I Olympiad Problem -- Revisiting the problem with 3 blocks and a pulley

[sysprog]

I was researching a problem that had once been posted here by someone else, and that had subsequently been posted (couched in somewhat different terms) here by me, and in doing the research, I ran across the following, which I think is the well-stated 'original version' of the problem.

The problem, in both versions posted here on PF, the first of which I responded to, and the second of which I was the OP for, and regarding which in the ensuing discussion I participated, elicited controversy between me and other participants in the threads in which it was discussed.

Specifically, in the first of the two threads, I held that, and in the second of them, I strongly questioned whether, there would be leftward movement of the hanging $m$ block that was equal to and simultaneous with the lateral movement of the big block.

Given the problem as presented in the image above, if someone thinks that my supposition that the hanging $m$ block would not at any time after release descend vertically wrt its original position, but would remain at all times vertical wrt to the pulley, is incorrect, then that someone will please explain why.

 

[jedishrfu]

I think that's a given since to allow more degrees of freedom as in a real-world example would make it a much harder problem to solve.

 

[sysprog]

I think that's a given since to allow more degrees of freedom as in a real-world example would make it a much harder problem to solve.

What are you saying is a given? Do you think that my supposition that the hanging $m$ block would not at any time after release descend vertically wrt its original position, but would remain at all times vertical wrt to the pulley, is incorrect? If you do think that, please explain why.

 

[jedishrfu]

I think the pulley wire is centered on the mass so that it descends vertically such that it is on the vertical line through the pulley center as is the mass originally.

Boy, this is hard to describe in words. from your picture I see the mass sliding down a wall and I infer the wire from pulley to block is vertical at all times.

 

[Dale]

Well, the Lagrangian is $L= (1/2 \ m \dot h^2 + 1/2 \ m \dot x^2)+1/2 \ M \dot x^2 + 1/2 \ m (\dot h + \dot x)^2 - m g h$

Should be straightforward to solve, but not tonight

 

[sysprog]

@Dale, I think that the reply of @jedishrfu agrees with my supposition that the descent of the hanging $m$ block is not vertical wrt the ground below it, but is vertical wrt the pulley above it. Do you agree with that supposition? If not, please say why not.

 

[Dale]

I would have to solve the equation before I could say. I will do so tomorrow

 

[jedishrfu]

@Dale The block is hanging precariously awaiting your answer... rather like the sword of Damocles.

 

[Baluncore]

The hanging block m will not begin to follow M until there is a horizontal component in the tension of the supporting string below the pulley. That requires a gap must form between M and the hanging block. For that reason the string cannot remain vertical.

 

[sysprog]

The hanging block m will not begin to follow M until there is a horizontal component in the tension of the supporting string below the pulley. That requires a gap must form between M and the hanging block. For that reason the string cannot remain vertical.

Doesn't the descent of the hanging $m$ block drive the leftward movement of the $M$ block and the pulley? Isn't it true that the tension on the string is what transmits all of the leftward force? Isn't the leftward force of that tension applied as much to the descending $m$ block as to the pulley from which it hangs? If your answer to any of those questions is no, please say why not.

 

[ergospherical]

The initial behaviour is clear from analysing the forces. At $t=0$, the string exerts a horizontal component of force on the $M$ block (via. the contact force at the pulley), but the string exerts only a vertical component of force on the $m$ block.

The horizontal displacement of the $M$ block is therefore $O(t^2)$ whilst the horizontal displacement of the hanging $m$ block is $O(t^3)$. They will indeed separate by a small horizontal gap at small time values.

 

[sysprog]

The initial behaviour is clear from analysing the forces. At $t=0$, the string exerts a horizontal component of force on the $M$ block (via. the contact force at the pulley), but the string exerts only a vertical component of force on the $m$ block.

The horizontal displacement of the $M$ block is therefore $O(t^2)$ whilst the horizontal displacement of the hanging $m$ block is $O(t^3)$. They will indeed separate by a small horizontal gap at small time values.

Why would it be that "the string exerts only a vertical component of force on the $m$ block", by which presumably you mean vertical wrt the ground, and not vertical wrt the pulley, when the pulley is impelled leftward along with the $M$ block to which it is attached? How can the hanging $m$ block move exactly as downward as it would if it were not hanging $-$ not connected to the string? What rightward force would for it 'zero out' the tension that is driving the leftward movement of the pulley and $M$ block?

