beta}{4}\right)+F\left(\frac{\beta}{2}-1\right).$$This is a different expression from the previous one. For $\beta =2$ this expression gives $N=\frac{1}{2}mg.$ I would choose the second expression as correct because the reference point for the torques etc. is at rest on the inertial frame of the horizontal surface.

 

[haruspex]

let's find the expression for $N$ with the torque equation relative to the point contact P as I started doing earlier. $$\begin{align} & F+N=mg+ma \nonumber \\ & F(2x)-mgx=\frac{4}{3}m(x^2+y^2)\frac{a}{x}\implies 2F-mg=\frac{4}{\beta}ma\implies ma=\frac{\beta}{4}(2F-mg).\nonumber \end{align}$$

One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.
The parallel axis theorem does not apply because the instantaneous centre of rotational acceleration is not at P.

Since the mass centre accelerates straight up, the instantaneous centre of rotational acceleration must lie on a horizontal line through it. Likewise, assuming P is accelerating purely horizontally (not lifting yet), it lies on a vertical line through P. That puts it half way up the side of the block, so distance x from the mass centre. Hence moment of inertia from the parallel axis theorem is $m\frac 13(4x^2+y^2)$.

Alternatively, we can treat the motion of the block as a mass centre velocity $v$ plus an angular velocity $\omega$. Correspondingly, a linear momentum $mv$ and an angular momentum $I_c\omega$. From the perspective of a fixed point at the initial location of P, the linear momentum of the mass centre adds an angular momentum $mvx$ for a total angular momentum $mvx+I_c\omega$.
Differentiating, the rate of gain of angular momentum is $max+I_c\alpha$.
The torque equation is therefore $2Fx-mgx=max+I_c\alpha=max+\frac 13m(x^2+y^2)\frac ax=m\frac ax \frac 13(4x^2+y^2)$, as before.

This leads to the same equation as taking moments about the mass centre. There is no fault in the Universe.

 

[Hak]

Sorry, but there are a few points that are not at all clear to me.

Below I explain why I think the model is applicable.

1. Is @kuruman's model applicable in the light of the errors subsequently highlighted? That is to say: is it not applicable at all, or are certain adjustments and modifications possible that make the model valid and applicable?

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

2. This assertion is not convincing and right for @haruspex. Indeed:

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.

As already mentioned, the inequality $N>0$ is not correct here. As seen in later posts, substituting (1) into (2) gives $N<0$. Since you correctly observe that the condition $N>0$ is in agreement with one's intuition, the correct condition $N<0$ is in disagreement with one's intuition. How to solve this? How to make it clear and correct?

No, in the actual problem the normal force will be >0.
Consider F=0. In your model, that only leaves $J_1$, a downward impulse, so P will have an initial downward movement, yet in the actual problem the normal force remains at $mg/2$.

OK. 3. What does this imply? How do you change @kuruman's arguments so that the model is correct and applicable?

Yes, I have been queasy about this point. This thread has become the bane of my existence with all the mistakes I have made. There is another issue that bothers me. We have derived the expression $$N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right]$$ from the force equation and the torque equation relative to the CM. For $\beta =2$ this expression gives $N=\frac{1}{3}(mg+F).$

Now let's find the expression for $N$ with the torque equation relative to the point contact P as I started doing earlier. $$\begin{align} & F+N=mg+ma \nonumber \\ & F(2x)-mgx=\frac{4}{3}m(x^2+y^2)\frac{a}{x}\implies 2F-mg=\frac{4}{\beta}ma\implies ma=\frac{\beta}{4}(2F-mg).\nonumber \end{align}$$Replace $ma$ in the top equation to get $$N=mg-F+\frac{\beta}{4}(2F-mg)=mg\left(1-\frac{\beta}{4}\right)+F\left(\frac{\beta}{2}-1\right).$$This is a different expression from the previous one. For $\beta =2$ this expression gives $N=\frac{1}{2}mg.$ I would choose the second expression as correct because the reference point for the torques etc. is at rest on the inertial frame of the horizontal surface.

4. I did not really understand the context of these equations. Or rather: you had already derived them correctly many posts ago. I don't understand how they can apply to the last model you set out (that of the rectangular block on a smooth surface and subject to the two pulses $J_1$ and $J_2$). The problem of the normal force remains to be solved: its sign from your analysis appears to be negative, whereas we expect it to be positive. Which equations to bring in to support it?
Also, there seems to me to be an error related to an oversight in some previous posts. In fact, you are led to think that the second equation is correct because, by substituting $\beta =2$ (the value you think is the threshold value), you obtain the known limit value $N = \frac{mg}{2}$. In truth, the value of $\beta$ is incorrect, because it returns $x^2 = 2y^2$, not $y^2 =2x^2$. You had already mistaken the threshold equation, whose correct expression is $y^2 =2x^2$, not $x^2 = 2y^2$. The correct value of $\beta$ is $\beta=1$, not $\beta=2$. Substituting $\beta=1$ into the first equation, you get $N=\frac{mg}{2}$, as expected, while the second one does not.

