Falling mass, additionally accelerated by a rope

• greypilgrim
greypilgrim
Thread moved from the technical forums to the schoolwork forums
Hi.

The following situation:

The pulleys are fixed to the floor/ceiling and massless, as are the ropes, and there is no friction. At ##t=0##, the masses are released from rest.

For the moment, I'll assume ##m_2\gg m_1##. So ##m_2## will accelerate at ##g## and the red rope at ##\frac{d_1}{d_2}g##. How do I now find the accelerations of motion for ##m_1##?

At the very beginning, as far as I can see ##m_1## experiences a downward force ##F_1=m_1 g+\frac{d_2}{d_1}m_2 g##. But I'm quite sure the tension in the red rope decreases, but how exactly?

Hi,

you want to write down a few equations. choose/define relevant variables and work towards ##N## equations with ##N## unknowns.
What about the tensions in the red rope at the (massless!) lower pulley ?
And tensions in the red and blue ropes at the upper pulley ?
And the in the blue rope is related to the acceleration of ##m_2##

greypilgrim said:
For the moment, I'll assume ##m_2\gg m_1##. So ##m_2## will accelerate at ##g## and the red rope at ##\frac{d_1}{d_2}g##. How do I now find the accelerations of motion for ##m_1##?

At the very beginning, as far as I can see ##m_1## experiences a downward force ##F_1=m_1 g+\frac{d_2}{d_1}m_2 g##. But I'm quite sure the tension in the red rope decreases, but how exactly?
If ##m_2 \gg m_1## then we can treat the downward acceleration of ##m_2## as being fixed. You already realize this.

This means that the downward acceleration of ##m_1## is also fixed. (Assuming ##d_2 > d_1## so that the rope does not go slack).

What is the resulting acceleration of ##m_1##?
What forces act on ##m_1##?
Which of those forces do you already know?
Does this allow you to calculate the remaining force?

If ##m_2## is not vastly greater than ##m_1## then you will have some simultaneous equations to solve as @BvU suggests.

jbriggs444 said:
This means that the downward acceleration of is also fixed.
By "fixed", do you mean constant?

greypilgrim said:
By "fixed", do you mean constant?
Constant yes. But I mean that it is "determined". Given what you already know, there is only one value it could possibly be.

But if the blue rope accelerates at ##g##, then the red one does so at ##\frac{d_1}{d_2}g##. And if my ##F_1## from #1 is correct and constant, then ##m_1## accelerates at ##g+\frac{d_2}{d_1}\frac{m_2}{m_1}g## which is bigger than ##\frac{d_1}{d_2}g## given that ##\frac{m_2}{m_1}\gg 1##.

This would mean that the rope goes slack immediately. But then again ##m_1## falls freely at ##g##, allowing the rope to tension again and so forth... What's wrong?

greypilgrim said:
But if the blue rope accelerates at ##g##, then the red one does so at ##\frac{d_1}{d_2}g##. And if my ##F_1## from #1 is correct and constant, then ##m_1## accelerates at ##g+\frac{d_2}{d_1}\frac{m_2}{m_1}g## which is bigger than ##\frac{d_1}{d_2}g## given that ##\frac{m_2}{m_1}\gg 1##.

This would mean that the rope goes slack immediately. But then again ##m_1## falls freely at ##g##, allowing the rope to tension again and so forth... What's wrong?
greypilgrim said:
But if the blue rope accelerates at ##g##, then the red one does so at ##\frac{d_1}{d_2}g##. And if my ##F_1## from #1 is correct and constant
Huh? You have given no valid computation for ##F_1##.

What you know is how ##m_1## moves. You can use that to calculate what ##F_1## must be.

I used the law of the lever on the wheel and axle and added this force to the gravitational force acting on ##m_1## (but I see now that this can't be right).

jbriggs444 said:
What you know is how m1 moves. You can use that to calculate what F1 must be.
Huh? That seems backwards to me. How would I know how ##m_1## moves without knowing ##F_1## first and calculate the acceleration out of that?

EDIT: Ah, I see. If there's no slack, ##m_1## obviously has to accelerate at the same rate as the blue rope.

Last edited:
jbriggs444
greypilgrim said:
EDIT: Ah, I see. If there's no slack, m1 obviously has to accelerate at the same rate as the blue rope.
Yep.

Things are not always as simple as they seem. When I see an analysis of the Indian Rope Trick, or a released balloon, I immediately fear the speed of sound in rope.

Okay, that makes my initial question trivial. I was way too focused on finding the forces first to realize that the acceleration can be calculated right away.

