System of two pulleys and three masses

• Tlurk
In summary, the conversation discusses a system with three masses connected by ropes and pulleys. The first part (a) involves finding the angle between the ropes and the middle mass in order for the system to be at rest. The second part (b) involves using the radius of the pulley, rope length, and distance between pulleys to find the position of the middle mass. The third part (c) involves exploring the relationship between the masses and the angle between the ropes. The conversation also suggests using symbols to represent the masses and distances in order to simplify the calculations.
Tlurk
Homework Statement
I have a system with two pulleys and three masses where we have two outer masses on each end of the connecting rope and one middle mass at the centre of the rope. The length of the rope, the space between the pulleys and the pulley size are chosen arbitrarily. At first, all masses are equal, but we're supposed to change the middle mass later.
a) Can the system be at rest?
b) If so, what is the position of the three masses?
c) What happens if we change the middle mass?
Relevant Equations
Newton's laws.
ΣF = 0
Basic trigonometry.
I used m1=m2=15,0kg
System:

FBD:

Note: I believe I have solved a) correctly and am more confused about b).
a) I started with drawing the FBD. Knowing that the net sum has to be zero for the system to be at rest I used the left mass to find the tension on the rope: S1 = G1 = 15kg*9,81m/s2 = 147,15 N. Now that I know the tension I moved over to the middle mass. I know G2 = 147,15 N and that |S1|=|S2|=|S3|=|S4|. As the system is symmetric on each side I didn't do anything with the x-component of S2 and S3 on the middle mass, as they cancel out.

I know that S2y + S3y = G2 or rather 2 * S2y = G2 to simplify as the system is symmetric. Therefore S2y = G2/2 = 147,15N / 2 = 73,575 N.

It looks like the system can be at rest, I just have to find the angle on the ropes on the middle mass.
I know that sin(θ) = S2y / S2.

θ = arcsin(S2y / S2) = arcsin(73,575N / 147,15N) = 30 degrees.

So the system is at rest when the angle between the mass and each rope is 30 degrees.b) I used the radius of pulley r=0,20m, the rope length s=3,0m and the length between pulleys (centre to centre) l=0,80m. Again, the system is symmetric so I will look at the left side. I will also use the connection points between the rope and the masses as their position. A simple diagram:

I can find the length h from a straight line between the pulley tops to the middle mass by doing the following:
tan(30deg) = h / 0,40m
h = 0,40m * tan(30deg) = 0,23. I can then find the rope length to the top of the pulley from the middle mass by:
sin(30deg) = h / c
c =h / sin(30deg) = 0,40m * tan(30deg) / sin(30deg) = 0,462m
Also, the rope length around the pulley is about a quarter circle: 0,40m*π / 4 = π / 10 = 0,314m.
The length from the leftmost point of the pulley and down to the left mass is 1,5m - 0,462m -0,314m = 0,724m ≈ 0,72m.

If we now make the coordinate system have its x-axis along the top of the pulleys, and the y-axis through the centre of the first pulley, the masses would have the following positions:
Left mass: (-0,20, -0,20-0,72) = (-0,20, -0,92)
Middle mass: (0,40, -0,23)
Right mass: (1,0, -0,92)

Is this correct? I feel like I'm missing a function to figure out how much rope goes around the pulley as I believe this will be more than a quarter circle if the middle mass weighs a lot more than the outer masses. But I'm not sure how to get started here.

As for c, I want to make sure I got b correct first, and I think it'll solve itself without being too difficult as the angle the system is at rest is only dependent on the masses and can also be found with arcsin(m2 / 2m1). Then I would just follow the process in b).

Note: Where the rope touches the pulley at point ##P## what is the relationship for the angle between that rope and a line joining ##P## to ##O## (##O## is the center of the pulley)? Try to redraw your diagram to represent this better so it becomes clearer to you.

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Tlurk and topsquark
I think you are making your life unnecessarily complicated. As @erobz remarked, your third drawing of the pulley detail is drawn incorrectly. For a pulley of finite size, the string will partially wrap around the pulley. Just imagine what the drawing would look like if the radius of each pulley were equal to the separation between the pulley centers!

I would use negligible zero radius for the two pulleys redraw the FBD and calculate the angle between the strings connected to the middle mass. Then I would use the given horizontal separation between masses and the angle to find the vertical position of the middle mass below the center-to-center separation of the pulleys.

Do yourself a favor and use symbols ##m_1## for the side masses, ##m_2## for the middle mass and ##L## for the separation between pulleys. In part (b) you can set ##m_1=m_2## and see what you get for the vertical position.

In part (c) you should check what happens to the angle when you change the ratio ##m_2/m_1## from zero to very large. If you have, say, 1 kg masses on the sides and the middle mass is that of a feather, the system will be at equilibrium and the angle between strings very close to 180°. However, if the middle mass is that of a battleship, the system cannot be at equilibrium. I would find at what ratio of masses equilibrium is untenable. Hint: somewhere between "feather" and "battleship".