 

[ergospherical]

Why would it be that "the string exerts only a vertical component of force on the $m$ block",

At $t=0$, hanging string is vertical.

 

[sysprog]

At $t=0$, hanging string is vertical.

And at $t>0$, is it not still vertical wrt the pulley, but no longer vertical wrt the (starting downward point on the) ground? If not, then how does the pulley acquire any impetus to move lefward?

 

[ergospherical]

how does the pulley acquire any impetus to move lefward?

String pushes the pulley, which is connected to $M$. Horizontal component of this force accounts for horizontal acceleration of $M$.

 

[sysprog]

String pushes the pulley, which is connected to $M$. Horizontal component of this force accounts for horizontal acceleration of $M$.

Wouldn't the same "horizontal component of this force" apply to the (leftward) horizontal acceleration of the hanging $m$ block? If not, why not? Are you denying that at t>0, the hanging $m$ block is still vertical wrt the pulley, but no longer vertical wrt the (starting downward point on the) ground? If you are denying that, please say why.

 

[A.T.]

I was researching a problem that had once been posted here by someone else, and that had subsequently been posted (couched in somewhat different terms) here by me, and in doing the research, I ran across the following, which I think is the well-stated 'original version' of the problem.

View attachment 296396

Given the problem as presented in the image above, if someone thinks that my supposition that the hanging $m$ block would not at any time after release descend vertically wrt its original position, but would remain at all times vertical wrt to the pulley, is incorrect, then that someone will please explain why.

The problem as presented asks about the "acceleration of the big block immediately after the system is released". At the time point of release the string is still vertical, so that can be assumed as given. Thats why the solution starts with "Initially..."

None of that says anything about velocity of the block later, at some time after release.

 

[PeroK]

Well, the Lagrangian is $L= (1/2 \ m \dot h^2 + 1/2 \ m \dot x^2)+1/2 \ M \dot x^2 + 1/2 \ m (\dot h + \dot x)^2 - m g h$

Should be straightforward to solve, but not tonight

See:

https://www.physicsforums.com/threa...imple-free-body-diagram.1011264/#post-6587523

 

[PeroK]

The initial behaviour is clear from analysing the forces. At $t=0$, the string exerts a horizontal component of force on the $M$ block (via. the contact force at the pulley), but the string exerts only a vertical component of force on the $m$ block.

The horizontal displacement of the $M$ block is therefore $O(t^2)$ whilst the horizontal displacement of the hanging $m$ block is $O(t^3)$. They will indeed separate by a small horizontal gap at small time values.

I did manage to generate some equations for an equilibrium angle:

https://www.physicsforums.com/threa...free-body-diagram.1011264/page-4#post-6588325

 

[A.T.]

Wouldn't the same "horizontal component of this force" apply to the (leftward) horizontal acceleration of the hanging $m$ block? If not, why not? Are you denying that at t>0, the hanging $m$ block is still vertical wrt the pulley, but no longer vertical wrt the (starting downward point on the) ground? If you are denying that, please say why.

A vertical rope cannot apply any horizontal force to the hanging block. So any horizontal acceleration of the hanging block implies the rope is not vertical anymore.

 

[sysprog]

A vertical rope cannot apply any horizontal force to the hanging block. So any horizontal acceleration of the hanging block implies the rope is not vertical anymore.

Given that the leftward acceleration of the pulley is driven by the downward acceleration of the hanging $m$ block, and not by an external or otherwise independent force, wouldn't the hanging $m$ block remain vertical wrt the pulley? Are you denying that at t>0, the hanging $m$ block is still vertical wrt the pulley, but no longer vertical wrt the (starting downward point on the) ground? If you are denying that, please say why.

 

[PeroK]

Given that the leftward acceleration of the pulley is driven by the downward acceleration of the hanging $m$ block, and not by an external or otherwise independent force, wouldn't the hanging $m$ block remain vertical wrt the pulley?

No. Because to move left, the hanging mass needs a leftwards force. That might be because it is attached to the large mass $M$. That's the simplest problem to solve.

If it's not attached to the large mass, it can only be pulled left by the string. And, by assumption, the string is incapable of exerting a lateral force. Therefore, the string itself must be at an angle in order to pull the hanging block leftwards.