One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.
The parallel axis theorem does not apply because the rotation is not about the corner P of the block.
We can treat the motion of the block as a mass centre velocity $v$ plus an angular velocity $\omega$. Correspondingly, a linear momentum $mv$ and an angular momentum $I_c\omega$. From the perspective of a fixed point at the initial location of P, the linear momentum of the mass centre adds an angular momentum $mvx$ for a total angular momentum $mvx+I_c\omega$.
Differentiating, the rate of gain of angular momentum is $max+I_c\alpha$.
The torque equation is therefore $2Fx-mgx=max+I_c\alpha=max+\frac 13m(x^2+y^2)\frac ax=m\frac ax \frac 13(4x^2+y^2)$.
This leads to the same equation as taking moments about the mass centre. There is no fault in the Universe.

5. OK, this expression is certainly correct, but it is just another way of calculating the same equation that was originally arrived at, no? How do we solve the problem of the applicability of the model and the normal reaction highlighted earlier?

Thank you for all.

 

[haruspex]

How do we solve the problem of the applicability of the model and the normal reaction highlighted earlier?

My position is that a) I do not understand the reasoning for the claim that the model is applicable, and b) it leads to answers that conflict with approaches that look sound to me.

 

[Hak]

it leads to answers that conflict with approaches that look sound to me.

I agree. For me, the striking fact is this: if there was a perfect match between arguments and answers, everything would have seemed to be in order, instead this could highlight the dangers and obstacles one might encounter in proposing problems such as this one formulated by the AAPT, which unintentionally created an exercise that was anything but simple and smooth. Tell me what you think. Thank you.

 

[haruspex]

I agree. For me, the striking fact is this: if there was a perfect match between arguments and answers, everything would have seemed to be in order, instead this could highlight the dangers and obstacles one might encounter in proposing problems such as this one formulated by the AAPT, which unintentionally created an exercise that was anything but simple and smooth. Tell me what you think. Thank you.

As shown in post #11, allowing for some thickness is not that complicated, but it did produce a surprising result. (In fact, it should not have been that surprising. It is well known that there is a 'sweet spot' along a stick where a blow at right angles will not jar the hand holding the stick at the far end, whereas a blow in the middle would jar the hand one way, and a blow at the tip will jar it the other way. The threshold ratio found in this thread is similar.)
The complications subsequently have come from trying different approaches in the hope of gaining greater understanding.

 

[Hak]

As shown in post #11, allowing for some thickness is not that complicated, but it did produce a surprising result. (In fact, it should not have been that surprising. It is well known that there is a 'sweet spot' along a stick where a blow at right angles will not jar the hand holding the stick at the far end, whereas a blow in the middle would jar the hand one way, and a blow at the tip will jar it the other way. The threshold ratio found in this thread is similar.)
The complications subsequently have come from trying different approaches in the hope of gaining greater understanding.

Thanks for your point of view. We await @kuruman for clarification regarding his model.

 

[kuruman]

One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.
The parallel axis theorem does not apply because the instantaneous centre of rotational acceleration is not at P.

I agree with all of the above and concede a degree of carelessness probably due to lateness in my time zone. Here is the improvement.

I define point of reference Q on the surface directly below corner P. Point Q is on an inertial frame.
The problem is asking us to find the normal force "shortly after" the right end leaves the surface. How short is "short"? Since no angle of inclination is given (after all by AIP's admission this has been a carelessly crafted problem), we assume that we are looking for the normal force at the moment the right end leaves the surface, otherwise we cannot answer the question. This means that points Q and P are still on top of each other.
Since Q and P are essentially the same point, the parallel axis theorem applies because at the particular instant that we consider, the stick is about to rotate about P/Q. We have seen that in this case $I_P=\dfrac{4m}{\beta}x^2.$

Now let us consider the following situation. We start with the stick lying on the surface with $F=0.$ The normal force is positioned at $x$ from either end, otherwise there would be a net torque. If we increase $F$ slowly, the point of application of $N$ will move towards the left until it reaches points P/Q. Further increase of $F$ will lift the right corner off the surface. The appropriate FBD for this situation is shown below where the surface has been replaced with a wedge at the corner.

