Double pulley Atwood machine (with 3 masses)

In summary: No.It might help to think about the three heights from the ground. Let these be y1, y2 and y. If the length of the string is s, can you write a relationship between y1, y2, y and...There is no relationship between y1, y2, y and a.
  • #1
Sho Kano
372
3

Homework Statement


EC21.png

I need help with number 4.
Also I am altogether unsure of this problem... if you catch any mistakes, point them out! :)

Homework Equations


F = ma

Here are the variables I used in my attempt at the solution
mx = m subscript 1, 2, or 3
±a1 = acceleration of either mass on smaller pulley
T1 = tension of string on smaller pully
±a2 = acceleration of either mass (one is a pulley) on larger pully
T2 = tension of string on larger pully

The Attempt at a Solution


[/B]
(a)

m2*g - T1 = m2*a1 ---> from the free body diagram of mass 2
T1 - m1*g = m1*a1 ---> from the free body diagram of mass 1
m3*g - T2 = m3*a2 ---> from the free body diagram of mass 3
T2 - m1*g - m2*g = (m1+m2)*a2 ---> from the free body diagram of small pulley

substituting T1 from the second equation above in the first equation
m2*g - m2*a1 = m1*a1 + m1*g
a1 = (m2*g - m1*g) / (m1+m2)
so if downward is the positive direction,
mass 1's acceleration = -a1
mass 2's acceleration = +a2

substituting T2 for the last two equations,
m3*g - m3*a2 = (m2+m1)*a2 + m1*g + m2*g
a2 = (m3*g - m1*g - m2*g) / (m3+m2+m1)
mass 3's acceleration is +a2

(b)

a1 = 0 now, because the numerator of a1 would end out equal to zero.
This means the smaller pulley's masses would not experience any acceleration

(c)

since we want a2 to be zero, the numerator of a2 must equal 0, so
m3 = m1 + m2
also,
m2>m1

(d)

I'm confused, the expression won't change for a1? It will still be ±a1 for either masses.
 
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  • #2
For part 4, you can simplify the problem - since m3 isn't moving you can forget about it. The problem then is what is the acceleration of two masses on a pulley?
Say you have m1 and m2, what is the net force? and then the acceleration?
 
  • #3
elegysix said:
For part 4, you can simplify the problem - since m3 isn't moving you can forget about it. The problem then is what is the acceleration of two masses on a pulley?
Say you have m1 and m2, what is the net force? and then the acceleration?
Fnet = m2*g - m1*g
Set Fnet = ma ---> total mass m = m1+m2
So, a = m2g - m1g / m2+m1
Which is the same answer I got for the original case.
 
  • #4
Sho Kano said:
±a1 = acceleration of either mass on smaller pulley
By doing that, you are making an incorrect assumption. It would be ok if you change that to be accelerations relative to the small pulley, but then your equation needs to change.
 
  • #5
haruspex said:
By doing that, you are making an incorrect assumption. It would be ok if you change that to be accelerations relative to the small pulley, but then your equation needs to change.
Do you mean that the pulley is the one accelerating, thus there is only one acceleration? Isn't m1 accelerating in the upwards direction, and m2 accelerating in the downwards direction?
 
  • #6
Sho Kano said:
Do you mean that the pulley is the one accelerating, thus there is only one acceleration? Isn't m1 accelerating in the upwards direction, and m2 accelerating in the downwards direction?
The pulley is accelerating, and m1 and m2 are also accelerating relative to the pulley.
Let the pulley's acceleration be a, and the absolute accelerations of the masses be a1, a2, taking up as positive for each. What is the algebraic relationship between the three accelerations?
 
  • #7
haruspex said:
The pulley is accelerating, and m1 and m2 are also accelerating relative to the pulley.
Let the pulley's acceleration be a, and the absolute accelerations of the masses be a1, a2, taking up as positive for each. What is the algebraic relationship between the three accelerations?
The magnitude of a, a1, a2 will all be the same.
I'm not quite sure what you mean by the algebraic relationship.
 
  • #8
Sho Kano said:
The magnitude of a, a1, a2 will all be the same.
Not so. Let's start with a simplification. Suppose the pulley does not accelerate. What is the algebraic relationship between a1and a2 now?
 
  • #9
Ok, a1 = -a2 or the other way around
I think the tangential acceleration of the pulley should be equal to a1 or a2
 
  • #10
Sho Kano said:
Ok, a1 = -a2 or the other way around
I think the tangential acceleration of the pulley should be equal to a1 or a2
Yes.
Now allow the pulley a vertical acceleration a. What is the relationship between a1, a2 and a?
 
  • #11
haruspex said:
Yes.
Now allow the pulley a vertical acceleration a. What is the relationship between a1, a2 and a?
-a1 = a2 = a ?
 