This is elementary: there can be no leftwards force on the hanging mass until the string is at an angle. The mass will swing out relative to the block as the large block moves leftwards. Eventually reaching or oscillating about some equilibrium angle.

It cannot be pulled leftwards by a vertical string.

 

[A.T.]

Are you denying that at t>0, the hanging $m$ block is still vertical wrt the pulley, but no longer vertical wrt the (starting downward point on the) ground? If you are denying that, please say why.

"Block is still vertical wrt the pulley" is a very confused statement. If you mean that the rope segment, that the block hangs on, remains vertical, then no. And I already explained why:

A vertical rope cannot apply any horizontal force to the hanging block. So any horizontal acceleration of the hanging block implies the rope is not vertical anymore.

If the hanging block is attached to M, such that it can slide down on a rail, the rail will provide the horizontal force on the block, and the rope will remain vertical.

 

[sysprog]

"Block is still vertical wrt the pulley" is a very confused statement. If you mean that the rope segment, that the block hangs on, remains vertical, then no. And I already explained why:

If the hanging block is attached to M, such that it can slide down on a rail, the rail will provide the horizontal force on the block, and the rope will remain vertical.

By what force would the rail be confronted that would require it to exert force? I don't see any rightward inertial force to oppose it $-$ I think that the hanging block would be dragged along by the tension in the string, immediately contemporaneously with the block moving downward and the pulley moving leftward.

 

[mohamed_a]

No. Because to move left, the hanging mass needs a leftwards force. That might be because it is attached to the large mass $M$. That's the simplest problem to solve.

If it's not attached to the large mass, it can only be pulled left by the string. And, by assumption, the string is incapable of exerting a lateral force. Therefore, the string itself must be at an angle in order to pull the hanging block leftwards.

This is elementary: there can be no leftwards force on the hanging mass until the string is at an angle. The mass will swing out relative to the block as the large block moves leftwards. Eventually reaching or oscillating about some equilibrium angle.

It cannot be pulled leftwards by a vertical string.

I know it isn't the best solution but could someone solve it using vector analysis of the forces in the question?

 

[A.T.]

I think that the hanging block would be dragged along by the tension in the string ...

Cannot happen if the string is vertical, because the string force has no horizontal component then.

... immediately contemporaneously with the block moving downward and the pulley moving leftward.

Not in a way that keeps the string vertical, otherwise you contradict your above assumption about non zero horizontal string force.

 

[Dale]

The hanging block m will not begin to follow M until there is a horizontal component in the tension of the supporting string below the pulley. That requires a gap must form between M and the hanging block. For that reason the string cannot remain vertical.

So I realized that my Lagrangian above assumed that the string was always vertical. Since that is being questioned then I should not make that assumption. So I have one extra degree of freedom and I have to rewrite my Lagrangian. Including that degree of freedom definitely complicates things.

So I have $x$ which is the horizontal position of the pulley, $r$ which is the amount of string between the pulley and the hanging mass, and $\theta$ which is the angle of the string with the vertical.

The position of the big mass is simply $x$ so its velocity is $\dot x$ and the corresponding KE is $\frac{1}{2} M \dot x^2$

The position of the little mass on top is $x+r$ so its velocity is $\dot x + \dot r$ and the corresponding KE is $\frac{1}{2} m (\dot r^2 + 2 \dot r \dot x + \dot x^2)$

The position of the hanging mass is the most complicated. It is $(r \sin(\theta) + x, - r \cos(\theta))$ so its velocity is $(\dot r \sin (\theta )+r \dot \theta \cos (\theta )+\dot x,-\dot r \cos (\theta )+r \dot \theta \sin (\theta) )$ and the corresponding KE is $\frac{1}{2}m( 2 \dot r \sin (\theta ) \dot x+\dot r^2+r^2 \dot \theta^2+2 r \dot \theta \cos (\theta) \dot x+\dot x^2 )$.

The potential energy is $-m g r \cos(\theta)$.