One can easily show that for this system to be in equilibrium, $N=F=\frac{1}{2}mg.$ Let $F_0$ be the value of $F$ such that the system is in equilibrium. It follows that $mg=2F_0$ and that the end of the stick will lift off the surface only if $F>F_0.$

We now write the force and torque equations using $F_0$ instead of $mg$ to prevent us from chasing our tails.
$$\begin{align} & N-2F_0+F=ma \implies N=ma+2F_0-F \nonumber \\
& F(2x)-2F_0x=\frac{4m}{\beta}x^2\frac{a}{x} \implies ma=\frac{\beta}{2}(F-F_0) \nonumber
\end{align}$$ Thus, $$N=\frac{\beta}{2}(F-F_0)+2F_0-F=\frac{\beta}{2}(F-F_0)+2F_0-2F+F=(F-F_0)\left(\frac{\beta}{2}-2\right)+F.$$ Now suppose that we start at equilibrium and increase the external force to $F_0+\delta F$. Then the normal force will be $$N=\left(\frac{\beta}{2}-2\right)\delta F+F_0+\delta F=F_0+\left(\frac{\beta}{2}-1\right)\delta F.$$We see immediately that for $\beta >2$, the normal force is positive for any $\delta F.$

Now let's look at the smallest possible value for $N$ which occurs when $\beta = 0$, i.e. we have a vertical thin stick. Then $N=F_0-\delta F.~$ This becomes zero when $\delta F=F_0$ which means that $F=2F_0=mg.$ We knew that all along. As we pull on the vertical stick harder and harder, the normal force will diminish until it goes to zero and the stick is lifted off the surface. Points P, Q and the CM will all be lined up while this happens.

What about a value of $\beta$ in the range $0<\beta<2~$? Let's be specific and pick $\beta =1$. Then $N=F_0-\frac{1}{2}\delta F.$ $N$ is zero when $\delta F=2F_0 \implies F=3F_0.$ In this case the contact is just lost and it doesn't matter whether the surface is smooth or rough when one considers the initial horizontal acceleration of point P.

What happens if $\beta = 1$ and $F>3F_0~$? I am only guessing that corner P will be lifted off the surface but, because the CM is not directly above point P, there will be some kind of pendulum motion depending on how force $F$ is anchored on the stick.

 

[haruspex]

Since Q and P are essentially the same point, the parallel axis theorem applies because at the particular instant that we consider, the stick is about to rotate about P/Q.

No, you are still missing it.

Call the mass centre G. Point P of the block is accelerating horizontally, while G is accelerating vertically. (PG is constant. If P is not accelerating then G is accelerating diagonally up, implying friction at P.) This is not a state of affairs that arises after some interval, it is how things are as soon as acceleration starts.

It follows that the instantaneous centre of rotational acceleration, C, is on a horizontal line through G and a vertical line through P=Q. The parallel axis theorem clearly applies at C.
(Note that all forces have the same torque about C as about P, and there is certainly no reason for P to be a more appropriate axis to consider than C.)

Taking C as the axis and using the parallel axis theorem produces the same answers as taking G as the axis. Taking P as the axis also works if you work from first principles, as in post #74, instead of assuming the parallel axis theorem applies there.

 

[Hak]

No, you are still missing it.

Call the mass centre G. Point P of the block is accelerating horizontally, while G is accelerating vertically. (PG is constant. If P is not accelerating then G is accelerating diagonally up, implying friction at P.) This is not a state of affairs that arises after some interval, it is how things are as soon as acceleration starts.

It follows that the instantaneous centre of rotational acceleration, C, is on a horizontal line through G and a vertical line through P=Q. The parallel axis theorem clearly applies at C.
(Note that all forces have the same torque about C as about P, and there is certainly no reason for P to be a more appropriate axis to consider than C.)

Taking C as the axis and using the parallel axis theorem produces the same answers as taking G as the axis. Taking P as the axis also works if you work from first principles, as in post #74, instead of assuming the parallel axis theorem applies there.

Sigh. This means that the model in post #80 cannot be applied, but must be modified to be valid, right?

 

[haruspex]

Sigh. This means that the model in post #80 cannot be applied, but must be modified to be valid, right?

Why the desire to fix the model? It is not needed.

 

[Hak]

Why the desire to fix the model? It is not needed.

Yes, you're right, it is not needed. However, I have no desire to fix the model, but I would like there to be a reasonable one applicable to the physical situation. Three were proposed by @kuruman, all three did not work because of some oversights and wrong physical assumptions, but that does not mean there is not one. An applicable model, with all the correct considerations you pointed out, must exist in my opinion. In my opinion, @kuruman's efforts should be rewarded by this. It may well be that I am wrong.

