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

[Crystal037]

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

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)

 

[haruspex]

a2=−a1+ar

Your diagram indicates you are taking a1and a2 as positive down, so why -a1?

 

[Charles Link]

What if $m_3 \approx 0$? Clearly the answer of $m_{eff}=m_2+m_3$ is incorrect.

 

[Crystal037]

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)

 

[Crystal037]

Your diagram indicates you are taking a1and a2 as positive down, so why -a1?

a2=a1-ar a3=a1+ar

 

[ehild]

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).

 

[haruspex]

a2=a1-ar a3=a1+ar

Maybe - delends which way you are taking as positive for ar.

 

[Crystal037]

Maybe - delends which way you are taking as positive for ar.

a2=ar-a1 a3=ar+a1

 

[Crystal037]

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)

 

[haruspex]

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?

 

[ehild]

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.

 

[Crystal037]

Sorry , I meant solving for a1.

But how will I find m(eff)

 

[Crystal037]

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)

 

[Crystal037]

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?

 

[Charles Link]

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.

 

[Crystal037]

Then how should I approach to find the m(eff)

 

[Charles Link]

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 $a_1$.

 

[Charles Link]

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.

 

[ehild]

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

 

[haruspex]

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.

 

[Charles Link]

@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?

 

[Crystal037]

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

 

[Crystal037]

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

 

[haruspex]

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?

 

[ehild]

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.

 

[Charles Link]

@Crystal037 Please see post 18.