Pulley attached to a pulley - Find the balance equation

[AHashemi]

Homework Statement

A string is hung over a fixed pulley, and a mass of m3 is suspended from one end of the string. The other end of the string supports a massless pulley, over which a second string is hung. This string has a m2 mass attached to one end, and a m1 mass attached to the other end.
Find an equation between m1, m2 and m3 that makes the m3 still.

Homework Equations

F=ma

The Attempt at a Solution

I took the upward as positive direction of y. and wrote these equations based on F=ma for each mass:
m3: $$T_1 - m_3g=m_3 a_3$$
m2: $$T_2 - m_2g=m_2 a_2$$
m1: $$T_2 - m_1g=m_1 a_1$$

but I can't find out how to make m3 still.

 

[Chestermiller]

If m3 has to be still, what can you say about m3's acceleration? What does this mean with regard to the downward acceleration of the lower pulley? How are the acclerations of m1 and m2 related?

Chet

 

[AHashemi]

If m3 has to be still, what can you say about m3's acceleration? What does this mean with regard to the downward acceleration of the lower pulley? How are the acclerations of m1 and m2 related?

Chet

emm... well a3 has to be equal to 0. by this we can say $$T_1=m_3g$$ and we know $$T_1=2T_2$$ so $$2T_2=m_3g$$ but I have no idea about how to find left pulley's acceleration which is necessary to find T1.
I've spend too much time on this. I'm sure I'm missing something.

 

[Chestermiller]

emm... well a3 has to be equal to 0. by this we can say $$T_1=m_3g$$ and we know $$T_1=2T_2$$ so $$2T_2=m_3g$$ but I have no idea about how to find left pulley's acceleration which is necessary to find T1.
I've spend too much time on this. but I'm sure I'm missing something.

If m3 is not accelerating, then the lower pulley, which is joined to m3 by an inextensible string is not accelerating either. So that pulley has to be stationary.

Chet

 

[AHashemi]

If m3 is not accelerating, then the lower pulley, which is joined to m3 by an inextensible string is not accelerating either. So that pulley has to be stationary.

Chet

Oh right.. that was a mistake.
But I can't understand how m1 and m2 are related to m3's acceleration. their net weight just has to be equal to m3. do their acceleration matter at all?

 

[Chestermiller]

Oh right.. that was a mistake.
But I can't understand how m1 and m2 are related to m3's acceleration. their net weight just has to be equal to m3. do their acceleration matter at all?

Sure. Their accelerations affect the tension (see your own equations). If the pulley is stationary, how is the upward acceleration of m1 related to the downward acceleration of m2?

Chet

 

[AHashemi]

Sure. Their accelerations affect the tension (see your own equations). If the pulley is stationary, how is the upward acceleration of m1 related to the downward acceleration of m2?
Chet

$$a_1=g(m_2-m_1)/(m_1+m_2)=-a_2$$
they have equal magnitude but opposite directions.

 

[Chestermiller]

So, based on this, what is T2 equal to?

 

[AHashemi]

So, based on this, what is T2 equal to?

$$T_2=m_1g((m_2-m_1)/(m_2+m_1)+1)$$
so
$$T_1=2m_1g((m_2-m_1)/(m_2+m_1)+1)$$
and because we need T1 to be equal to m3g we can say:
$$m_3g=2m_1g((m_2-m_1)/(m_2+m_1)+1)$$
which is:
$$m_3=2m_1((m_2-m_1)/(m_2+m_1)+1)$$
Oh! that's right! also m1=m2=m results m3=2m fits into this!
Thanks!

 

[Chestermiller]

$$T_2=m_1g((m_2-m_1)/(m_2+m_1)+1)$$
so
$$T_1=2m_1g((m_2-m_1)/(m_2+m_1)+1)$$
and because we need T1 to be equal to m3g we can say:
$$m_3g=2m_1g((m_2-m_1)/(m_2+m_1)+1)$$
which is:
$$m_3=2m_1((m_2-m_1)/(m_2+m_1)+1)$$
Oh! that's right! also m1=m2=m results m3=2m fits into this!
Thanks!

Good job, but just for aesthetic purposes, why don't you reduce that thing in parenthesis to a common denominator?

 

[AHashemi]

Good job, but just for aesthetic purposes, why don't you reduce that thing in parenthesis to a common denominator?

$$m_3=2m_1(2m_2/(m_2+m_1))$$

Thanks again for your help.

 

[Chestermiller]

$$m_3=2m_1(2m_2/(m_2+m_1))$$

Thanks again for your help.

Isn't that the same as $$m_3=\frac{4m_1m_2}{(m_1+m_2)}$$