Solve Mobile Pulley System: Acceleration of 3 Masses

[arildno]

Let's see:
T_1-m_1g=m_1a_1 (eq.1)

T_2-m_2g=m_2(-a_1+a_2) (eq.2)

T_2-m_3g=m_3(-a_1-a_2) (eq.3)
T_1=2T_2 (eq.4)

Sure, ehild

Or, as I wrote above (I managed to remove a post where I had corrected the signs)

Here, my a_2 is the relative acceleration to the moving pulley.

 

[CAF123]

What you are lacking then, is a similar eq. as in 8 for the other mass, and in addition, either the requirement that the two lower tensions must be equal to each other, or the tricky condition that the accelerations a_2 and a_3, for you absolute quantities must have a relation to each other on the requirement that the rope about mobile pulley remains of constant length.

Acc of 2 relative to the fixed pulley, a_2F = a_2M + a_MF = -a2 –a1
Acc of 3 relative to the fixed pulley a_3F = a_3M + a_MF = a3 –a1.
No slip implies a_3M=-a_2M
a_2F = -a2 – a1 and a_3F = a2 – a1.
F is an inertial frame so should I sub these into my eqns in e.g 3ma2 + mg = T_2 and ma3 + mg = T_2 to obtain another eqn?

 

[arildno]

Hmm..seems I specified 4 equations, rather than 3.
The kinematic relation between the two accelerations is the one that is lacking; it represents a distinct constraint requirement the rest does not provide for (and which is embedded in the Lagrangian equations of motion).

What you were lacking was the kinematic constraint condition between mass 2 and mass 3's accelerations.

 

[CAF123]

I denoted the distance of the 4m, 3m , m masses and the fixed and mobile pulley from the ground to obtain constraint equations in the Lagrangian approach. I.e denote x,y,z as position of 4m,3m and m masses while h and H denote mobile and fixed pulleys. So x + h + l1 = 2H and y + z + l2 = 2h. Was this what you were referring to ?

 

[arildno]

7. "The net upward force on the mobile pulley is equal to the sum of the tensions 3ma2+3mg and ma3+mg. So 4ma1 + 4mg = 3ma2+3mg+mg+ma3 giving 4a1 - 3a2 -a3 = 0. (1st EQN)"
True.
8. "3ma2 + 3mg = mg + ma3 so 3a2 - a3 = -2g. (2nd EQN)"
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7. is the FORCE law for the moving pulley, where the tensions has been written in terms of accelerations.
This is a DYNAMIC relation T_1=T_2+T_3
8. Equates the tensions acting on m_2 and m_3, that is, a DYNAMIC relation between the accelerations
T_2=T_3
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You still need the KINEMATIC relationship between a_2 and a_3, that the rope on the moving pulley remains of equal length.

The problem with how you write the equations is that it is very messy the way you substitute too fast, bmaking reading very difficult.

Why not KEEP F=ma for the longest possible time, prior to eliminate the tensions?

You are, basically, trying to Lagrangianize the Newtonian formulation of the problem.

 

[ehild]

Or, as I wrote above (I managed to remove a post where I had corrected the signs)

Here, my a_2 is the relative acceleration to the moving pulley.
T_1-m_1g=m_1a_1 (eq.1)

T_2-m_2g=m_2(-a_1+a_2) (eq.2)

T_2-m_3g=m_3(-a_1-a_2) (eq.3)
T_1=2T_2 (eq.4)

These equations are correct and enough, just solve.

ehild

 

[arildno]

Let me set up the 10 relevant equations for absolute accelerations and tensions, and show how your equations are gained from them:

