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

In summary: Also, I screwed up in post #11, when I corrected the equations you wrote in Post #1. I should have just added a minus sign to your ars. So, the equations in #1 are correct as you wrote them. But, what is the sign for ar? The basic idea to solve this problem is to realize that the actual pull on m1 is the same as if the two pulleys were replaced by a single mass ## m_{eff}##. So, if you could find ## m_{eff} ##, then you would have written the correct equation for ## a_1 ## in Post #1. So, our strategy is to find the correct ## m
  • #1
Crystal037
167
7
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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  • #2
Crystal037 said:
a2=−a1+ar
Your diagram indicates you are taking a1and a2 as positive down, so why -a1?
 
  • #3
What if ## m_3 \approx 0 ##? Clearly the answer of ## m_{eff}=m_2+m_3 ## is incorrect.
 
  • #4
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)
 
  • #5
haruspex said:
Your diagram indicates you are taking a1and a2 as positive down, so why -a1?
a2=a1-ar a3=a1+ar
 
  • #6
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).
 
  • #7
Crystal037 said:
a2=a1-ar a3=a1+ar
Maybe - delends which way you are taking as positive for ar.
 
  • #8
haruspex said:
Maybe - delends which way you are taking as positive for ar.
a2=ar-a1 a3=ar+a1
 
  • #9
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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  • #26

1. How is effective mass calculated?

Effective mass is calculated by taking into account the mass of the object and the forces acting on it. It is a measure of the object's inertia and is often used in systems with pulleys and ropes. To find the effective mass, you must consider the mass of the object and the forces acting on it, such as tension and gravity.

2. How does effective mass affect acceleration?

The effective mass of an object affects its acceleration by determining how much inertia the object has. In simpler terms, the larger the effective mass, the more difficult it is to accelerate the object. This is why objects with larger effective mass require more force to move.

3. Can effective mass be negative?

No, effective mass cannot be negative. It is a measure of an object's inertia and must always be a positive value. Negative effective mass would not make physical sense and would not accurately represent the object's behavior in a system.

4. How is effective mass used in finding accelerations of 3 masses connected by pulleys?

In a system with 3 masses connected by pulleys, the effective mass is used to determine the overall inertia of the system. By considering the effective mass and the forces acting on each object, the acceleration of the system can be calculated using Newton's second law of motion (F=ma).

5. Can effective mass change in a system with pulleys?

Yes, effective mass can change in a system with pulleys if the forces acting on the objects change. For example, if the tension in a rope changes, the effective mass of the object connected to that rope will also change. This is why it is important to consider all forces and their effects on the effective mass when calculating accelerations in a system with pulleys.

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