Problem with pulleys - two fixed and one movable

[brotherbobby]

Homework Statement

Consider the setup shown in the figure below. Two weights, with masses $m_1$ and $m_2$, hang on the outside of a three-pulley system, while a weight of mass M hangs on the middle pulley. We assume the pulleys and the connecting rope have negligible mass, so their kinetic and potential energies are also negligible. We will suppose for now that all three pulleys have the same radius R, but this will turn out to be of no importance. Calculate the accelerations of $m_1,m_2$ and $M$.

Relevant Equations

Newton's law : Net force $\displaystyle{\Sigma F = ma}$

[The problem involves the use of Lagrange's equations for conservative systems. I am first trying to solve it using Newton's mechanics, similar to that of a school student]

Diagram : The diagram of the problem is shown to the right.

Solution : The tension is the same all over the (same) rope - let's assume it to be some value $T$.

The masses $m_1$ and $m_2$ accelerate up with an acceleration of $2a$ if we assume that the mass $M$ moves down with an acceleration of $a$. This is because the movable pulley has two sides that move down with $a$. This has to be "accounted for" by just one rope connected to masses $m_1$ and $m_2$, which have to move up with $2a$.

Writing the equations of motion for each mass :

(1) $\underline{m_1}$ : $T - m_1g = 2m_1 a\qquad (1)$

(2) $\underline{m_2}$ : $T - m_2 g = 2m_2 a \qquad (2)$

(3) $\underline{M}$ : $Mg-2T = Ma\qquad (3)$

Immediately I notice something's wrong. I have two unknowns, viz. $T,a$ but three equations where the masses are independent of each other.

Request : Where am I going wrong? A hint would be very appreciated.

 

[jbriggs444]

Request : Where am I going wrong? A hint would be very appreciated.

You have assumed that the left hand mass $m_1$ and the right hand mass $m_2$ accelerate downward at the same rate, $a$.

What if they accelerate at two different rates, $a_1$ and $a_2$?

 

[A.T.]

The masses m1 and m2 accelerate up with an acceleration of 2a if we assume that the mass M moves down with an acceleration of a.

This makes no sense. If m1 and m2 are different masses but rope force on them is the same, how can they have the same acceleration?

This is because the movable pulley has two sides that move down with a. This has to be "accounted for" by just one rope connected to masses m1 and m2, which have to move up with 2a.

Think again. Even if m1 = m2, so a1 = a2, then a1 = a2 = a, not = 2*a. The center pulley doesn't even rotate in that case.

 

[Steve4Physics]

Request : Where am I going wrong? A hint would be very appreciated.

@brotherbobby, if you haven’t yet cracked it, here’s a short exercise which might help you see what's going on.

Let $L$ be the total length of the 2 internal string-sections – shown in red in the diagram.

In each of the following cases, what is:
i) the change in the total length of the 2 internal string-sections, $\Delta L$;
ii) the height-change of the centre mass, $\Delta Y$.

I’ve even done the first one for you!

a) $\Delta y_1 = +3$ cm and $\Delta y_2 = +3$ cm (i.e. both $m_1$ and $m_2$ risea a distance of 3 cm).
Answer:
i) $\Delta L = 6$ cm.
ii) $\Delta Y = -3$ cm.
Make sure you understand why!

b) $\Delta y_1 = +3$ cm and $\Delta y_2 = -3$ cm.

c) $\Delta y_1 = +3$ cm and $\Delta y_2 = 0$.

d) $\Delta y_1 = +3$ cm and $\Delta y_2 = 1$ cm.

e) $\Delta y_1 = +3$ cm and $\Delta y_2 = -1$ cm.

Can you see the pattern? Can you express $\Delta Y$ as a simple function of $\Delta y_1$ and $\Delta y_2$? ?

M’s acceleration in terms of the accelerations of $m_1$ and $m_2$ then immediately follows.

 

[kuruman]

Another approach is shown on the right. Consider the invariant length of the string, $L$. You can write is as $$L=x_1+2X+x_2+C$$ where $C$ is the (constant) total length of string in contact with the pulleys. When you take the second derivative of this, you have a fourth equation. Add it to your collection and you will have four equations and four unknowns, $T$, $\ddot x_1$, $\ddot X$ and $\ddot x_2.$

You could use $x_1$, $X$ and $x_2$ as your generalized coordinates when you work out the Lagrangian.