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Tlurk, topsquark, Lnewqban and 2 others
Hi @Tlurk

You apparently haven’t given the original question in full, so we are guessing if your answers are appropriate.

As noted by @kuruman, your working could be a lot simpler if you used symbols. (Also, this minimises potential arithmetic and rounding errors.)

If this is not clear (and since your answer/working for part a) is correct) let me explain:

To check if the system can be at rest we need to show that there is some configuration where the net force on each mass is zero.

If such a configuration exists:
##T = m_1g## (forces on an outer mass)
##2T \sin \theta = m_2g## (vertical forces on the centre mass)

It is then trivial (with ##m_1=m_2##) to show that ##\theta = 30^o## satisfies the equations. (In addition, it is useful to know that this answer is valid for any 3 equal masses and is independent of g.)

Tlurk and topsquark
So I redid a) using symbols, I'm not sure why I didn't do that in the first place so thanks for reminding me:

This matches my previous answer, but I can now change the masses without doing it over again.

For part b) I added a length of the rope R, made the Y-axis go through the centre of the left pulley, and the X-axis go through the centre of both pulleys. Now the length of the rope in total on the left half of the system is $$\frac{1}{2}R$$ and I can use the imaginary triangle drawn in the system overview in the figure above to find the length of the hypotenuse c:
$$\sin{\beta} = \frac{\frac{1}{2}L}{c}$$
$$c = \frac{L}{2\sin{\beta}}$$
The length of the rope from the left pulley (in other words the left pulley's Y-coordinate) is then given by:
$$\frac{1}{2}R - c$$ which can be written as $$\frac{1}{2}R - \frac{L}{2\sin{\beta}}$$ We find the angle beta or 1/2 alfa from a), so we don't have any unknowns variables here. R and L is of course chosen by us.

Now to find the height of m2 I can use that $$\tan(\beta) = \frac{\frac{1}{2}L}{h}$$
$$h = \frac{L}{2\tan{\beta}}$$
And we know h will be the negative y-coordinate for m2.

I think I got this down with a negligible radius at least, but I have to look some more into solving it with a larger radius for the pulleys, thanks a lot for the help so far!
As for erobz' answer, I realize the rope will be perpendicular to the line from P to O which gives me yet another right-angle triangle to work with. I have to think some more about this one.

erobz
Tlurk said:
So I redid a) using symbols, I'm not sure why I didn't do that in the first place so thanks for reminding me:
View attachment 314767
This matches my previous answer, but I can now change the masses without doing it over again.

For part b) I added a length of the rope R, made the Y-axis go through the centre of the left pulley, and the X-axis go through the centre of both pulleys. Now the length of the rope in total on the left half of the system is $$\frac{1}{2}R$$ and I can use the imaginary triangle drawn in the system overview in the figure above to find the length of the hypotenuse c:
$$\sin{\beta} = \frac{\frac{1}{2}L}{c}$$
$$c = \frac{L}{2\sin{\beta}}$$
The length of the rope from the left pulley (in other words the left pulley's Y-coordinate) is then given by:
$$\frac{1}{2}R - c$$ which can be written as $$\frac{1}{2}R - \frac{L}{2\sin{\beta}}$$ We find the angle beta or 1/2 alfa from a), so we don't have any unknowns variables here. R and L is of course chosen by us.

Now to find the height of m2 I can use that $$\tan(\beta) = \frac{\frac{1}{2}L}{h}$$
$$h = \frac{L}{2\tan{\beta}}$$
And we know h will be the negative y-coordinate for m2.

I think I got this down with a negligible radius at least, but I have to look some more into solving it with a larger radius for the pulleys, thanks a lot for the help so far!
As for erobz' answer, I realize the rope will be perpendicular to the line from P to O which gives me yet another right-angle triangle to work with. I have to think some more about this one.
Why did your pulleys disappear in part (b)? The intermediate objective is to determine the angle of wrap around the pulley (you kind of need them represented in the analysis), then use it to find the length of the rope ( call it ##l_2##) from the point of tangency on the pulley to ##m_2##. The length ##l_2## will be a function of ##L, r, \theta ## ( or ##\beta## - the angle ##\theta## I'm referring to is what you have as ##90-\beta## ). Then there will be more constraints that apply after you find that.

EDIT: I didn't catch that you were just testing the water with an approximation there. Good idea. Now it's on to the main event!

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Tlurk
erobz said:
Why did your pulleys disappear in part (b)? The intermediate objective is to determine the angle of wrap around the pulley (you kind of need them represented in the analysis), then use it to find the length of the rope ( call it ##l_2##) from the point of tangency on the pulley to ##m_2##. The length ##l_2## will be a function of ##L, r, \theta ## ( or ##\beta## - the angle ##\theta## I'm referring to is what you have as ##90-\beta## ). Then there will be more constraints that apply after you find that.