$$T_{1,left}-m_{1}g=m_{1}a_{1,abs} (a)$$
(a) is Newton's second law of motion for mass 1 (assumed to be on left side of rope)
$$T_{1,right}=T_{1,left}=T_{1} (b)$$
Newton's 2.law for 1st rope, which yields constant tension T_1 throughout the massless rope
$$T_{1}-T_{2}-T_{3}=0 (c)$$
is Newton's second law of motion for the moving pulley
$$T_{2}-m_{2}g=ma_{2,abs} (d)$$
(d) is Newton's 2.law for mass 2.
$$T_{3}-m_{3}g=ma_{3,abs} (e)$$
(e) is Newton's 2.law for mass 3.
$$T_{2}=T_{3} (f)$$
This is the result of Newton's 2.law for the massless rope on the moving pulley.
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Now, apart from these applications of Newton's laws you need KINEMATIC requirements, that fulfills that the ropes remain of constant length:
$$a_{pulley}=-a_{1} (g)$$
This ensures that the rope along the fixed pulley remains of constant length.
$$a_{3,abs}=a_{pulley}+a_{3,rel} (h)$$
Introduces the concept of relative acceleration of mass 3.
$$a_{2,abs}=a_{pulley}+a_{2,rel} (i)$$
Introduces the concept of relative acceleration of mass 2.
$$a_{3,rel}=-a_{2,rel} (j)$$
Ensures that the rope around the moving pulley remains of constant length.
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Now, we have meticuluously written down all 10 equations of possible relevance; the problem when you did this was to jumble them together a bit too fast, so that you forgot some, and repeated others.

That was a result of your desire to Lagrangianize Newton a bit too fast.

 

[CAF123]

I am nearing the solution, I just have a negative error somewhere in my calculations:
For clarity, I will keep to your notation.
We have $T_2 - m_2g = ma_{2,abs}$ and $T_2 -m_3g = ma_{3,abs}$.**

In the fixed reference frame of the pulley: $ma_{2,abs} = a_{pulley } + a_{2,rel} = -a_1 + a_2$ and $ma_{3,abs} = a_{pulley} + a_{3,rel} = -a_1 -a_2$ using the fact that $a_{2,rel} = -a_{3,rel}$
Sub these expressions into ** to get $-4ma_2 - 2ma_1 + 2mg = 0$ which would be a third eqn independent of the rest, but i think there is a sign error in it.

Thanks.

 

[arildno]

I am nearing the solution, I just have a negative error somewhere in my calculations:
For clarity, I will keep to your notation.
We have $T_2 - m_2g = ma_{2,abs}$ and $T_2 -m_3g = ma_{3,abs}$.**

In the fixed reference frame of the pulley: $ma_{2,abs} = a_{pulley } + a_{2,rel} = -a_1 + a_2$ and $ma_{3,abs} = a_{pulley} + a_{3,rel} = -a_1 -a_2$ using the fact that $a_{2,rel} = -a_{3,rel}$
Sub these expressions into ** to get $-4ma_2 - 2ma_1 + 2mg = 0$ which would be a third eqn independent of the rest, but i think there is a sign error in it.

Thanks.

Be VERY careful of your a_2 here!
It seems right to me at a first glance, but mixing together a_2,abs and a_2,rel will spell disaster..

 

[arildno]

As you've probably realized, the tricky kinematic condition on the moving pulley when you express it in absolute accelerations was:
$$a_{2,abs}+a_{3,abs}=-2a_{1,abs}$$

The exercise is a good one which shows readily how much simpler in general it is to use Lagrangian mechanics directly, but also that it is much simpler to transform the clunky Newtonian formulation into the equivalent Lagrangian, than vice versa. This latter holds because "N to L" is one of direct simplification, whereas "L to N" is one of direct complexification..

 

[CAF123]

As you've probably realized, the tricky kinematic condition on the moving pulley when you express it in absolute accelerations was:
$$a_{2,abs}+a_{3,abs}=-2a_{1,abs}$$

$a_{2,abs} = a_{2,rel} - a_{1,abs}$ and $a_{3,abs} = -a_{2,rel} - a_{1,abs}$ so $a_{2,abs} + a_{3,abs} = -2a_{1,abs}$ as required.

 

[arildno]

Yup, since you had solved it, I assume you had that relation.
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It is not altogether intuitive to get that in a Newtonian formulation without intermediate use of the relative accelerations, yet that is, I believe, yet another simplifying feature of the Lagrangian formalism: As long as your constraints are properly set up, such an equation falls out easily enough.

 

[CAF123]

Yup, since you had solved it, I assume you had that relation.
----
It is not altogether intuitive to get that in a Newtonian formulation without intermediate use of the relative accelerations, yet that is, I believe, yet another simplifying feature of the Lagrangian formalism: As long as your constraints are properly set up, such an equation falls out easily enough.

I agree, I have only really begun using the Lagrangian approach and I see how nicely it simplifies problems.