 

[brotherbobby]

You have assumed that the left hand mass $m_1$ and the right hand mass $m_2$ accelerate downward at the same rate, $a$.

What if they accelerate at two different rates, $a_1$ and $a_2$?

That is what I have a problem with. If it is the same (inextensible) rope, how can different parts of it move with different accelerations?

For the two mass Atwood-Machine with a single fixed pulley, our argument was that the two masses go down (and up) with the same acceleration because it is the same rope (string) that cannot extend (stretch). It's (fixed) length $l$ was a constraint. Given $x_1+x_2=l\Rightarrow \ddot{x}_1=-\ddot{x}_2$.

Here there's a movable pulley and one more mass - but the argument should remain the same. Same rope so same acceleration.

Of course I must be wrong, for, to spill the beans, I have solved the problem using the Lagrangian method. Only, I don't see what is wrong in my argument here.

Many thanks for your interest.

 

[brotherbobby]

This makes no sense. If m1 and m2 are different masses but rope force on them is the same, how can they have the same acceleration?

Think again. Even if m1 = m2, so a1 = a2, then a1 = a2 = a, not = 2*a. The center pulley doesn't even rotate in that case.

Because the force of the rope is not the only force that acts on them. There's the force of gravity (weight) too, which is different for the two masses. To write it out, for masses labelled by $i$, the acceleration of each mass $a=\dfrac{\Sigma F_i}{m_i}$, which can be the same if the forces and masses are in the same proportion.

 

[jbriggs444]

View attachment 364044That is what I have a problem with. If it is the same (inextensible) rope, how can different parts of it move with different accelerations?

You may be thinking of a fixed array of pulleys. In that case, the weight on one end moves with an acceleration opposite to the weight on the other. Not the same acceleration. Opposite accelerations.

But that is not the situation here. Here one of the pulleys is free to move. Now instead of two weights that can move we have three. These weights can move in various ways constrained by the total length of the rope in between.

If we, for instance, freeze the weight on the left then the weight in the middle can move up and down while the weight on the right moves down and up at twice the [reversed] rate.

If we freeze the weight in the middle then the weight on the left can move up and down while the weight on the right moves down and up at the same [reversed] rate.

If we freeze the weight on the right the weight in the middle can move up and down while the weight on the left moves down and up at twice the [reversed] rate.

If we freeze the center pulley so that it is not free to rotate but is still free to move up and down then indeed the accelerations of the left hand and right hand weights would be identical and exactly opposite to the acceleration of the central weight.

In the real world, all three weights are free to move. All three will move [barring some exact coincidence]. There is no guarantee that they will all move at the same rate.

 

[Steve4Physics]

That is what I have a problem with. If it is the same (inextensible) rope, how can different parts of it move with different accelerations?

Can I add this. It may be helpful to consider the simple case where $m_2$ is held fixed but $m_1$ and $M$ are still free to move. The rope around the right-hand pulley is now stationary, so its acceleration is zero. But the other parts of the rope can have non-zero acceleration.

The key is to understand what happens to the central pulley.

 

[erobz]

I second @Steve4Physics suggestion. Try a simpler scenario like this first:

Will blocks $A,B$ experience the same accelerations under the inextensible rope condition( in a dynamic system)?

 

[kuruman]

Same rope so same acceleration.

Same as what? Referring to post #5, the "fixed string length constraint" is $$\ddot x_1+2\ddot X+\ddot x_2=0.$$ In the special case $m_1=m_2$ it follows by symmetry that $\ddot x_1=\ddot x_2$ in which case $\ddot X = -\ddot x_1=-\ddot x_2.$

This problem can be solved quite easily (and more quickly) without Lagrangians. It helps to write the masses in terms of the middle mass, i.e. $m_1=\alpha M~;~~m_2=\beta M$. I have the solution ready to post when it would be permissible to "spill the beans." Until then, I post a figure that shows the three accelerations for $~0.01 <m_1/M < 100~$ and $m_2=M$ and $g=9.81~\text{m/s}^2.$

Note that the abscissa is logarithmic. The graph shows what everybody has been saying, "The three accelerations depend on the choice of masses."

 

[A.T.]

Because the force of the rope is not the only force that acts on them. There's the force of gravity (weight) too, which is different for the two masses.