So ##a_1=\frac{d_1}{d_2}g## and since ##F_1=m_1 a_1=m_1 g+T_r##, the tension in the red rope is
$$T_r=\frac{d_1}{d_2}m_1 g-m_1 g=\left(\frac{d_1}{d_2}-1\right)m_1 g\ .$$

However, now the law of the lever
$$T_r\cdot d_1=T_b\cdot d_2=m_2 g\cdot d_2$$
seems to fail completely. Is this an artifact of ##m_2\rightarrow\infty## or does that generally not hold for accelerated systems?

greypilgrim said:
$$T_b\cdot d_2=m_2 g\cdot d_2$$
Since ##m_2## is accelerating there must be a net force on it, so ##T_b\neq m_2g##.

greypilgrim said:
I used the law of the lever on the wheel and axle and added this force to the gravitational force acting on ##m_1## (but I see now that this can't be right).
Falling m2 pulls down on also falling m1 via the lever (mechanical advantage of d2/d1).
Therefore, m2 must accelerate at a rate lesser than g, while m1 must accelerate at a rate greater than g.

As the height difference between both masses decrease as they fall, the combined (m1+m2) center of mass must be descending at a value of acceleration that is within the range of values at which m1 and m2 are accelerating.

haruspex said:
Since ##m_2## is accelerating there must be a net force on it, so ##T_b\neq m_2g##.
Ah, I see. I think I'm running into trouble here because of my assumption that if ##m_2\gg m_1##, then ##m_2## is "practically" free falling, which would mean that ##T_b\approx 0## (right?), but on the other hand still having ##m_1>0## which means that ##T_b=\frac{d_1}{d_2}T_r>0## (but still small, so so it kind of works out).

I guess I should try and tackle the full problem without approximations.

Lnewqban
greypilgrim said:
I guess I should try and tackle the full problem without approximations.
Good idea.

Okay. For consistency, let ##T_1=T_r## and ##T_2=T_b##. I'll assume no slack, which means ##a_1>g## (otherwise, both masses are free falling). Then I get the four equations:
\begin{aligned} m_1 a_1&=m_1 g+T_1 \\ m_2 a_2&=m_2 g-T_2 \\ \frac{a_1}{d_1}&=\frac{a_2}{d_2} \\ T_1 d_1&=T_2 d_2 \end{aligned}
I used a CAS to solve this for me:
\begin{aligned} a_1&=d_1 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2} \\ a_2&=d_2 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2} \\ T_1&=d_2 m_1 m_2 g\cdot\frac{d_1-d_2}{d_1^2 m_1+d_2 ^2 m_2} \\ T_2&=d_1 m_1 m_2 g\cdot\frac{d_1-d_2}{d_1^2 m_1+d_2 ^2 m_2} \end{aligned}
Now in the limit ##\frac{m_1}{m_2}\rightarrow 0##, this simplifies to:
\begin{aligned} a_1&=\frac{d_1}{d_2}g \\ a_2&=g \\ T_1&=\left(\frac{d_1}{d_2}-1\right)m_1 g \\ T_2&=\frac{d_1}{d_2^2}\left(d_1-d_2\right)m_1 g=\frac{d_1}{d_2}\left(\frac{d_1}{d_2}-1\right)m_1 g=\frac{d_1}{d_2}T_1 \end{aligned}
That looks good so far. Going back to the general solution, the condition ##a_1>g##, is:
$$d_1 g\cdot\frac{d_1 m_1+d_2 m_2}{d_1^2 m_1+d_2 ^2 m_2}>g$$
And this simplifies to ##d_1>d_2##, which was to be expected.

Is all that correct?

Last edited:
Oh and another question, more mathematical: If I take the limit ##m_1\rightarrow 0## instead of ##\frac{m_1}{m_2}\rightarrow 0##, I get the same accelerations, but ##T_1=T_2=0##. Is this some zeroth vs. first order approximation thing? But why do I get the same accelerations then?

greypilgrim said:
Oh and another question, more mathematical: If I take the limit ##m_1\rightarrow 0## instead of ##\frac{m_1}{m_2}\rightarrow 0##, I get the same accelerations, but ##T_1=T_2=0##. Is this some zeroth vs. first order approximation thing? But why do I get the same accelerations then?
If ##m_1## approaches zero while ##m_2## remains finite and approximately constant then ##\frac{m_1}{m_2}## also approaches zero. So naturally you get the same limiting accelerations either way.

If you get a finite limiting acceleration and you are considering a mass ##m_1## that approaches zero then of course the tension ##T_1## will approach zero. A non-zero limiting tension would produce an infinite limiting acceleration of ##m_1##. Which you've calculated is not the case.

I solved the equations for ##a_1## first and then saw what happens when ##m_2>>m_1##. To do this, I simplified the algebra (no need for a CAS) by replacing ##m_2=\mu m_1## and ##d_1=d_2 \delta.## Note that scaling parameters ##\mu## and ##\delta## are both greater than 1. The bottom two of the starting equations in post #16 reduce to ##~T_2=T_1\delta~## and ##~a_2=a_1/\delta.## These can be substituted back into the top two equations to yield a system of two equations and two unknowns. One then gets ##~a_1=f_1(\mu,\delta)g~## and ##~T_1=f_2(\mu,\delta)m_1g##.

One can then investigate the behavior of dimensionless numerical constants ##f_1(\mu,\delta)## and ##f_2(\mu,\delta)## in the limit ##\mu >>1## at fixed ##\delta##. The result may be surprising.

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