  • #12
Sho Kano said:
-a1 = a2 = a ?
No.
It might help to think about the three heights from the ground. Let these be y1, y2 and y. If the length of the string is s, can you write a relationship between y1, y2, y and s?
 
  • #13
Haha, I think I might be onto something.
I overlooked the free body diagram of the smaller pulley itself.

It would look something like this right?
T2 - 2*T1 = 0

Meaning haruspex, the acceleration of the pulley is the negative of the acceleration of m3?
 
  • #14
Sho Kano said:
Haha, I think I might be onto something.
I overlooked the free body diagram of the smaller pulley itself.

It would look something like this right?
T2 - 2*T1 = 0
Since the pulley has no mass, that is not going to help in finding the acceleration of the pulley. The relationship between the three accelerations is a purely kinematic one, nothing to do with masses or forces. Please try to answer my question in post #12.
 
  • #15
haruspex said:
Since the pulley has no mass, that is not going to help in finding the acceleration of the pulley. The relationship between the three accelerations is a purely kinematic one, nothing to do with masses or forces. Please try to answer my question in post #12.
Sorry, I edited my post, the vertical acceleration of the pulley is equal to the negative of the acceleration of m3.
 
  • #16
Sho Kano said:
Sorry, I edited my post, the vertical acceleration of the pulley is equal to the negative of the acceleration of m3.
That is true, but it does not give us the relationship between that acceleration and the accelerations a1 and a2. The relationship we need results from the fact that the string connecting m1 and m2 has a fixed length.
 
  • #17
haruspex, I've tried to answer post 12, but it seems like a pretty hard question to answer (to me).
Here is an attempt,
For m1, it has both a1 and a2 contributing to its height.
y1 = y0 + y_due m2 accelerating + y_due to m3 accelerating
I can substitute by using kinematics equation y = 1/2at^2, not sure where to go from here.

I think that,
y1 = y0 + y2 + y3 ---> [y3 = y]
is that right?
 
  • #18
Sho Kano said:
haruspex, I've tried to answer post 12, but it seems like a pretty hard question to answer (to me).
Here is an attempt,
For m1, it has both a1 and a2 contributing to its height.
y1 = y0 + y_due m2 accelerating + y_due to m3 accelerating
I can substitute by using kinematics equation y = 1/2at^2, not sure where to go from here.

I think that,
y1 = y0 + y2 + y3 ---> [y3 = y]
is that right?
No, you're making it a hard question. Forget about accelerations for now. This is just a question about the relationship between four distances that comes straight from the geometry.
Treat the pulley as a point. If m1 is height y1 above the ground, m2 is height y2 above the ground, the pulley above them is height y above the ground, and a string of length s runs from m1 up to the pulley and back down to m2, there is a very simple algebraic relationship between those four distances. What is it?
 
  • #19
haruspex said:
Yes.
Now allow the pulley a vertical acceleration a. What is the relationship between a1, a2 and a?
I was confused on which acceleration of the pulley you meant (Tangential due to m2 or Vertical due to m3). This is what you mean right?

acceleration of m1 = a1+a
acceleration of m2 = -a1+a
With respect to the ground
 
  • #20
haruspex said:
No, you're making it a hard question. Forget about accelerations for now. This is just a question about the relationship between four distances that comes straight from the geometry.
Treat the pulley as a point. If m1 is height y1 above the ground, m2 is height y2 above the ground, the pulley above them is height y above the ground, and a string of length s runs from m1 up to the pulley and back down to m2, there is a very simple algebraic relationship between those four distances. What is it?
y1 = y2
y = y1 + 0.5*s
 
  • #21
Sho Kano said:
I was confused on which acceleration of the pulley you meant (Tangential due to m2 or Vertical due to m3). This is what you mean right?

acceleration of m1 = a1+a
acceleration of m2 = -a1+a
With respect to the ground
Well, I did specify in post #10 that I was asking about the vertical acceleration of the pulley.
Your equations just above are correct if you are now defining a1 as the upward acceleration of m1 relative to the pulley. But I think that is not how you originally defined it. If you think you did, then refer back to my post #4: if a1 is a relative acceleration then your Fnet=ma equations for m1 and m2 will need to change.
 
  • #22
Sho Kano said:
y1 = y2
y = y1 + 0.5*s
y1 need not equal y2, but other than that you are getting closer.
 
