Olympiad dynamics problem with 3 masses and a pulley

[Glenn G]

Homework Statement

Mechanics question, olympiad style

Relevant Equations

f = ma

Hello all, this is an adaptation of a question i saw some time ago (can't find the original now). There are two forces acting on two masses both 1kg. The masses are joined by an inextendable rope and going over a frictionless pulley of negigible mass. In blue I have written in some working out to show that IF the pulley was tethered to a wall (say) the difference in accelerations of the two masses was the same as the difference in the two forces SO as I'd imagine if force f1 was larger than f2 then m1 would be accelerating to the right and m2 would be accelerating to the left. IN this situation imagine this was in outer space where everything is free to move AND in this case the pulley is attached to a third mass m3 via another rope.

The main question IS what is happening to mass 3? Clearly there must be a tension force that m3 is experiencing causing it to accelerate. Does m3 experience the sum of the two forces (f1 + f2) OR the average of the two forces (f1 + f2)/2 and of course everything is connected together and so if the acceleration of m3 is greater than m1 or m2 then it is going to smash into the pulley but if it's acceleration is less than that of either m1 or m2 then it is going to lag behind BUT if the cables can't extend then this can't happen. My guess is that m3 has to have the same acceleration as the larger acceleration out of m1 and m2. If m1 and m2 have different accelerations then that doesn't bother me as the pulley can rotate so the cable either side of the pulley can change length (initially).

ANYWAY I have realized that my mechanics is lacking here and would appreciate help with this analysis please.

cheers,
Glenn.

ps in blue are bits of analysis that I think would be true if the pulley was tethered not sure how this all applies IF everything is dynamic and in 'outer space'.

 

[kuruman]

You need to reconsider your equations. First of all I would not substitute m = 1 kg in the equations. Just leave it as a symbol and you can put in the numbers at the very end. This makes the equations less confusing to the reader. More importantly, you have not used the fact that the rope is inextensible. Let $x_1$ be the length of the rope from $m_1$ to the pulley, $x_2$ be the length of the rope from $m_2$ to the pulley, and $L$ the total length of the rope. Then, $L=x_1+x_2=\text{constant}.$ Now if you take second derivatives, you get $\ddot {x}_1+\ddot {x}_2=0$.

Note that these are not the accelerations $a_1$ and $a_2$ because the entire assembly of three masses accelerates so that its center of mass has acceleration $A_{cm}=\dfrac{f_1+f_2}{m_1+m_2+m_3}.$ I suggest that you find $a_1$ and $a_2$ as if mass $m_3$ were free to move.

 

[nasu]

What is the tatement of the problem? And what are the conditions? Is the pulley fixed or is moving? Is this a view from the top of a horizontal setup?

 

[kuruman]

What is the tatement of the problem? And what are the conditions? Is the pulley fixed or is moving? Is this a view from the top of a horizontal setup?

According to OP,

$\dots~$ this is an adaptation of a question i saw some time ago (can't find the original now).

which I interpret to mean that the original question is unavailable and that what we have here is a reconstruction from OP's memory. I would assume that gravity is not an issue in this case, i.e. that the assembly is on a horizontal frictionless surface or in free space and constrained to move in one dimension..

 

[nasu]

He may not remember the original question but if he attempted a solution he should have a clear picture of what he is trying to solve. It may not be identical to the lost original but at least he should state what problem he is trying to solve.

 

[kuruman]

I assumed one is looking for the accelerations of the three masses given the pulling forces f₁ and f₂. It doesn't really matter because OP's main question is not what the initial Olympiad problem asked but,

The main question IS what is happening to mass 3?

So I think that question can be answered if one finds the acceleration of each mass given the external pulling forces f₁ and f₂.

 

[swampwiz]

I remember this problem from my Engineering Mechanics course. The root pulley does not move for the branch pulley having weights of 10 & 15 if the other side is 24 - which is completely unintuitive!

