Mobile pulley system

1. Oct 5, 2013

CAF123

1. The problem statement, all variables and given/known data
A light smooth pulley is attached to a support a fixed height above the ground. An inextensible string passes over the pulley and carries a mass 4m on one side. The other end of the string supports a similar mobile pulley; over this passes a second string, carrying masses of 3m and m on its two ends.
Deduce the acceleration of the three masses

2. Relevant equations
N2, Lagrange Equations

3. The attempt at a solution
I solved this already and got the correct answer using Lagrange's Equations. Now I try again with Newton's Eqns. The final set of three equations I obtain from Newtonian analysis are satisfied by the accelerations I got when I solved via Lagrange's Eqns. However, I cannot actually solve the three equations to obtain a unique solution (they are a linearly dependent set). What I need is another equation which I cannot get.
No picture was given, so I created my own and I will describe what I did.
Forces on 4m mass give eqn: T - 4mg = 4ma1, a1 the acceleration of mass 4m. This is also the tension on the other side of the same pulley. Since this pulley is the fixed pulley, it has an upward force equal to 8ma1 + 8mg

The string then passes down to the mobile pulley. Upward force on this is 4ma1 + 4mg. Consider the 3m mass: Net force is 3ma2 + 3mg and that of the m mass is ma3 + mg.
Now I used these to construct suitable equations. 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). The tension on either side of the mobile pulley must be the same => 3ma2 + 3mg = mg + ma3 so 3a2 - a3 = -2g. (2nd EQN)

Finally, the net upward force on the fixed pulley is equal to twice the tensions on either side of the mobile pulley => 8ma1 + 8mg = 2(3ma2 + 3mg + 3ma2 + 3mg) giving 8a1 - 12a2 = 4g. (3rd EQN)

This set is linearly dependent ,so I have a free parameter
Is there another eqn I can get somewhere?

Also, now that I think about it, why is my 1st EQN valid? This pulley is mobile so its upward force need not equal the sum f two tensions below it?

Many thanks.
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Last edited: Oct 5, 2013
2. Oct 5, 2013

arildno

Note that the tension in the rope by which the light (massless), mobile pulley is drawn must equal twice the tension wound about it, in order for that pulley's acceleration to be finite.

Perhaps that was the equation you were lacking?

You should have 4 equations, for unknowns a_1 and T_1 (related to the fixed pulley), and a_2 and T_2 (related to the mobile pulley)

3. Oct 5, 2013

CAF123

Hi arildno,
I think I might already have taken this into account in the eqn labelled 1st EQN in the OP?
As I noted in the OP, I am not quite sure why the tension has to be twice the tension wound around it. Would this not imply the mobile pulley is in equilibrium?

4. Oct 5, 2013

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)

Edited away stupid g's

Last edited: Oct 5, 2013
5. Oct 5, 2013

arildno

Nope. The pulley is massless (m=0), yet is subject to Newton's laws of motion.
Thus, if there is an imbalance of forces working on it, it will have infinite acceleration.

6. Oct 5, 2013

arildno

Oops, why did I put in those g-factors???

7. Oct 5, 2013

arildno

Dividing (2) and (3) with respective masses, adding them together for a_2-removal, and using (4) for T_2 removal should yield (5):
$$\frac{T_{1}}{4}\frac{m_{2}+m_{3}}{m_{2}m_{3}}-g=a_{1} (5)$$
This can then be used with (1)
EDIT:
$$\frac{T_{1}}{4}\frac{m_{2}+m_{3}}{m_{2}m_{3}}-g=-a_{1} (5)$$

Last edited: Oct 5, 2013
8. Oct 5, 2013

CAF123

Okay, this makes sense. But I don't understand your eqns (2) and (3) yet, in particular as to why you have (a1+a2) and (a1-a2) on the right hand sides.

9. Oct 5, 2013

arildno

10. Oct 5, 2013

arildno

You're right, it should be (-a_1+a_2) and (-a_1-a_2)

Relative to their common acceleration -a_1, they have opposite accelerations.

11. Oct 5, 2013

arildno

This is now corrected to quoted edit.

12. Oct 5, 2013

arildno

We therefore have, from (5):
$$T_{1}=\frac{4m_{2}m_{3}}{m_{2}+m_{3}}g-\frac{4m_{2}m_{3}}{m_{2}+m_{3}}a_{1}$$

13. Oct 5, 2013

arildno

We gain, by insertion:
$$\frac{4m_{2}m_{3}-m_{1}(m_{2}+m_{3})}{m_{2}+m_{3}}g=\frac{4m_{2}m_{3}+m_{1}(m_{2}+m_{3})}{m_{2}+m_{3}}a_{1}$$, that is
$$a_{1}=\frac{4m_{2}m_{3}-m_{1}(m_{2}+m_{3})}{4m_{2}m_{3}+m_{1}(m_{2}+m_{3})}g$$

Last edited: Oct 5, 2013
14. Oct 5, 2013

CAF123

Sorry what I meant was why is there two terms on the RHS. On the m2 mass, there exists a gravitational force downwards and a tension force upwards. This gives T_2 - m_2g = m_2(a_2) only, no? Similarly for the other mass: T_2 - m_3g = m_3(a_3).

If I rearrange for T_2 in the second and sub in the first, gives me a_3 - 3a_2 = 2g, which is my 2nd EQN in the OP.

15. Oct 5, 2013

arildno

"T_2 - m_2g = m_2(a_2) only, no? Similarly for the other mass: T_2 - m_3g = m_3(a_3)."

Only in the INERTIAL frame relative to the fixed pulley.

Your a_2 can be written as a_2=-a_1+a*, where a* is the relative acceleration of mass_2 to the moving pulley, which is -a_1.
Similarly, because the rope around the mobile pulley must remain fixed in length,
your a_3 can be written as a_3=-a_1-a*

16. Oct 5, 2013

arildno

Note that the terms -m_2a_1 and -m_3a_1 can be regarded as contributions from the fictitious forces in a non-inertial frame of reference.

17. Oct 5, 2013

CAF123

Should that be a* = -a_2? The acceleration of the m_2 relative to the fixed pulley (a2F) is the (vector)sum of the acceleration of m_2 relative to the mobile pulley(a2M) + the acceleration of the mobile pulley relative to the fixed pulley(aMF)

a2F = a_2, a2M =(+/-)a_2 and I am unsure of aMF. How did you determine the signs here?

18. Oct 5, 2013

arildno

When it comes to the sign of a_2, i.e the relative accelerations masses 2 and 3 have relative to the mobile pulley, the ONLY thing that matters is that we have OPPOSITE signs for masses 2 and 3.
This is also the case for a_1: if we call the mass 1 acceleration for a_1, the mobile pulley must have -a_1 as its acceleration.
We could have chosen oppositely, it does not matter.
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The total correct sign distribution is ensured that the tensions work opposite to gravity, and that opposite accelerations have opposite signs.

19. Oct 5, 2013

CAF123

Just to check a couple of things: So what you did was reexpress all the accelerations of the masses relative to the fixed pulley, yes? In the equation T_2 - m_2g = m_2a_2, this is valid for the acceleration of m_2 relative to the mobile pulley?

20. Oct 5, 2013

arildno

Yes to the first!

No to the second, it is not, because the moving pulley system is NON-INERTIAL.
Thus, the correct representation relative to the moving system accelerating with -a_1 is:
-(-m_2a_1)+T_2-m_2g_2=m_2a_2,
where a_2 is the acceleration of m_2 relative to the moving pulley.
The first term is the fictitious force associated with the noninertial frame's acceleration, relative to an inertial frame.

Note that THIS force law is equivalent to the one I gave.