  • #23
haruspex said:
Well, I did specify in post #10 that I was asking about the vertical acceleration of the pulley.
Your equations just above are correct if you are now defining a1 as the upward acceleration of m1 relative to the pulley. But I think that is not how you originally defined it. If you think you did, then refer back to my post #4: if a1 is a relative acceleration then your Fnet=ma equations for m1 and m2 will need to change.
Oh do you mean that I specified downwards as the positive direction in the problem statement? My mistake.
So the equations would look like this?

acceleration of m1 = -a1 + -a
acceleration of m2 = a1 + -a
With respect to the ground
 
  • #24
haruspex said:
y1 need not equal y2, but other than that you are getting closer.
For the case which they are not equal
s = (y-y2) + (y-y1)
s = 2y - y2 - y1
 
  • #25
Sho Kano said:
For the case which they are not equal
s = (y-y2) + (y-y1)
s = 2y - y2 - y1
Right. Now, to get accelerations from distances, differentiate twice with respect to time. What equation does that give you for a, a1 and a2?
 
  • #26
haruspex said:
Right. Now, to get accelerations from distances, differentiate twice with respect to time. What equation does that give you for a, a1 and a2?
0 = 2a - a2 - a1 ?
 
  • #27
Sho Kano said:
0 = 2a - a2 - a1 ?
Yes!
 
  • #28
haruspex said:
Yes!
Interesting, does that mean that the acceleration of the pulley or m3 is not simply attained by solving a net force equation for acceleration?
 
  • #29
Sho Kano said:
Interesting, does that mean that the acceleration of the pulley or m3 is not simply attained by solving a net force equation for acceleration?
No, it just gives you a direct algebraic relationship between the three accelerations. You still need to consider the four free body diagrams with their respective Fnet=ma equations. That will give you five equations altogether, and you have five unknowns - three accelerations and two tensions.
 
  • #30
haruspex said:
No, it just gives you a direct algebraic relationship between the three accelerations. You still need to consider the four free body diagrams with their respective Fnet=ma equations. That will give you five equations altogether, and you have five unknowns - three accelerations and two tensions.
a2 = (4*m1*m2*g + m2*m3*g - 3*m1*m3*g) / (4*m1*m2 + m2*m3 - m1*m3)
Any chance you know this is the acceleration for m2 off the top of your head?
 
  • #31
Sho Kano said:
a2 = (4*m1*m2*g + m2*m3*g - 3*m1*m3*g) / (4*m1*m2 + m2*m3 - m1*m3)
Any chance you know this is the acceleration for m2 off the top of your head?
That is not what I get. The negative term in the denominator looks wrong. With m2 sufficiently small, your denominator could go to zero, giving an infinite acceleration.
(It is useful to get into the habit of thinking up sanity checks like that.)
Please post your working.
 
  • #32
haruspex said:
That is not what I get. The negative term in the denominator looks wrong. With m2 sufficiently small, your denominator could go to zero, giving an infinite acceleration.
(It is useful to get into the habit of thinking up sanity checks like that.)
Please post your working.
4 Equations.JPG
Equation 1.JPG
Equation 2.JPG
Combined Equation.JPG

Read from left to right :) Seems like I got a different answer, must have made some mistakes the first time around.
 
  • #33
Sho Kano said:
View attachment 98395 View attachment 98396 View attachment 98397 View attachment 98398
Read from left to right :) Seems like I got a different answer, must have made some mistakes the first time around.
As I posted, it does not make sense to have a denominator that could be zero with some combination of masses.
You need to be consistent with signs. The algebraic relationship we found between the three accelerations was using the convention that up is positive. If we stick with that convention, in the F=ma equation for each mass we should have the tension as positive and gravitation as negative, so they should all be of the form miai=Ti-mig.

Oh, and please do not post handwritten working as images. It is almost always very difficult to read. Please take the trouble to type it in, preferably using superscripts and subscripts as appropriate.
 
  • #34
haruspex said:
As I posted, it does not make sense to have a denominator that could be zero with some combination of masses.
You need to be consistent with signs. The algebraic relationship we found between the three accelerations was using the convention that up is positive. If we stick with that convention, in the F=ma equation for each mass we should have the tension as positive and gravitation as negative, so they should all be of the form miai=Ti-mig.

Oh, and please do not post handwritten working as images. It is almost always very difficult to read. Please take the trouble to type it in, preferably using superscripts and subscripts as appropriate.
Sorry for the wait, I'll get back to this problem tomorrow.
 
  • #35
haruspex said:
As I posted, it does not make sense to have a denominator that could be zero with some combination of masses.
You need to be consistent with signs. The algebraic relationship we found between the three accelerations was using the convention that up is positive. If we stick with that convention, in the F=ma equation for each mass we should have the tension as positive and gravitation as negative, so they should all be of the form miai=Ti-mig.

Oh, and please do not post handwritten working as images. It is almost always very difficult to read. Please take the trouble to type it in, preferably using superscripts and subscripts as appropriate.
Using my convention, the relationship that I got is now

a = -a1-a2 / 2 ; So I can use this without changing my equations right?
 

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