To simplify this a bit, I will assume that $\theta$ is small, so $\cos(\theta)=1$ and $\sin(\theta)=\theta$. So the new Lagrangian with the extra degree of freedom is $$L=m r \left(g+\dot \theta \dot x\right)+\frac{1}{2} (2 m+M) \dot x^2+m (\theta+1) \dot r
\dot x+m \dot r^2+\frac{1}{2} m r^2 \dot \theta^2 $$ I would not mind at all if someone could check this. Even with the small angle approximation it is a pretty hairy Lagrangian.

There is a conserved energy and a conserved momentum conjugate to $x$, which I won't write down. The Euler Lagrange equations are: $$ -(2 m+M) \ddot x-m (\theta +1) \ddot r-2 m \dot r \dot \theta -m r \ddot \theta =0$$ $$ m \left(g-2 \ddot r+r \dot \theta ^2-\theta \ddot x-\ddot x\right)=0 $$ $$ -m r \left(2 \dot r \dot \theta +r(t) \ddot \theta +\ddot x\right)=0 $$

This doesn't appear to have an analytical solution, so I will do a numerical solution with $g=10$, $m=1$, and $M=5$ with $r(0)=1$ and all other initial conditions set to 0. This gives the following plot for $\theta$

So we see here that $\theta$ starts at 0 (since that was my initial condition) and then is always positive. That means that the string is not vertical except at the beginning, and then only because that was my initial condition.

 

[PeroK]

This doesn't appear to have an analytical solution, so I will do a numerical solution with $g=10$, $m=1$, and $M=5$ with $r(0)=1$ and all other initial conditions set to 0. This gives the following plot for $\theta$

So we see here that $\theta$ starts at 0 (since that was my initial condition) and then is always positive. That means that the string is not vertical except at the beginning, and then only because that was my initial condition.

For those values of $m$ and $M$, I got an equilibrium angle of about $0.09$ radians and acceleration of $M$ at equilibrium of $\ddot x = -0.09g$. And $\ddot r = 0.55g$.

Are you able to confirm that?

 

[Dale]

No, I got more than 0.5 radians at long times and -0.09 g was closer to my initial acceleration than my long-term acceleration. It didn't really seem to have an equilibrium.

I suspect that the small angle approximation is not valid. Did you use that?

 

[PeroK]

No, I got more than 0.5 radians at long times

That seems like a lot. Nearly $30$ degrees?

 

[Dale]

That seems like a lot. Nearly $30$ degrees?

Yes, I agree. That is why I don't trust the small angle approximation. If I remove that approximation (I am not writing down that Lagrangian) then for $g=1$, $m=1$, $M=5$, and $r(0)=1$ I get $\theta(100)=0.084$ and $\ddot x(100)=-0.084$, and $\ddot r(100)=0.544$, which seem more reasonable and reasonably close to your values. I suspect you were not using the small angle approximation.

 

[PeroK]

Yes, I agree. That is why I don't trust the small angle approximation. If I remove that approximation (I am not writing down that Lagrangian) then for $g=1$, $m=1$, $M=5$, and $r(0)=1$ I get $\theta(100)=0.084$ and $\ddot x(100)=-0.084$, which seem more reasonable.

I'll check the Lagrangian tomorrow.

 

[Dale]

@Dale, I think that the reply of @jedishrfu agrees with my supposition that the descent of the hanging $m$ block is not vertical wrt the ground below it, but is vertical wrt the pulley above it. Do you agree with that supposition? If not, please say why not.

So I do not agree with that supposition after $t=0$. As far as why not, that is what the math said. It is entirely possible that I made a mistake in the math, but whether I used the small angle approximation or not I still got that $\theta > 0$ for all $t>0$

 

[PeroK]

The Euler Lagrange equations are: $$ -(2 m+M) \ddot x-m (\theta +1) \ddot r-2 m \dot r \dot \theta -m r \ddot \theta =0$$ $$ m \left(g-2 \ddot r+r \dot \theta ^2-\theta \ddot x-\ddot x\right)=0 $$ $$ -m r \left(2 \dot r \dot \theta +r(t) \ddot \theta +\ddot x\right)=0 $$

I agree with those equations. I'm not sure how best to analyse them.

 

[A.T.]

Here a simulation with M/m = 5



Done in Algoodo:
http://www.algodoo.com/

 

[sysprog]

Here a simulation with M/m = 5



Done in Algoodo:
http://www.algodoo.com/

I would like to see the input to the simulation; not just the resulting simulation $-$ please post something that includes the full set of parameters.