 

[haruspex]

a reasonable one applicable to the physical situation

post #11.

 

[Hak]

post #11.

Yes, obviously, I've already seen this. I was referring to @kuruman's efforts, but still nothing ... If @kuruman still feels like coming up with a convincing model that can fit the situation, that's fine, otherwise, if he wants to give up, we are content with this excellent discussion. Thank you very much.

 

[haruspex]

Yes, obviously, I've already seen this. I was referring to @kuruman's efforts, but still nothing ... If @kuruman still feels like coming up with a convincing model that can fit the situation, that's fine, otherwise, if he wants to give up, we are content with this excellent discussion. Thank you very much.

I haven't checked, but #80 can probably be salvaged by using the correct moment of inertia, $4x^2+y^2$ instead of $4x^2+4y^2$.

 

[kuruman]

OK, let's redo post #80 using the torque equation relative to the CM. The force equation is the same and yields
$$N=ma+2F_0-F.$$ From post #11
$$F-N=\frac 13m( 1+\frac{y^2}{x^2})a=\frac{ma}{\beta} \implies ma=\beta(F-N). $$ Then $$N=\beta(F-N)+2F_0-F \implies N=\frac{1}{1+\beta}[(\beta-1)F+2F_0].$$ In the range $1<\beta<3$, the RHS is clearly positive.
In the range $0<\beta<1$ we replace $F$ with $F_0+\delta F$ as in post #80 to get $$\begin{align} & N = \frac{1}{1+\beta}[(\beta-1)(F_0+\delta F)+2F_0]\nonumber \\
& =\frac{1}{1+\beta}[(1+\beta)F_0-(1-\beta)\delta F)]\nonumber \\
&=F_0-\frac{1-\beta}{1+\beta}\delta F. \nonumber
\end{align}$$This can be used to find for what increase $\delta F$ the normal force goes to zero.

 

[Hak]

Now let's look at the smallest possible value for $N$ which occurs when $\beta = 0$, i.e. we have a vertical thin stick. Then $N=F_0-\delta F.~$ This becomes zero when $\delta F=F_0$ which means that $F=2F_0=mg.$ We knew that all along. As we pull on the vertical stick harder and harder, the normal force will diminish until it goes to zero and the stick is lifted off the surface. Points P, Q and the CM will all be lined up while this happens.

The above conditions, which you made in post #80, remain unchanged in post #88 below. In that respect, everything runs smoothly.

OK, let's redo post #80 using the torque equation relative to the CM. The force equation is the same and yields
$$N=ma+2F_0-F.$$ From post #11
$$F-N=\frac 13m( 1+\frac{y^2}{x^2})a=\frac{ma}{\beta} \implies ma=\beta(F-N). $$ Then $$N=\beta(F-N)+2F_0-F \implies N=\frac{1}{1+\beta}[(\beta-1)F+2F_0].$$ In the range $1<\beta<3$, the RHS is clearly positive.
In the range $0<\beta<1$ we replace $F$ with $F_0+\delta F$ as in post #80 to get $$\begin{align} & N = \frac{1}{1+\beta}[(\beta-1)(F_0+\delta F)+2F_0]\nonumber \\
& =\frac{1}{1+\beta}[(1+\beta)F_0-(1-\beta)\delta F)]\nonumber \\
&=F_0-\frac{1-\beta}{1+\beta}\delta F. \nonumber
\end{align}$$This can be used to find for what increase $\delta F$ the normal force goes to zero.

But what happens if $\beta = 1$? I get: $N = F_0$. This means that $N$ goes to $0$ when $F_0$ goes to $0$, not when $F=3F_0$, right?

What about a value of $\beta$ in the range $0<\beta<2~$? Let's be specific and pick $\beta =1$. Then $N=F_0-\frac{1}{2}\delta F.$ $N$ is zero when $\delta F=2F_0 \implies F=3F_0.$ In this case the contact is just lost and it doesn't matter whether the surface is smooth or rough when one considers the initial horizontal acceleration of point P.

Is the condition in bold preserved or not, given the new terms?

What happens if $\beta = 1$ and $F>3F_0~$? I am only guessing that corner P will be lifted off the surface but, because the CM is not directly above point P, there will be some kind of pendulum motion depending on how force $F$ is anchored on the stick.

Does this still happen in the new model? And if it happens, when does it happen? In the case $F_0 >0$?

Thanks.

 

[Hak]

This can be used to find for what increase $\delta F$ the normal force goes to zero.

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?

 

[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.