EDIT: I didn't catch that you were just testing the water with an approximation there. Good idea. Now it's on to the main event!
Based on the statement of the problem, it seems to me that one needs not worry about the radii of the pulleys because they are not given or assumed known . That said, it is clear that the statement of the problem is @Tlurk's interpretation of what the problem gives and what it asks. No physics problem begins with "I have a system with two pulleys and ##\dots~##"

To @Tlurk : Can you post the statement of the problem exactly as given to you?

Tlurk and Steve4Physics
Tlurk said:
And we know h will be the negative y-coordinate for ##m_2##.
How does knowing ##h## help you figure out the threshold value of ##m_2## above which there can be no equilibrium?

kuruman said:
Based on the statement of the problem, it seems to me that one needs not worry about the radii of the pulleys because they are not given or assumed known . That said, it is clear that the statement of the problem is @Tlurk's interpretation of what the problem gives and what it asks. No physics problem begins with "I have a system with two pulleys and ##\dots~##"

To @Tlurk : Can you post the statement of the problem exactly as given to you?
Yeah, I think the problem statement is missing some relevant information. If you notice they say in the analysis they use ## r = 0.2 \, \rm{m}## and ##s= 3 \, \rm{m}##
Tlurk said:
b) I used the radius of pulley r=0,20m, the rope length s=3,0m and the length between pulleys (centre to centre) l=0,80m. Again, the system is symmetric so I will look at the left side. I will also use the connection points between the rope and the masses as their position. A simple diagram:
I assumed that it is just relevant information left out of the problem statement. ( in order to get numerical coordinates of the masses in (b) you need that information )

kuruman said:
To @Tlurk : Can you post the statement of the problem exactly as given to you?
The problem is not in English, but translated it reads:
The figure shows two pulleys and three masses. You choose the value for the mass m, the starting position for the masses and the distance between the pulleys. Disregard the friction in the pulleys and the air resistance. Is it possible to make all the masses stay in place (system in equilibrium)? If so, what is the position of each mass? What happens if you change the weight of the middle mass? Draw a FBD for every mass and use Newton's laws to analyze the problem.

That is absolutely everything that's given.
erobz said:
Yeah, I think the problem statement is missing some relevant information. If you notice they say in the analysis they use ## r = 0.2 \, \rm{m}## and ##s= 3 \, \rm{m}##

I assumed that it is just relevant information left out of the problem statement. ( in order to get numerical coordinates of the masses in (b) you need that information )
The values I used were just something I came up with to solve it numerically. But as I quickly discovered it was better to use symbols instead. I believe I have to take the pulley radius into consideration because the position of the outer masses is dependent on it, right? Anyways, now I've started working on finding the length around the pulley, so I will at least give it a try even if it's not needed.

Lnewqban
So, it seems to me you can only answer (b) algebraically then (there are no values for anything except ##\theta## from part (a) - just parameters). The problem statement has left out the radius of the pulley or the length of the rope.

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Welcome, @Tlurk !

The question that makes absolutely no sense to me is :
" If so, what is the position of each mass?".

The spatial location of masses ##m_1## can be anywhere vertically.
The location of mass ##m_2## only depends on the distance between pulleys, as the value of the symmetrical angle of the rope is unique for each combination of ##2m_1/m_2##.

Thank you for posting the problem as given. The problem specifically says that you can choose the mass so choose ##m##, the center-to-center separation so choose ##L## and the initial position of the masses. It seems to me that this means "choose the entire length of the rope". Any number greater than ##L## will do. But once you specify that, you will be expected to find the distance of each mass from the ceiling in terms of ##L##, ##m_1## and ##m_2##. I would start the rope horizontal with the side masses hanging symmetrically and then put the middle mass in place and look at the vertical displacements.

The problem says nothing about choosing the radius of a pulley which to me means assume that it is negligible. Then the horizontal coordinate of each mass will be fixed at (##-\frac{L}{2}##), (##0##) and (##+\frac{L}{2}##).

An interesting variation would be to make the three masses unequal.

Lnewqban said:
The spatial location of masses m1 can be anywhere vertically.
The full statement of the problem gives the choice of initial position. See post #13 about figuring out displacements in an "experiment" where you start with no mass in the middle and then hooking it on.

erobz said:
So, it seems to me you can only answer (b) algebraically then (there are no values for anything - just parameters). The problem statement has left out the radius of the pulley or the length of the rope.
I do believe that we're supposed to test a few different values to see how the system moves, eventually graphing the position, velocity and acceleration of each mass as that's what we've been doing to simpler systems so far. But as it's been quite a while since the last time I've had any physics subjects so I decided it would be easier to start in equilibrium (which I also feel the first question is leading us towards). Please do tell me if you think I should have a different approach.