 

[PeroK]

To simplify this a bit, I will assume that $\theta$ is small, so $\cos(\theta)=1$ and $\sin(\theta)=\theta$. So the new Lagrangian with the extra degree of freedom is $$L=m r \left(g+\dot \theta \dot x\right)+\frac{1}{2} (2 m+M) \dot x^2+m (\theta+1) \dot r
\dot x+m \dot r^2+\frac{1}{2} m r^2 \dot \theta^2 $$ I would not mind at all if someone could check this. Even with the small angle approximation it is a pretty hairy Lagrangian.

The Euler Lagrange equations are: $$ -(2 m+M) \ddot x-m (\theta +1) \ddot r-2 m \dot r \dot \theta -m r \ddot \theta =0$$ $$ m \left(g-2 \ddot r+r \dot \theta ^2-\theta \ddot x-\ddot x\right)=0 $$ $$ -m r \left(2 \dot r \dot \theta +r(t) \ddot \theta +\ddot x\right)=0 $$

It's probably necessary to delay the small angle approximation until at least we have the E-L equations. This gives:
$$2\ddot r + (1+\sin \theta)\ddot x -r\dot \theta^2 - g\cos \theta = 0$$$$(2m + M)\ddot x + m[(1+\sin \theta)\ddot r + 2\dot r \dot \theta \cos \theta + r \ddot \theta \cos \theta - r\dot \theta^2 \sin \theta] = 0$$$$r \ddot \theta + \ddot x \cos \theta + 2\dot r \dot \theta + g\sin \theta = 0$$

 

[PeroK]

If we look for an "equilibrium" solution of constant $\theta$ the equations (with $\mu = \frac M m$) are:
$$2\ddot r + (1+\sin \theta)\ddot x = g\cos \theta$$$$(2 + \mu)\ddot x + (1+\sin \theta)\ddot r = 0$$$$\ddot x \cos \theta = -g\sin \theta$$And the solution is:
$$\ddot x = -(\tan \theta)g$$$$\ddot r = \frac{1 + \sin \theta}{2\cos \theta}g$$Where $\theta$ can be obtained numerically from the equation:$$\mu = \frac{1 - \sin \theta}{2\tan \theta}$$E.g. if $\mu = 5$, then:
$$\theta \approx 0.09$$The small angle approximation is, of course$$\theta = \frac{1}{2\mu + 1}$$I can't prove, however, that the solution tends asymptotically to that equilibrium angle.

 

[PeroK]

Here's an idea. If we look at the first E-L equation:
$$2\ddot r + (1+\sin \theta)\ddot x -r\dot \theta^2 - g\cos \theta = 0$$All the terms are finite as $t \to \infty$, except possibly the term $r\dot \theta^2$. So, this must be asymptotically finite too. But, $r$ increases without limit, so $\dot \theta$ must tend to zero.

 

[jbriggs444]

I would like to see the input to the simulation; not just the resulting simulation $-$ please post something that includes the full set of parameters.

It has the right qualitative behavior. The falling block never moves rightward. It does (eventually) move leftward. The big block accelerates away leftward starting immediately.

 

[A.T.]

I would like to see the input to the simulation; not just the resulting simulation $-$ please post something that includes the full set of parameters.

I attached the Algodoo file. You can rename it to "block_pulley.phz" and open it in the free software. There you can check and change all the parameters:

http://www.algodoo.com/

Here an updated animation with a grid.

 

[Baluncore]

The big block accelerates away leftward starting immediately.

I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways. As a pendulum, the period of oscillation is increasing as the string lengthens, and it can only swing through half a cycle before falling m collides with big M.
Given sufficient string will that collision ever occur before it runs out of string? Or will the period lengthen at a rate that prevents the collision until after it runs out of string?

 

[PeroK]

I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways. As a pendulum, the period of oscillation is increasing as the string lengthens, and it can only swing through half a cycle before falling m collides with big M.
Given sufficient string will that collision ever occur before it runs out of string? Or will the period lengthen at a rate that prevents the collision until after it runs out of string?

My analysis shows that it tends to a fixed angle determined by $\frac M m$. It doesn't oscillate. See also AT's simulation.

 

[sysprog]

There you can check and change all the parameters.

How do I check the parameters?