Consider two rockets of different mass, but the same thrust. Will they lift off with the same acceleration? Obviously not. So why do you assume that m1 and m2 have the same acceleration, even when they are different masses pulled by the same tension?

 

[A.T.]

View attachment 364044That is what I have a problem with. If it is the same (inextensible) rope, how can different parts of it move with different accelerations?

When the pulleys are not fixed relative to eachother, the shape of the rope changes. Some rope pieces move along different spatial paths than others. So there is no reason to assume the same acceleration magnitude for them all.

 

[kuruman]

When the pulleys are not fixed relative to eachother, the shape of the rope changes. Some rope pieces move along different spatial paths than others. So there is no reason to assume the same acceleration magnitude for them all.

Here we are interested in the accelerations of the masses relative to the inertial lab frame, e.g. the fixed axis of one of the pulleys attached to the ceiling. It looks like @brotherbobby has missed the point that the accelerations of the two string sections on either side of the middle pulley have equal magnitudes and opposite directions relative to the axis of the middle pulley. We want the accelerations relative to the inertial lab frame.

 

[A.T.]

the accelerations of the two string sections on either side of the middle pulley have equal magnitudes and opposite directions relative to the axis of the middle pulley.

For all mass combinations?

Consider m1 = m2. The center pulley doesn't rotate. The inner ropes are static relative to the axis of the middle pulley. But the outer ropes are accelerating up relative to the axis of the middle pulley.

 

[jbriggs444]

For all mass combinations?

Consider m1 = m2. The center pulley doesn't rotate. The inner ropes are static relative to the axis of the middle pulley. But the outer ropes are accelerating up relative to the axis of the middle pulley.

In this situation, depending on the value of $M$, the outer ropes could accelerate up, down or not at all.

 

[kuruman]

For all mass combinations?

Certainly not. I meant in general. As I indicated in post #11,

In the special case $m_1=m_2$ it follows by symmetry that $\ddot x_1=\ddot x_2$ in which case $\ddot X = -\ddot x_1=-\ddot x_2.$

The accelerations for $m_1=m_2=m$ are shown in the graph below left. The acceleration of the inner strings (green line) has equal magnitude and opposite direction to the acceleration of the outer strings (red line) for all mass ratios $m/M$. There is an equilibrium point where all accelerations are zero at $(m/M)=\frac{1}{2}.$ This picture is analogous to the standard two-mass Atwood machine (below right) with the exception that the equilibrium point is shifted to a mass ratio of 1.

Another interesting situation is the selection of $m_1$ and $m_2$ such that the middle mass does not accelerate and the middle pulley is essentially an idler pulley. The graph for that is shown below.

 

[Lnewqban]

Aren't the relative accelerations given in the OP's diagram?
I see 2a for m1 and m2, as a for M.

 

[kuruman]

Aren't the relative accelerations given in the OP's diagram?
I see 2a for m1 and m2, as a for M.

The OP's diagram shows the accelerations relative to the inertial lab frame in the special case $m_1=m_2$, not in general. A graph of that is shown in post #17 with the title "m1 = m2 = m."

 

[Lnewqban]

The OP's diagram shows the accelerations relative to the inertial lab frame in the special case $m_1=m_2$, not in general. A graph of that is shown in post #17 with the title "m1 = m2 = m."

Understood, but it seems not to be a special case because it reads "The diagram of the problem is shown to the right."
The OP may clarify further.

 

[kuruman]

Understood, but it seems not to be a special case because it reads "The diagram of the problem is shown to the right."
The OP may clarify further.

That's what it reads because the OP does not realize that the diagram illustrates the special case of equal masses. The OP assumes that the constancy of the string's length implies that the accelerations of the side masses must have equal magnitudes. That is correct only if the side masses are equal but not in general. See first paragraph in post #11.

 

[Lnewqban]

Thank you, @kuruman

 

[A.T.]

the accelerations of the two string sections on either side of the middle pulley have equal magnitudes and opposite directions relative to the axis of the middle pulley.

I think I misunderstood this part. You simply say that in the rest frame of the axis of a pulley, both rope segments at that pulley have the same acceleration magnitude. That is of course true for all pulleys.

But since here there is no common rest frame for all the pulley axes, one cannot assume that all rope segments have the same acceleration magnitude.

I think the OPs misconception here comes from starring at a static diagram of the rope, and imaging the rope flowing along that fixed shape. But with relative motion between pulley axes, that shape continously changes.