# Energy: What will be the minimum ratio of m/M here

• PhysicsBoi1908
In summary: I am having a hard time creating equations. I am not able to determine how the equations can be formed for this question.

## Homework Statement

The objects are set free from the scenario shown in the attached image. Their masses have been given and they are under the effect of gravity. The length of the string has been shown in the figure. If the two blocks cross each other, then we need to determine the minimum value of m/M.

## Homework Equations

I am having a hard time creating equations. I am not able to determine how the equations can be formed for this question.

## The Attempt at a Solution

I tried using constraint, however, that was not of much help. I drew the diagram at intermediate and jotted down some equations with trigonometric ratios in them, but I wasn't able to make any conclusion.

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Delta2
Try drawing the two FBDs as a starting point...

I did, we will have 2Tsinθ on the object with mass M in the upward direction, I then drew an intermediate diagram too, but it is not working out.

PhysicsBoi1908 said:
I did, we will have 2Tsinθ on the object with mass M in the upward direction, I then drew an intermediate diagram too, but it is not working out.
Can you please Upload a copy of your FBDs (in PDF or JPEG format), and write out the equations that you have written for the problem? That will help us to help you. Thanks.

This is the constraint diagram that I drew, this resulted in me writing a trigonometric equation that resulted in an error. I think I am making a mistake at the part where we are required to determine m/M such that they both "cross" each other, but I don't know for sure.

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If I understand it correctly, and I think I do, I think it is kind of simple. When they cross, the lengths will be equal on all of the lengths. You can even compute this length, and it is ## \frac{3}{2} L ##, because you have ## L_{total}=6L ##. In this case, when the diagram is symmetric, what happens if ## m>M ##? I think that is all this requires. The tension is symmetric around the pulleys. And it is not necessary to compute the tension or even the distance ## \frac{3}{2}L ##. Am I missing something? ## \\ ## The other thing I'm thinking though is that it requires a potential energy assessment, but scratch that=let's assume they cross each other and remain crossed.

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@PhysicsBoi1908 -- Are the radii of the pulleys specified as zero, or at least implied? Also, do they say more explicitly what is meant by "crossing" of the masses? When they just touch top-to-bottom (in which case we need the vertical dimensions of each mass), or when their COMs cross? These questions change the geometry of the drawing...

That was bothering me too, however, I don't think that we will have to consider the pulley to have any radius. As for the objects crossing, we need to determine the minimum ratio of m/M, that leads me to think that we just need to have the objects touching each other, because if either of the masses will continue moving, then its mass will have to be relatively higher, but again, I don't know if I am right or wrong.

It might interest you that from energy considerations, the results give that ##\frac{m }{M} \geq 2 \sqrt{\frac{3}{5}}-1 \approx .55 ## if I computed it correctly, for ## m ## to ever be able to get even with ## M ##. It's difficult to assess the academic level that this problem is, and what exactly they are looking for. From a balance of forces, the answer is almost obvious, but there are a couple of ways to interpret it. ## \\ ## For the energy case, it would not remain balanced in that position=the heavier mass would fall back down. ## \\ ## The solution using energy considerations actually makes it a more interesting problem, but it does require the student to know that potential energy ## U=mgh ##.

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PhysicsBoi1908 and Delta2
Are we supposing that the final diagram will be symmetric because both of them will have the same velocity?

For the balanced force case, the masses come to a stop if the ratio is the minimum=the answer for that case is ## \frac{m}{M}=1 ##. ## \\ ## For the energy consideration case, when ## \frac{m}{M} ## is at the minimum, they will both be slowing down and become stationary as they meet, so that the kinetic energy of each is zero. The only energy that needs to be computed there is the change in gravitational potential for each. The more I look at this problem, the more I think they might be looking for the student to consider the energy of the system. The statement of the problem just says "if the two blocks cross each other". It doesn't say that they need to remain crossed.

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Delta2
Yes! Even I arrived at m/M=1, but that isn't the answer, I just had an idea, I will try it out and report back in a few minutes.

Charles Link
Can you please confirm if what I have done is correct? After this, I aim at writing x and l individually in terms of l via trigonometry, however, I am not able to establish the relation between angles that the string will make with each other.
I have attached an image, please check it.

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Any ideas on how we can relate x and y to l individually? The relation: x+y= 3½ is a relation having both x and y.

You are starting to get the idea, but it stil needs some additional computing. What is the height of the two as they meet? I get that the mass ## m ## drops by a distance of ## \frac{\sqrt{5}}{2} L ##. (Compute the y distance when the hypotenuse is ## \frac{3}{2} L ## ).## \\ ## Then you also need to compute how much mass ## M ## rises, etc. That is actually straightforward. It starts out down below by ## \sqrt{3} L ##.

How did the mass m drop by √5/2? Moreover, if it did, then we can say that since x+y=√3l, y=√3l-√5/2

See my post 6 above. Hypotenuse=(3/2) L.

Oh ok, I will try using it.

Plunging √5/2 into the equation m/M= (√3l-x)/x gives the following result:

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I corrected post 15=it needs an ## L ## multiplied by ## \frac{\sqrt{5}}{2} ##, i.e. ## x=\frac{\sqrt{5}}{2} L ##. You almost have the result. Check your arithmetic/algebra. I get ## \frac{m}{M}=2 \sqrt{\frac{3}{5}}-1 ##.

Yes, I got it, however, I am not able to understand why the length of string=3l/2, of course, it is something rather obvious, but how can we prove it?

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The total length of the string is ## 6 L=2L+2L+L+L ##. You have 4 equal lengths, two on the left side and two on the right, going to the pulleys.

