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

[guv]

Homework Statement

This US Physics Olympiad problem 2018A Question
A uniform stick of mass $m$ is originally on a horizontal surface. One end is attached to a vertical rope,
which pulls up with a constant tension force $F$ so that the center of the mass of the stick moves upward
with acceleration $a < g$. The normal force $N$ of the ground on the other end of the stick shortly after
the right end of the stick leaves the surface satisfies $mg > N > mg/2$

Relevant Equations

$F_{net} = ma$
$\tau_{net} = I \alpha$

The question is not directly related to the problem itself, but an odd discovery that when using left end of the stick as axis of rotation versus using the CM as axis of rotation, two different normal force expressions are found. Note that the solutions are for the initial moment when the stick is still horizontal. I am looking for suggestions to reconcile the two different answers, thanks!

The link to the problem set is here: https://www.aapt.org/physicsteam/2019/upload/Fma-2018-A.pdf It's question 9.

Using left end as axis of rotation:
$$ma=F+N-mg$$
$$\alpha * L/2= a$$
$$Fl-mgl/2=\alpha * \frac{1}{3} ml^2=\frac{2}{3} mal$$
$$F=\frac{2}{3}ma+mg/2=mg+ma-N$$
$$N=\frac{1}{3} ma+mg/2$$

Using CM as axis of rotation:
$$Fl/2-Nl/2=\frac{1}{12} ml^2 \alpha$$
$$a=\alpha l/2$$
$$F-N=ma/6$$
$$F+N-mg=ma$$
$$F=ma+mg-N = ma/6 + N$$
$$2N = \frac{5}{6}ma+mg$$
$$N=\frac{5}{12} ma + mg/2$$

The difference between the two normal force is $1/12 ma$.

 

[kuruman]

The difference between the two normal force is $1/12 ma$.

Check your algebra for the calculation about the CM because the algebra is in error.

 

[hutchphd]

Why must they be equal?? What misbehavior do you see here? The stick will rotate because of the imbalance, as you should expect.

 

[kuruman]

Why must they be equal?? What misbehavior do you see here? The stick will rotate because of the imbalance, as you should expect.

The normal force exerted by the surface on the stick must have a value that does not depend on how one writes the torque equation. OP has obtained two different values that depend on whether torques are calculated about the point of contact or about the CM.

 

[TSny]

Using CM as axis of rotation:
$$Fl/2-Nl/2=\frac{1}{12} ml^2 \alpha$$
$$a=\alpha l/2$$
$$F-N=ma/6$$

Check the last equation above.
EDIT: Oops, I see @kuruman already noted this.

 

[hutchphd]

I would love to see a free body diagram.

 

[kuruman]

I would love to see a free body diagram.

 

[haruspex]

View attachment 332117

The diagram at the link shows the stick still horizontal and possessed of some thickness.
The thickness bothers me. It does not say a thin stick, but neither does it call it a rectangular block. Allowing an arbitrary thickness, I can still show N<mg, but not N>mg/2, not even N>0.

 

[pbuk]

I can still show N<mg, but not N>mg/2, not even N>0.

Then there must be something wrong in your calculations: you don't even need to do any calculations to answer this question:

Imagine a horizontal unform stick which is not accelerating, but is supported at one end by a point and at the other by vertical string. Clearly the supporting force at both ends is $\frac {mg} 2$. If we now pull up on the string gently we create a torque at the point so the normal force at that end increases so we have $N > \frac {mg} 2$. The only answer that admits that is B so we don't need to bother with an upper limit.

 

[guv]

Check your algebra for the calculation about the CM because the algebra is in error.

Yeap, now I can see what's wrong. Thanks everyone!

 

[haruspex]

pull up on the string gently

The only indication of gentility is a<g.

Let the stick be length 2x, depth 2y.
$N+F=m(g+a)$
Taking moments about the mass centre,
$Fx-Nx=I\alpha=\frac 13m( x^2+y^2)a/x$
$F-N=\frac 13m( 1+\frac{y^2}{x^2})a$
We can choose to eliminate F or a.

$2N=mg+\frac 13ma(2-\frac{y^2}{x^2})$
So for N>mg/2 we need $y<x\sqrt 2$.

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.

There may well be an error in my algebra, but I can't spot it.

 

[kuruman]

There may well be an error in my algebra, but I can't spot it.

There is no algebra error, but I believe there is an error in the moment of inertia about the CM that you used.

Assuming a $(2x)\times(2y)$ thin plate, the surface density is $\sigma = \dfrac{m}{4xy}$. The moment of inertia about the $y$-axis passing through the CM is $$I_{yy}=\int dm~x'^2=\frac{m}{4xy}\int_{-y}^{y} dy'\int_{-x}^{x} x'^2~dx'=\frac{m}{4xy}(2y)(\frac{1}{3}2x^3)=\frac{1}{3}x^3.$$The "depth" doesn't matter.