 

[A.T.]

How do I check the parameters?

In Windows you right click on objects. Then under Material you have mass and friction.

 

[sysprog]

I thought of falling m as a pendulum bob, that started swinging, driven by big M moving sideways.

If falling $m$ doesn't move leftward exactly when and as much as the pulley does, but at some $t>0$ time later it does move leftward (unlike as in @A.T.'s simulation, in which it tilts so as to maintain topwise orthogonality wrt the string, but is somehow constrained to not swing leftward, despite the leftward movement of the pulley from which it is hanging), then when does it do that, and if then, why then? If it's not the case that it never moves leftward, then why wouldn't it move leftward immediately as $M$ and the pully do, given that the pulley moves leftward only as driven by the descent of falling $m$?

 

[PeroK]

If falling $m$ doesn't move leftward exactly when and as much as the pulley does, but at some $t>0$ time later it does move leftward (unlike as in @A.T.'s simulation, in which it tilts so as to maintain topwise orthogonality wrt the string, but is somehow constrained to not swing leftward, despite the leftward movement of the pulley from which it is hanging), then when does it do that, and if then, why then? If it's not the case that it never moves leftward, then why wouldn't it move leftward immediately as $M$ and the pully do, given that the pulley moves leftward only as driven by the descent of falling $m$?

Gravity acts on the hanging mass as soon as it is released. There is a non-zero acceleration at time $t= 0$. If we drew a graph of the acceleration it would start at some value such as $g/3$ or something like that.

The force on the top mass is also instantaneous, so it moves with a non-zero acceleration at time $t=0$.

And, the lateral force on the pulley is instantaneous, so the large block moves with a non-zero acceleration at time $t = 0$.

The difference with the hanging mass is that there is zero acceleration at time $t=0$ and the acceleration must continuously increase from zero. It cannot, like the mass M, suddenly take off with non-zero acceleration.

If we look at the acceleration again after a very short time, then the side mass has non-zero acceleration leftwards: but it is still (much) less than the acceleration with which block $M$ began.

Technically, of course, the hanging mass is moving leftwards at any time $t > 0$, but with less initial acceleration and hence less velocity than the mass $M$.

And, in fact, the above analysis shows that it gradually catches up with the leftwards acceleration of $M$ until an equilibrium angle is reached.

 

[A.T.]

If it's not the case that it never moves leftward,

It moves leftward. Look at the animation and note where it ends up relative to the vertical grid line it started on.

then why wouldn't it move leftward immediately

For the hanging mass:

- initial horizontal velocity is zero.
- initial horizontal acceleration is zero (because the string is vertical)

But the above doesn't mean that all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero. So it starts out moving horizontally very gently.

 

[jbriggs444]

But the above doesn't mean that the all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero.

Note that it is mathematically possible to have an displacement over time graph with in which all derivatives are zero but in which displacement is nonetheless only zero exactly at $t=0$.

As I recall position $s = e^{-1/t^2} - 1$ for $t \neq 0$ and $s=0$ for $t = 0$ is a continuous and infinitely differentiable function satisfying those properties.

 

[sysprog]

It moves leftward. Look at the animation and note where it ends up relative to the vertical grid line it started on.For the hanging mass:

- initial horizontal velocity is zero.
- initial horizontal acceleration is zero (because the string is vertical)

But the above doesn't mean that all the further derivatives of horizontal position (jerk, snap, crackle, pop) are also zero. So it starts out moving horizontally very gently.

Thanks for the correction. I still have to finish mulling over over the consequences $-$ I think that I will probably soon wind up owing you, and @jbriggs444, @PeroK, @Dale, @berkeman, @PeterDonis, @Baluncore, @jedishrfu, and others, a big apology for my hard-headedness, and a big thank you for your patience and efforts.

I recognize and acknowledge that a higher derivative can be nonzero when a lower one is zero, e.g. jounce/snap nonzero when velocity, acceleration, and jerk are all zero:

Let $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$.
Then $f'(x) = 4x^3 - 12x^2 + 12x - 4$,
$f''(x) = 12x^2 - 24x + 12$,
$f'''(x) = 24x - 24$,
$f''''(x) = 24$.
So $f(1) = f'(1) = f''(1) = f'''(1) = 0$, but $f''''(1) = 24$.