Using Effective Mass to find the accelerations of 3 masses connected by pulleys

AI Thread Summary
The discussion revolves around calculating the effective mass in a system of three masses connected by pulleys, where the effective mass is crucial for understanding the dynamics of the system. Participants debate the correct formulation of the effective mass, with a consensus emerging that it should be calculated as m(eff) = 4m2m3/(m2+m3) rather than simply m2 + m3. The relationship between the accelerations of the masses is established through a series of equations, emphasizing the importance of defining the direction of acceleration correctly. The conversation highlights the need to solve for the acceleration a1 first to determine the effective mass accurately. Ultimately, the effective mass reflects the combined influence of the masses on the system's acceleration, rather than a straightforward sum.
Crystal037
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Homework Statement
Three blocks of masses m1, m2 and m3 are connected as shown in the figure. All the surfaces are frictionless and the string and pulleys are light. Find the acceleration of the block of mass m1.
Relevant Equations
2T=m1a1
m3g-T=m3a3
T-m2g=m2a2
a2+a3=2a1
IMG_20200506_051517.jpg

Since 2nd pulley accelerates downward with the same acceleration of m1, and because the string around pulley has a constant length, it must be that a2=−a1+ar and a3=−a1−ar, where ar is the relative acceleration between the pulley and m2.
From the above 4 equations, we are supposed to determine the effective mass pulling on m1 by comparing to the case where the 2nd pulley system is replaced with a single hanging mass. In my opinion the effective mass should be m2+m3 but it is given as
m(eff)= 4m2m3/(m2+m3)

I know that
m(eff)*g - 2T=m(eff)a1
[m(eff) + m1]a1 = m(eff) *g
But how to find m(eff)
 
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Crystal037 said:
a2=−a1+ar
Your diagram indicates you are taking a1and a2 as positive down, so why -a1?
 
What if ## m_3 \approx 0 ##? Clearly the answer of ## m_{eff}=m_2+m_3 ## is incorrect.
 
Charles Link said:
What if ## m_3 \approx 0 ##? Clearly the answer of ## m_{eff}=m_2+m_3 ## is incorrect.
Then how is it 4*m2*m3/(m2+m3)
 
haruspex said:
Your diagram indicates you are taking a1and a2 as positive down, so why -a1?
a2=a1-ar a3=a1+ar
 
Crystal037 said:
Relevant Equations::
2T=m1a1
m3g-T=m3a3
T-m2g=m2a2
a2+a3=2a1

From the above 4 equations, we are supposed to determine the effective mass pulling on m1 by comparing to the case where the 2nd pulley system is replaced with a single hanging mass.

I know that
m(eff)*g - 2T=m(eff)a1
[m(eff) + m1]a1 = m(eff) *g
But how to find m(eff)
Solve the system of four equations for m1 and then use the last two equations to determine m(eff).
 
Crystal037 said:
a2=a1-ar a3=a1+ar
Maybe - delends which way you are taking as positive for ar.
 
haruspex said:
Maybe - delends which way you are taking as positive for ar.
a2=ar-a1 a3=ar+a1
 
ehild said:
Solve the system of four equations for m1 and then use the last two equations to determine m(eff).
What do you mean by solve for m1, the mass is already given and we'll find a1 using m(eff)
 
  • #10
Crystal037 said:
a2=ar-a1 a3=ar+a1
No, what you had in post #5 was probably fine, but you need to define which way you are taking as positive for ar, up or down?
 
  • #11
Crystal037 said:
What do you mean by solve for m1, the mass is already given and we'll find a1 using m(eff)
Sorry , I meant solving for a1.
 
  • #12
ehild said:
Sorry , I meant solving for a1.
But how will I find m(eff)
 
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  • #13
haruspex said:
No, what you had in post #5 was probably fine, but you need to define which way you are taking as positive for ar, up or down?
But how will I use those to find m(eff)
 
  • #14
Charles Link said:
What if ## m_3 \approx 0 ##? Clearly the answer of ## m_{eff}=m_2+m_3 ## is incorrect.
Then the force will be m2g why it isn't correct?
 
  • #15
If you let ##m_3 \rightarrow 0 ##, then ## m_2 ## will accelerate with value ## g ## downward, (pulling the almost massless ##m_3 ## upward), and ##m_2 ## will not pull downward on the system.
 
  • #16
Then how should I approach to find the m(eff)
 
  • #17
Find the ## m_{eff} ## that when attached to the system, it causes an acceleration of ## a_1 ##. Have you computed ## a_1 ## yet? If you attach ## m_{eff} ## to the system, then ## a_1=(m_{eff}g)/(m_1+m_{eff} )##. Set that equal to your ##a_1##.
 
  • #18
One correction for you: In your set of equations in the OP, the third one should read ##m_2g-T=m_2 a_2 ##. If you solve those for ##a_1 ##, you should get a correct answer. I solved them, and it did lead to a correct ## m_{eff} ##.## \\ ## I'd be happy to check your expression for ## a_1 ##, if necessary. ## \\ ## The 4th equation is actually the trickiest of the bunch, and you have that one correct.## \\ ## Edit: One other item, is that for your ## a_r ## equations, you need a plus sign on both ## a_1 ##'s, but you seemed to correct that by writing the 4th equation correctly.
 
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  • #19
Crystal037 said:
But how will I find m(eff)
Have you found a1? If yes, you get m(eff) from your equation [m(eff) + m1]a1 = m(eff) *g in Post #1
 
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  • #20
Crystal037 said:
But how will I use those to find m(eff)
By solving the actual set-up to find T then using that in the equations you wrote in post #1.
But I ask again, which way are you defining as positive for ar?

Btw, it is worth explaining why the effective mass is not m2+m3. The common mass centre of m2 and m3 will accelerate downwards, so it exerts less pull on the upper cord than ( m2+m3)g.
 
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  • #21
@Crystal037 You posted a very interesting problem, and you were very close to having solved it. We would enjoy your feedback=did you get the correct solution?
 
  • #22
ehild said:
Have you found a1? If yes, you get m(eff) from your equation [m(eff) + m1]a1 = m(eff) *g in Post #1
But I thought that we should first get m(eff) and then use the m(eff) to find a1
 
  • #23
haruspex said:
By solving the actual set-up to find T then using that in the equations you wrote in post #1.
But I ask again, which way are you defining as positive for ar?

Btw, it is worth explaining why the effective mass is not m2+m3. The common mass centre of m2 and m3 will accelerate downwards, so it exerts less pull on the upper cord than ( m2+m3)g.
I am defining the ar as vertically downwards
 
  • #24
Crystal037 said:
I am defining the ar as vertically downwards
Then what you had in post #5 is still wrong.
Suppose you were to hold m1 still so that a1 is zero, but ar is positive. Would a2 be positive or negative?
What if ar is zero? What would be the relationship between a1 and a2 then?
 
  • #25
Crystal037 said:
But I thought that we should first get m(eff) and then use the m(eff) to find a1
You have enough number of equations to get all accelerations, tensions, and even an effective mass. You decide, in what order you calculate them. You can eliminate all unknowns but the effective mass if yo like.
 
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