 

[haruspex]

There is no algebra error, but I believe there is an error in the moment of inertia about the CM that you used.

Assuming a $(2x)\times(2y)$ thin plate, the surface density is $\sigma = \dfrac{m}{4xy}$. The moment of inertia about the $y$-axis passing through the CM is $$I_{yy}=\int dm~x'^2=\frac{m}{4xy}\int_{-y}^{y} dy'\int_{-x}^{x} x'^2~dx'=\frac{m}{4xy}(2y)(\frac{1}{3}2x^3)=\frac{1}{3}x^3.$$The "depth" doesn't matter.

Maybe I confused matters by using y and “depth" for the vertical height shown in the diagram. I assumed thickness into the page was irrelevant.

 

[kuruman]

Ah, I see, you called "depth" what I would call "height". Yes, thickness into the page screen is irrelevant.

Can you explain your torque equation $$Fx-Nx=I\alpha=\frac 13m( x^2+y^2)a/x$$ Quantity $x$ on the RHS is the fixed half-length of the "stick". On the LHS it is apparently the lever arm which varies from $x$ when the stick is horizontal to zero when it is vertical.

 

[haruspex]

which varies from x when the stick is horizontal to zero when it is vertical

The question specifies "shortly after the right end leaves the ground". I.e. it is still almost horizontal.

 

[kuruman]

The question specifies "shortly after the right end leaves the ground". I.e. it is still almost horizontal.

I think I got it now. I calculate moments about the pivot corner instead of the CM. When vertical force $F$ is applied, just before the edge of the rectangle lifts off the surface, the point of application of the normal force has shifted to the pivot corner. In that case
$F+N=mg~$ and $~F(2x)-mgx=0.$
From these two equations it follows that $~F=N=\frac{1}{2} mg~$ just before tipping occurs. The condition for tipping to occur is $~F>\frac{1}{2} mg.$ If tipping occurs,
$F+N=mg+ma$
$F(2x)-mgx=\dfrac{4}{3}m(x^2+y^2)\dfrac{a}{x}.$

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.

Upper limit for the normal force
To find the upper limit for $N$, we solve the torque equation for $F$
$F=\frac{1}{2} mg+\dfrac{2}{3}\dfrac{(x^2+y^2)}{x^2}ma.$
Then we substitute in the force equation, solve for $N$ and look for its upper limit.
$$N=\frac{1}{2}mg+ma\left[1-\frac{2(x^2+y^2)}{3x^2}\right].$$The upper limit of the term in the square brackets is maximum in the thin-stick limit $y\rightarrow 0$. Also, the problem states that $a<g$ so when we set $a=g$, the upper limit for the normal force is $$N=\frac{1}{2}mg+\frac{1}{3}mg=\frac{5}{6}mg.$$ We conclude that $\frac{1}{2}mg<N<\frac{5}{6}mg.$ Answer (B) comes closest to satisfying this condition.

However, I don't think that the intention of the problem's author was this detail of analysis and that a "very thin" stick was intended to be the case despite the drawing. I note that this contest allows 75 min. to answer 25 questions, meaning 3 minutes per problem on average with no time left for checking one's work. The thickish stick is certainly not a 3-minute problem.

 

[haruspex]

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.

Sure, but that's the thin stick limit. My concern is the case $y>\sqrt 2x$, which is not ruled out by the problem statement.
I find this result rather surprising, as must @pbuk.

 

[pbuk]

My concern is the case $y>\sqrt 2x$, which is not ruled out by the problem statement.

I think it is ruled out: would you describe something that is $\sqrt 2$ times as thick as it is long a 'stick'?

I find this result rather surprising, as must @pbuk.

No, I am not surprised that if you pull up hard enough on one edge of something the shape of a tea chest the opposite edge might leave the ground. But that is a completly different situation to pulling up gently on one end of a stick.

 

[haruspex]

I think it is ruled out: would you describe something that is 2 times as thick as it is long a 'stick'?

You are missing the point. The fact that the result is only true provided $y<\sqrt 2x$ means that such a detailed analysis is necessary. The reasoning in post #9 is not as reliable as it seems.
The epithet "thin" was added by the OP. It does not appear in the original.

It seems to me the author of the problem failed to understand the subtlety and should have described it as thin, showing it as such in the diagram.

I am not surprised that if you pull up hard enough on one edge of something the shape of a tea chest the opposite edge might leave the ground.

It's not merely that it leaves the ground, but that it may do so instantly, despite the applied force being bounded (not a jerk that instantaneously provides momentum). I find that surprising. The author accidentally created a rather interesting problem.