 

[kuruman]

I remember this problem from my Engineering Mechanics course. The root pulley does not move for the branch pulley having weights of 10 & 15 if the other side is 24 - which is completely unintuitive!

Can you explain which is the "root" pulley and which is the "branch" pulley. The hand drawn picture in post #1 shows only one pulley.

 

[swampwiz]

Can you explain which is the "root" pulley and which is the "branch" pulley. The hand drawn picture in post #1 shows only one pulley.

My bad, this is not exactly the same problem - the one I am thinking about has instead of the mass on the left a pulley with a mass hanging. The new pulley would be the root pulley, while the existing one would be the branch.

 

[Orodruin]

More importantly, you have not used the fact that the rope is inextensible. Let x1 be the length of the rope from m1 to the pulley, x2 be the length of the rope from m2 to the pulley, and L the total length of the rope. Then, L=x1+x2=constant. Now if you take second derivatives, you get x¨1+x¨2=0.

I would not use these coordinates because the pulley is not fixed. Because of that, if you introduce coordinates like this, then $\ddot x_i$ will not be the acceleration of mass $i$ unless the pulley does not accelerate (which it will unless $m_3$ is infinite). Instead, I would use $x_i$ as the position of the masses in an inertial frame. This of course means that all three accelerations will form part of the string equation. Using that and a few FBDs, OP should be able to obtain sufficient information to solve the problem.

 

[kuruman]

I would not use these coordinates because the pulley is not fixed.

When I sat down to solve this problem I used coordinates as you suggested for the exact same reasons. I was only trying to direct OP to the correct way of thinking about the constraint of the inextensible rope.

 

[Steve4Physics]

Correct me if I'm wrong but ...

Because $m_1=m_2$, from the symmetry we can tell that the system’s centre of mass is a constant distance from $m_3$.

This means $m_3$'s acceleration is the same as that of the system's centre of mass (as already calculated by @kuruman in Post #2).

 

[Orodruin]

Correct me if I'm wrong but ...

Because $m_1=m_2$, from the symmetry we can tell that the system’s centre of mass is a constant distance from $m_3$.

This means $m_3$'s acceleration is the same as that of the system's centre of mass (as already calculated by @kuruman in Post #2).

You're not wrong. However, the general case with different masses is at least as interesting.

 

[Steve4Physics]

You're not wrong. However, the general case with different masses is at least as interesting.

Yes, the general case provides a nice problem.

I suspect that the original question (with $m_1=m_2$) was written by design, so a quick solution was available in a time-limited competition.

 

[Orodruin]

Yes, the general case provides a nice problem.

I suspect that the original question (with $m_1=m_2$) was written by design, so a quick solution was available in a time-limited competition.

It is also a good sanity check to verify that the acceleration of $m_3$ indeed becomes that of the CoM when $m_1 = m_2$ even in the general expression.

Edit: Other good sanity checks include sending one mass to infinity - effectively putting its acceleration to zero by hand.

 

[kuruman]

An interesting variant in the case of different masses is to ask, "what should the ratio $f_2/f_1$ be so that all three masses have the same acceleration?

Spoiler

$\dfrac{f_2}{f_1}=\dfrac{2m_2+m_3}{2m_1+m_3}$

 

[bob012345]

It is interesting to note the general solution is that of a fixed pulley if $m_3 →∞$ and also the tension in the rope goes to 0 as $m_3 →0$ i.e. the masses act independently.

 

[Orodruin]

It is interesting to note the general solution is that of a fixed pulley if m3→∞

I mentioned this in the edit of #15. Sending any of the masses to infinity will effectively send the acceleration of that mass to zero. If doing it to mass 1 or 2, you obtain what is essentially a system with a rope fixed in one end attached to a pulley and a mass in the opposite end. If you fix mass 3 you get a fixed pulley.

also the tension in the rope goes to 0 as m3→0 i.e. the masses act independently.

This also forces one to consider the applicability of results. The masses act independently - until the distance between the masses (plus half the circumference of the pulley) reaches the length of the rope.