I will try to solve it algebraically as you said. I think I need to if I want to import some of the functions I find underways to graph the movement as well. But I'm getting ahead of myself now and need to understand this on paper first.

Lnewqban said:
The question that makes absolutely no sense to me is :
" If so, what is the position of each mass?".

The spatial location of masses ##m_1## can be anywhere vertically.
The location of mass ##m_2## only depends on the distance between pulleys, as the value of the symmetrical angle of the rope is unique for each combination of ##2m_1/m_2##.
Yes, that's what I thought as well. The way the problem is given is why I've tried to introduce a rope length and pulley radius myself, to figure out the exact position of the masses.

kuruman said:
The problem says nothing about choosing the radius of a pulley which to me means assume that it is negligible.
That's probably the case.

Tlurk said:
##\dots~## velocity and acceleration of each mass ##\dots~##
What velocity and acceleration? Is this system not at equilibrium?

erobz
Tlurk said:
I do believe that we're supposed to test a few different values to see how the system moves, eventually graphing the position, velocity and acceleration of each mass as that's what we've been doing to simpler systems so far. But as it's been quite a while since the last time I've had any physics subjects so I decided it would be easier to start in equilibrium (which I also feel the first question is leading us towards). Please do tell me if you think I should have a different approach.

I will try to solve it algebraically as you said. I think I need to if I want to import some of the functions I find underways to graph the movement as well. But I'm getting ahead of myself now and need to understand this on paper first.
I expect you mean how the equilibrium position moves with different masses, as @kuruman is properly concerned about the interpretation here.

You can parameterize the radius too, It's just little extra trigonometric gymnastics.

kuruman said:
What velocity and acceleration? Is this system not at equilibrium?
Well, yes, if the initial position of the middle mass makes the angle $$\beta = \arccos{\frac{m_2}{2m_1}}$$ but not if the initial position makes a different angle with the same masses. But as I said I'm getting to far ahead of myself now, and need to understand this fully at equilibrium first, as originally asked.
kuruman said:
How does knowing ##h## help you figure out the threshold value of ##m_2## above which there can be no equilibrium?
I'm not really sure what you mean here. I guess ##m_2 < 2m_1## has to be true to make the $$arccos{\frac{m_2}{2m_1}}$$ valid which h depends on.

kuruman said:
What velocity and acceleration? Is this system not at equilibrium?
Just a little scope creep!

Tlurk said:
...
Yes, that's what I thought as well. The way the problem is given is why I've tried to introduce a rope length and pulley radius myself, to figure out the exact position of the masses.
Consider that in problems like this, including non-moving ideal pulleys (no mass, no friction, no translation), their function is limited to modify the direction in which the force is transferred by the rope or string.

You could simplify this particular problem to a summation of vectors representing all acting forces and their lines of action (respect to th ecentral body or mass), which reach a natural state of equilibrium.

Note that the vertical component of each of the two non-vertical actual force or rope tension must be of a magnitude equal to half the actual vertical force (weight of mass 2).

And, as suggested by @kuruman , the asymmetrical case:

[Edit: Incorrect diagram replaced with shown one]

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erobz
Lnewqban said:
And, as suggested by @kuruman , the asymmetrical case:

View attachment 314786
I think you mean m3<m1 for β>α, m2<m1+m3.
I can’t see that β>60° is of special interest.

haruspex said:
I think you mean m3<m1 for β>α, m2<m1+m3.
I can’t see that β>60° is of special interest.
Thank you, @haruspex
The posted diagram was incorrect.
I am editing it.

1. What is a system of two pulleys and three masses?

A system of two pulleys and three masses is a mechanical setup that involves two pulleys connected by a rope or belt, with three masses attached to the rope. The masses can be different weights and are used to demonstrate principles of force and motion.

2. How does a system of two pulleys and three masses work?

In this system, one end of the rope is attached to a fixed point, while the other end is threaded through the two pulleys and attached to the three masses. As one mass moves down, the other two masses will move up, demonstrating the transfer of force through the rope and pulleys.

3. What is the purpose of using a system of two pulleys and three masses?

This system is commonly used in physics experiments to demonstrate concepts such as Newton's laws of motion, mechanical advantage, and conservation of energy. It allows for a simple and visual representation of these principles.

4. How do the masses affect the movement of the system?

The masses attached to the rope will affect the movement of the system by exerting different amounts of force. Heavier masses will require more force to move, while lighter masses will require less force. This demonstrates the relationship between mass and force.

5. Can a system of two pulleys and three masses be used for practical applications?

Yes, this system is commonly used in everyday objects such as elevators and cranes. By using multiple pulleys and masses, it allows for the lifting of heavy objects with less force, making it more efficient and practical.

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