Oh, ok, I will try asking my teacher if the answer is correct and report back. Thank you for your help.

Charles Link
Additional comment: The clue is actually in the title of the thread that I am just looking at now="Energy". Looks like we might have interpreted it correctly.

Delta2 and PhysicsBoi1908
Seems like it , I have sent our answer to my teacher, he will respond when he will be free, hopefully within a few hours

Charles Link
If you try to form the differential equations of motion for this problem, they are going to be non linear cause as you already pointed out they ll contain term of ##\sin\theta(t)## and ##\tan\theta(t)## or possible square roots, so the only easy way I see is by energy methods as already Charles pointed out.

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PhysicsBoi1908
Yes, I tried forming such equations, but it only made the question more complex, the energy method is much nicer.

Delta2
My sir hasn't replied yet, however, my friend has pointed out a mistake, according to what we have calculated, the ratio m/M is nearly 0.5, this signifies that M is greater than m, however, if that was to be the case, then the object of mass M will not go up, or will it because of initial tension?

It will go up, simply because the body of mass m will go down (the only vertical force that is acting at start is its weight, there is no vertical component of tension initially) and by going down it will pull the strings towards its side*, so the body of mass M will go up. The whole point is whether the body of mass M will go up enough so that they meet.

*Unless of course the strings can be stretched or broken but I am sure we are not examining this case here.

Charles Link and PhysicsBoi1908
This is what I thought, but my friend is not ready to accept it. Moreover, if m was to be heavier, then M will go up no matter what and will hence touch the mass m, the force acting on it will not be able to stop it since the angle that it is making will continue to decline, and so will the component of the force dragging it upwards. Thanks anyways.

My sir just replied and our answer was indeed correct! Thank you very much https://www.physicsforums.com/members/delta.189563/ and Charles Link for your help. I really appreciate how I was led to the answer instead of it being given out to me. Thank you very much!

Delta2 and Charles Link
Well thanks too but I didn't too much @Charles Link did all the job here.

The differential equations of motion for this problem would be like

for body of mass m:
$$mg-2T(t)\sin\theta_1(t)=L\frac{d^2\tan\theta_1(t)}{dt^2}$$

for body of mass M
$$2T(t)\sin\theta_2(t)-Mg=L\frac{d^2\tan\theta_2(t)}{dt^2}$$

##\theta_1(t)## is the angle made by the horizontal and the strings that are connected to body of mass m. ##\theta_2(t)## is the angle by the horizontal and the strings that are connected to body of mass M.
Need one more equation that relates ##\theta_1## and ##\theta_2## but I don't seem to find an obvious one. Also not sure if the tension T(t) is constant or time varying...

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PhysicsBoi1908 said:
My sir just replied and our answer was indeed correct! Thank you very much https://www.physicsforums.com/members/delta.189563/ and Charles Link for your help. I really appreciate how I was led to the answer instead of it being given out to me. Thank you very much!
@PhysicsBoi1908 I'm very glad we got the right answer. And thank you for posting the question. It really is one of the better ones we have encountered. At first sight, it almost looked like it was a problem with a very trivial solution of ## \frac{m}{M} \geq 1 ##, but it turned out to be very interesting.

Delta² said:
Well thanks too but I didn't too much @Charles Link did all the job here.

The differential equations of motion for this problem would be like

for body of mass m:
$$mg-2T(t)\sin\theta_1(t)=L\frac{d^2\tan\theta_1(t)}{dt^2}$$

for body of mass M
$$2T(t)\sin\theta_2(t)-Mg=L\frac{d^2\tan\theta_2(t)}{dt^2}$$

##\theta_1(t)## is the angle made by the horizontal and the strings that are connected to body of mass m. ##\theta_2(t)## is the angle by the horizontal and the strings that are connected to body of mass M.
Need one more equation that relates ##\theta_1## and ##\theta_2## but I don't seem to find an obvious one. Also not sure if the tension T(t) is constant or time varying...
Using energy equations, I think you can write an equation for the velocity of each mass as a function of y . The equation for each mass would have the form ## \frac{dy}{dt}=f(y) ##, which could be written as ## \frac{dy}{f(y)}=dt ##. Perhaps it even has a closed form solution. I haven't worked it yet to see. ## \\ ## Edit: Letting ## y ## the the distance that ## m ## moves, and ## y_1 ## the distance that ## M ## moves, and ## s ## the cable length from ## M ## to the pulley, we have ## s=3L-\sqrt{L^2+y^2} ##, and ## y_1=\sqrt{3}L-\sqrt{s^2-L^2} ##. ## \\ ## Finally, ## K.E.=\frac{1}{2}m(\frac{dy}{dt})^2+\frac{1}{2}M(\frac{dy_1}{dt})^2=mgy-Mgy_1 ##. Let me work it a couple more steps to see if I can get a single equation of ## \frac{dy}{dt}=f(y) ##. ## \\ ## Additional edit: Yes, ## y_1 ## can be expressed completely in terms of ## y ##, but it's not a simple expression. Thereby, we have ## g(\frac{dy}{dt})=f(y) ##, but ## f(y) ## is not a simple function, and ## g(\frac{dy}{dt}) ## is also not simple. ## \\ ## Additional edit: If anything at all useful comes out of this last bit of algebra and calculus, perhaps it is the almost obvious result that if ## \frac{dy}{dt}=0 ##, then ## \frac{dy_1}{dt}=0 ## as well.

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Delta2
yes I knew from start that no matter how you decide to make the ODEs for this problem they would be non linear. @Charles Link the functions g and f you mention are not linear, they contain squares and square roots of their arguments if I understood correctly.

Charles Link