 

[pbuk]

It seems to me the author of the problem failed to understand the subtlety and should have described it as thin, showing it as such in the diagram.

This is the stick depicted in the original question: it is actually quite thick:

This is the shape you are talking about:

That does not seem to me to be a "subtlety", rather it seems to be a completely different problem.

 

[Lnewqban]

Could we simply state that $N=m(g-a)$?

 

[kuruman]

Could we simply state that $N=m(g-a)$?

View attachment 332336

No, because upward force $F$ is also acting at the other end of the stick. So Newton's second law is $F+N-mg=ma$.

 

[haruspex]

This is the stick depicted in the original question: it is actually quite thick:

View attachment 332328

This is the shape you are talking about:

View attachment 332329

That does not seem to me to be a "subtlety", rather it seems to be a completely different problem.

The subtlety is that the thickness matters. I embarked on the full analysis thinking to show it did not, and was very surprised to find this $\sqrt 2$ ratio threshold.
Now, there is no way to guess that value. Once you accept that the ratio could matter, it is not enough to say "oh, the diagram shows $y<x/3$ (or whatever), so we can take it as arbitrarily thin". To prove $N>mg/2$, it is necessary to show that the threshold is clearly more than in the diagram, and to assume the diagram is reasonably accurate.

 

[kuruman]

Sure, but that's the thin stick limit. My concern is the case $y>\sqrt 2x$, which is not ruled out by the problem statement.
I find this result rather surprising, as must @pbuk.

OK I'll do it your way using moments about the CM but stick to my approach for finding the lower limit of the normal force.

$F+N=mg+ma$ (force equation)
$F-N=\dfrac{x^2+y^2}{3x^2} ma$ (torque equation)

Use the torque equation to find an expression for $ma$ and substitute in the force equation
$ma=\dfrac{3x^2}{x^2+y^2}(F-N)=\beta(F-N)~~~\left(\beta\equiv \dfrac{3x^2}{x^2+y^2}\right).$

$F+N=mg+\beta(F-N)$
$N(\beta+1)=mg+F(\beta-1)$

In the specific case $x^2=2y^2$, $\beta =2$ and $N=\frac{1}{3}mg +\frac{1}{3}F.$

Regardless of the geometry (value of $\beta$), the fact remains that, at equilibrium $F=N=\frac{1}{2}mg$. Therefore, for the bottom to lift off the surface, $F$ needs to be greater than the limiting value of half the weight. The lower limit for the normal force for a given value of $\beta$ is obtained by replacing $F$ with $\frac{1}{2}mg$. Then
$N_{\text{lower lim.}}=\dfrac{1}{\beta+1}mg+\dfrac{\beta-1}{2(\beta+1)}mg=\frac{1}{2}mg.$
This says that the lower limit of the normal force does not depend on the geometry and is equal to the value of the normal force when the stick is supported at its two ends in horizontal equilibrium, that is half the weight.

 

[haruspex]

The lower limit for the normal force for a given value of $\beta$ is obtained by replacing $F$ with $\frac{1}{2}mg$.

Even when $\beta<1$, making $F(\beta-1)$ negative?

In the specific case $x^2=2y^2$, $\beta =2$ and $N=\frac{1}{3}mg +\frac{1}{3}F.$

The threshold I calculated was at $2x^2=y^2$, giving $\beta=1$.

 

[Lnewqban]

No, because upward force $F$ is also acting at the other end of the stick. So Newton's second law is $F+N-mg=ma$.

Thank you, professor.
I should have written $N=0.5[m(g-a)]$
Still incorrect?

 

[haruspex]

I should have written $N=0.5[m(g-a)]$

How do you get that?

 

[Lnewqban]

How do you get that?

Please, see diagram in post #21.

 

[haruspex]

Please, see diagram in post #21.

That diagram is not a FBD. If you want to mix forces and accelerations in one diagram you need to discriminate them.
$a$ is an acceleration, but what is $g$ here? Nothing is accelerating at g.
In the usual formulation, the FBD would show forces F and N up, mg down. The resulting acceleration is $a$ up, giving $F+N-mg=ma, F+N=m(a+g)$.
Another view is to use the frame of reference of a free falling object. That makes the force of gravity go away, but adds $g$ upwards to all observed accelerations, again leading to $F+N=m(a+g)$.

But this does not lead to $N=\frac 12m(a+g)$. The stick is not only rising at an increasing rate but also rotating at such, so there must be a net torque (anticlockwise in the diagram). Hence $F>N$.

 

[kuruman]

The threshold I calculated was at $2x^2=y^2$, giving $\beta=1$.

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 $\frac{1}{2}mg.$

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.